TITLE: Boundary Conditions Invariant Under Conformal Transformations in Electrostatics?
QUESTION [7 upvotes]: in two dimensional electrostatics it is assumed that the whole physical system is translationally invariant in one direction. Here, the two-dimensional Laplace equation $$\Delta \phi(x,y) = \frac{\sigma(x,y)}{\epsilon_0}$$ holds in free space and solutions can be mapped on solutions if space is transformed conformally - the metric changes from $$g\rightarrow \Omega^2 g\ .$$
This approach has a lot of nice applications like the possibility to calculate the field around a complicated conductor or, in aerodynamics, the flow around an airfoil as discussed earlier.
Aubry et al. have in a number of papers (see e.g. Interaction between Plasmonic Nanoparticles Revisited with Transformation Optics (PRL) and Plasmonic Light-Harvesting Devices over the Whole Visible Spectrum (Nano Lett.)) used this technique to calculate the eigenfunctions of "crescend" structures and coupled cylinders:
(taken from the Nano Lett.)
I think this is a very nice approach but I am not sure why it can be done in this way. My problem is that as far as I know usually only Neumann and Dirichlet boundary conditions are invariant under conformal transformations. In electrodynamics, however, we have that both $E_t = -d\phi(\mathbf{t})$ (tangential) and $\epsilon\cdot E_n = -\epsilon\cdot d\phi(\mathbf{n})$ (normal) are continuous at a boundary where the permittivity $\epsilon$ changes.
In the papers I could not find a discussion of this issue so I might ask here:

Are the boundary conditions for the electrostatic potential $\phi$ invariant under conformal transformations?

Thank you in advance
Sincerely

REPLY [4 votes]: Under a conformal transformation $f$  we have that the coordinate $x_i \rightarrow f_i(x)$, the potential $\phi \rightarrow \phi$ and therefore the derivative $\partial_i \phi \rightarrow \frac{\partial f_j}{\partial x_i} \partial_j \phi$. A tangent vector $t$ to some surface $S$ gets mapped to a tangent vector to $f(S)$ as always, and because the transformation is conformal the normal to a surface $n$ gets mapped to a vector proportional to the normal $n'$ on the surface $f(S)$ since $t\cdot n = 0$ is preserved by conformal transformation.
The usual boundary conditions  $\phi(S) = \phi_0$ and that $\phi(S)$ is continuous are preserved, since $\phi$ transforms trivially. A different condition is that $\epsilon_{+}n\cdot \partial \phi_{+} - \epsilon_{-}n\cdot \partial \phi_{-} = 0$, where $\phi_{\pm}$ is the limit of $\phi$ approaching the surface from either direction and  $\epsilon_{\pm}$ are constants. This gets mapped to the equation $\Omega^2\times\left(\epsilon_{+}n'\cdot \partial \phi_{+} - \epsilon_{-}n'\cdot \partial \phi_{-}\right) = 0$ so that is preserved as well (where $\Omega$ is the scale factor of $f$). The same occurs for the part of the $\partial_j \phi$ tangent to the the surface as well.  A sheet of charge $\sigma$ produces the b. c. $\epsilon_{+}n\cdot \partial \phi_{+} - \epsilon_{-}n\cdot \partial \phi_{-} = \sigma$ which is preserved as well with the correct transformation $\sigma \rightarrow \Omega^2 \sigma$.
So all the usual b. c. of electrostatics work well with conformal transformations.


text:
\appendix
\section{Proof of Theorem \ref{prop:main}.}\label{section:proof}

\begin{proof}

We present the proof for each part of the theorem, starting with 1).

1) We prove this result by induction. By assumption $s_i(0), x_i(0),$ $r_i(0) \in [0,1]$ and $s_i(0) + x_i(0) + r_i(0) = 1$ for all $i \in [n] $. Recall from Assumption \ref{assm:1} that $s_i(0)>0$ and $(1-h\sum^n_{j=1} \beta_{ij}) > 0$. Therefore, from \eqref{eq:sir_si}, we have $s_i(1) \geq (1-h\sum^n_{j=1} \beta_{ij}) s_i(0) > 0$. Since $ h \big[-s_i(0) \sum^n_{j=1} \beta_{ij} x_j(0) \big]\leq 0$, $s(1) \leq s(0)\leq 1$. By Assumption~\ref{assm:1} and \eqref{eq:sir_xi}, $x_i(1) \geq (1-h\gamma_i) x_i(0) \geq 0$.
Since $x_j(0) \leq 1$ for all $ j\in [n]$, by \eqref{eq:sir_xi} and Assumption \ref{assm:1}, $x_i(1)\leq x_i(0) + s_i(0)h\sum^n_{j=1} \beta_{ij}\leq x_i(0) + s_i(0)\leq 1$.
By \eqref{eq:sir_ri} and the non-negativity of $h$ and $\gamma_i$, and since $x_i(0)\geq 0$, we have $r_i(1) \geq r_i(0) \geq 0$. By \eqref{eq:sir_xi} and Assumption \ref{assm:1}, we have $r_i(1) \leq r_i(0) + x_i(0) \leq 1$.
Adding up \eqref{eq:sir_si}-\eqref{eq:sir_ri}, gives that $s_i(1) + x_i(1) + r_i(1) = s_i(0) + x_i(0) + r_i(0)$, which by assumption equals~1.

Now assume for an arbitrary $k$, $s_i(k), x_i(k), r_i(k) \in [0,1]$ and $s_i(k) + x_i(k) + r_i(k) = 1$. Following the exact same arguments as for the base case except replacing $0$ with $k$ and $1$ with $k+1$, it can be shown that $s_i(k+1), x_i(k+1), r_i(k+1) \in [0,1]$ and $s_i(k+1) + x_i(k+1) + r_i(k+1) = 1$. Therefore, by induction, $s_i(k), x_i(k), r_i(k) \in [0,1]$
and
$s_i(k) + x_i(k) + r_i(k) = 1$
for all $k\geq0$ and $i\in [n]$. In addition, if $s_i(0) > 0$, the above argument implies that $s_i(k)>0$ for all $i\in [n]$ and for every finite $k$. As a result, the matrix $A_k$ remains irreducible for all finite $k$.

2) By 1) and the non-negativity of the $\beta_{ij}$'s, it follows that $ h \big[-s_i(k) \sum^n_{j=1} \beta_{ij} x_j(k) \big]\leq 0$ for all $k\geq 0$. Therefore, from \eqref{eq:sir_si}, we have $ s_i(k+1) \leq s_i(k)$.

3) Since the rate of change of $s(k)$, $- h \diag(s(k)) B  x(k)$, is non-positive for all $k\geq0$ and $s(k) $ is lower bounded by zero, by  1), we conclude that $\lim_{k\rightarrow \infty}s(k)$ exists. Therefore
\begin{equation}\label{eq:lim_schange}
    \lim_{k\rightarrow \infty}- h \diag(s(k)) B x(k) = \0,
\end{equation}
where $\0$ is the vector of dimension $n$ with all entries equal to $0$.
Therefore, $\lim_{k\rightarrow \infty} x(k+1) - x(k) = \lim_{k\rightarrow \infty}-h\diag(\gamma) x(k)$. Thus, by the assumption that $h \gamma_i>0$ for all $i \in [n]$ \rev{and 1), we have that}
$$\rev{\lim_{k\rightarrow \infty} x_i(k+1) - x_i(k) = \lim_{k\rightarrow \infty}-h \gamma_i x_i(k)\leq 0,}$$
\rev{where equality holds only if $\lim_{k \to \infty}x_i(k)=0$ and if the inequality is strict, $x_i(k)$ is decreasing. Therefore,}  $\lim_{k \to \infty} x_i(k) = 0$, for all $i \in [n]$.

4) Recall that by Assumption \ref{assm:1}, $A_k$ is an irreducible non-negative matrix and thus by the Perron-Frobenius Theorem for irreducible non-negative matrices \cite[Theorem~2.7 and Lemma~2.4]{varga}, $\lambda^{A_k}_{\max} = \rho(A_k)$, where $\rho(\cdot)$ indicates the spectral radius. By \cite[Theorem~2.7]{varga}, $\rho(A_k)$ increases when any entry of $A_k$ increases. Therefore, by 2) and since $A_k$ is defined as $\mathbf{I}_n + h \diag(s^k) B - h \diag(\gamma)$, we have that
$$\rho(A_k) \geq \rho(A_{k+1}),$$
which implies
$$\lambda^{A_k}_{\max} \geq \lambda^{A_{k+1}}_{\max}.$$

5) There are two possible types of equilibria for the SIR model: i) $\lim_{k\rightarrow \infty}s(k) = \0$, or ii)~$\lim_{k\rightarrow \infty}s(k) = s^* \neq \0$. We explore the two cases separately.

i) If $\lim_{k\rightarrow \infty}s(k) = \0$, then the rate of change of $x(k)$ \rev{will become}

$-h\diag(\gamma)x(k)$. Therefore, by

the definition of \rev{$A_{k}$ and the assumption that $h\gamma_i < 1, \forall i \in [n]$},

there exists a $\bar{k}$ such that $\lambda^{A_{k}}_{\max}< 1$ for all $k \geq \bar{k}$.

ii) If

$\lim_{k\rightarrow \infty}s(k) = s^* \neq \0$, then, by 3), for any $(s(0),x(0),r(0))$ the system dynamics converge to some equilibria of the form $(s^*,\0,\1-s^*)$, for some nonzero $s^*$.
Define
\begin{equation}\label{eq:eps}
    \epsilon_s(k):= s(k) - s^* \text{ and } \epsilon_x(k):= x(k) - \0.
\end{equation}By 2) and 1), respectively, we know that $\epsilon_s(k)\geq0$ and $\epsilon_x(k)\geq 0$ for all $k\geq 0$. Furthermore, we know that $\epsilon_s(k+1)\leq \epsilon_s(k)$ for all $k\geq 0$, $\lim_{k\rightarrow \infty}\epsilon_s(k) = \0$, and $\lim_{k\rightarrow \infty}\epsilon_x(k) = \0$.

Let $\lambda^{A^*}_{\max}$ be the maximum eigenvalue
of $\mathbf{I}_n + h \diag(s^*) B - h \diag(\gamma)$ with corresponding normalized left eigenvector $w^*$, that is,
\begin{equation}\label{eq:eigen}
     w^{*\top}(\mathbf{I}_n + h \diag(s^*) B - h \diag(\gamma)) = \lambda^{A^*}_{\max} w^{*\top}.
\end{equation}
\rev{Note that since $\mathbf{I}_n + h \diag(s^*) B - h \diag(\gamma)$ is irreducible, $w^*_i>0$, for all $i \in [n]$, by the Perron-Frobenius Theorem.}

\rev{Assume on the contrary that $\lambda^{A^*}_{\max}\geq 1$}. Left multiplying the equation for $x(k+1)$ in \eqref{eq:sir_dt_x} by $w^{*\top}$ and using
\eqref{eq:eps} and \eqref{eq:eigen}, we obtain
\begin{align*}
    & w^{*\top}\epsilon_x(k+1)
    \\ & \quad = w^{*\top} [\mathbf{I}_n + h \diag(\epsilon_s(k) + s^*) B - h \diag(\gamma)] \epsilon_x(k) \\
    & \quad =\lambda^{A^*}_{\max}w^{*\top}\epsilon_x(k)  + w^{*\top} h \diag(\epsilon_s(k)) B \epsilon_x(k) \\
    & \quad \rev{\geq} w^{*\top}\epsilon_x(k)  + w^{*\top} h \diag(\epsilon_s(k)) B \epsilon_x(k),
\end{align*}
where the last equality holds since
$\lambda^{A^*}_{\max}\rev{\geq}1$.
Thus, $$w^{*\top}(\epsilon_x(k+1)-\epsilon_x(k)) \rev{\geq} w^{*\top} h \diag(\epsilon_s(k)) B \epsilon_x(k) \geq 0.$$
\rev{If either inequality is strict, clearly, $\lim_{k\rightarrow \infty}x(k) \neq \0$, which is a contradiction to 3).
If equality holds, $w^{*\top}(\epsilon_x(k+1)-\epsilon_x(k))=0$ implies $\epsilon_x(k+1)=\epsilon_x(k)$, since $w^*_i>0$, for all $i \in [n]$.
Then, since by assumption, $x_i(0) > 0$ for some~$i$, $\lim_{k\rightarrow \infty}x(k) \neq \0$, again a contradiction to 3). }

Therefore, there exists a $\bar{k}$ such that $\lambda^{A_{k}}_{\max}< 1$ for all $k \geq \bar{k}$.

6) Since, by 5), there exists a $\bar{k}$ such that $\lambda^{A_{k}}_{\max} < 1$ for all $k \geq \bar{k}$, and we know that $\lambda^{A_{k}}_{\max} = \rho(A_{k})\geq 0$ by Assumption~\ref{assm:1}, we have
\begin{equation}\label{eq:conv}
    \lim_{k \to \infty} \frac{\|x(k+1)\|}{\|x(k)\|} = \frac{\|A_k x(k)\|}{\|x(k)\|} \leq \lambda^{A_{k}}_{\max}<1.
\end{equation}
Therefore, for $k \geq \bar{k}$, $x(k)$ converges linearly to $\0$.
\end{proof}

config: arxiv


text:
\section{Quasi-invariant surfaces and Conley decomposition}\label{s.conley}

We introduce here two important tools that will be used in the next sections: \emph{quasi-invariant surfaces} (defined below) and decompositions of our configuration space $\Sigma$ given by the Fundamental Theorem of the Dynamical Systems \cite{conleyfundthm}. These two objects are strongly related, as the mentioned decompositions of $\Sigma$ are given by families of surfaces that may  only fail to be \emph{quasi-invariant} by small technical details. On the other hand some \emph{quasi-invariant surfaces} that we will employ are not associated to these decompositions. Let us start by defining quasi-invariant surfaces for a
homeomorphism $f:\Sigma\to\Sigma$.

When we talk about a {\em subsurface} $S\subseteq\Sigma$ we understand it to be compact and topologically embedded in $\Sigma$, not necessarily connected. As such, the boundary $\partial S$ is a union of disjoint simple closed curves, which is non empty unless $S=\Sigma$, and the complement $\Sigma\setminus S$ is also a subsurface.

Given an homeomorphism $f:\Sigma\to\Sigma$
we say that $S$ is \emph{quasi-invariant} under $f$ if there exists a disjoint family of annuli $A_{\gamma}$, one for each boundary component $\gamma\subseteq\partial S$ so that:
\begin{enumerate}

\item $A_{\gamma}\subseteq\Sigma\setminus \Omega(f)$,
\item $A_{\gamma}\cap S=\gamma$,
\item $f(S)\cup f^{-1}(S)\subseteq S\bigcup(\cup_{\gamma} A_{\gamma})$,
\item Either $f(\gamma)$ or $f^{-1}(\gamma)$ is contained in $A_{\gamma}\setminus S$.

\end{enumerate}

If $S\subseteq\Sigma$ is a subsurface, we denote by $\textrm{Fill}(S)$ the union of $S$ and all the connected components of $\Sigma\setminus S$ that are topological disks. We say that $S$ is {\em filled} when $\textrm{Fill}(S)=S$.

\begin{rem}\label{r.quasiconexas} Notice that if $S$ is quasi-invariant then its connected components and $\text{Fill}(S)$ are also quasi-invariant.
\end{rem}

Given a connected subsurface $S\subset\Sigma$ let $\textrm{i}_*:\pi_1(S)\to \pi_1(\Sigma)$ be the morphism induced by the inclusion map. We say that a connected subsurface $S$ is:

\begin{itemize}

\item \emph{trivial} whenever $\textrm{Im}(\textrm{i}_*)$ is given by the trivial element,

\item \emph{annular} whenever $\textrm{Im}(\textrm{i}_*)$ is isomorphic to $\Z$,

\item \emph{curved} whenever $\textrm{Im}(\textrm{i}_*)$ is not trivial nor isomorphic to $\Z$.
\end{itemize}

Next we want to introduce some important lemmas concerning quasi-invariant surfaces that will be used in the following sections. We start by introducing the relevant notations. For a subsurface $S$ of $\Sigma$ we denote by $H_1(S)$ the image in $H_1(\Sigma;\R) = \R^{2g}$ of the homology map $H_1(S;\R)\to H_1(\Sigma;\R)$ induced by the inclusion. Notice that $H_1(S)=H_1(\textrm{Fill}(S))$.

Every time we consider a closed oriented surface $\Sigma$ of negative Euler characteristic, we shall fix a Poincar\'{e} covering $\pi:\D\to\Sigma$. Recall that in this situation, given any map $f:\Sigma\to\Sigma$ in the homotopy class of the identity, we have a unique lift $F:\D\to\D$ of $f$ which extends as the identity map to $\partial\D$. We call it the \emph{canonical lift} of $f$.

\begin{lemma} \label{l.qisurface}
Let $f\in \mathcal{A}_0(\Sigma)$ and $S\subset \Sigma$ be a connected surface that is quasi-invariant under $f$. Then:
\begin{enumerate}

\item $S\cap\Omega(f)$ is $f$-invariant. Moreover, it is a union of basic pieces,

\item If $S$ is non-trivial and $\tilde{S}$ is a connected component of $\pi^{-1}(S)$ we have that $\tilde{S}\cap\pi^{-1}(\Omega(f))$ is $F$-invariant,

\item If $S$ is trivial, then there is a closed curve $\alpha$ in $\Sigma$ such that $\rho_{\Lambda}(f) = \{[\alpha]\}$ for every basic piece $\Lambda$ contained in $S$,

\item If $S$ is non-trivial and $\Lambda\subseteq S$ is a basic piece, we have $\rho_{\Lambda}(f) \subset H_1(S)$.

\end{enumerate}

\end{lemma}

\begin{proof}

Consider $(A_{\gamma})$ the disjoint family of closed annuli given by the quasi-invariance of $S$ and define $$S_1:=S\cup\left(\bigcup_{\gamma\subseteq\partial S}A_{\gamma}\right).$$ Since $S_1\setminus S$ is contained in the complement of $\Omega(f)$ and $f(S)\cup f^{-1}(S)\subseteq S_1$ we deduce that $f(S\cap\Omega(f))$ and $f^{-1}(S\cap\Omega(f))$ are contained in $S$ proving the first part of point (1). The second part follows from the first and the transitivity of basic pieces, noting that $S\cap\Omega(f)$ is both open and closed in $\Omega(f)$.

For point (2) assume that $S$ is non trivial and consider $\tilde{S}$ a connected component of $\pi^{-1}(S)$. Let $\tilde{S}_1$ be the connected component of $\pi^{-1}(S_1)$ containing $\tilde{S}$. Since $F$ lifts $f$ we know that $F^{\pm 1}(\tilde{S})$ must be contained in some lift of $S_1$. On the other hand, $F$ restricts to the identity on $\partial\mathbb{D}$ and the different lifts of $S_1$ can be identified by their accumulation points on $\partial\mathbb{D}$, so  we get that $F^{\pm 1}(\tilde{S})\subseteq\tilde{S}_1$. Observe that $\tilde{S}_1\setminus\tilde{S}\subseteq \pi^{-1}(\cup_{\gamma} A_{\gamma})$, which does not meet $\pi^{-1}(\Omega(f))$, and this concludes the proof of point (2) by the same argument used for point (1).

We proceed with the proof of point (3). Consider $f_t$ a homotopy between $f_0=Id$ and $f_1=f$, and let $F_t$ be the lift of $f_t$ to the universal cover $\mathbb{D}$ that begins at $F_0 = Id$. According to our notation, we have $F_1=F$. Denote $D:=\text{Fill}(S)$ and $D_1:=\text{Fill}(S_1)$. Under the assumption for point (3)  both $D$ and $D_1$ are disks. Notice also that $f(D)\subseteq D_1$. Let $\tilde{D}\subseteq\tilde{D}_1$ be lifts of $D_0$ and $D_1$. Then $F(\tilde{D})$ is contained in $g\tilde{D}_1$ for some  $g\in\pi_1(\Sigma)$, and we take $\alpha$ a closed curve representing $g$. If $\Lambda\subset S$ is a basic piece and $x\in\Lambda$, we consider a lift $\tilde{x}\in \tilde{S}$ and notice that $\{F_t(\tilde{x})\}_{t\in[0,1]}$ can be homotoped to $\alpha$ with endpoints varying inside fundamental domains of the covering (to be more specific, inside $\tilde{D}$ and $g\tilde{D}_1$ respectively). As this holds for any $x\in\Lambda$ (including $f^k(x)$ for $k\geq 1$) we see that $\rho_{\Lambda}(f) = \{[\alpha]\}$ as desired.

Finally we turn to point (4). We can assume that $S$ is filled, since $H_1(\text{Fill}(S))=H_1(S)$ and the filling of a quasi-invariant surface is also quasi-invariant. Let $\pi_S:\hat{\Sigma}_S\to\Sigma$ be the covering space that corresponds to the subgroup $\textrm{i}_*\pi_1(S)\leq\pi_1(\Sigma)$. Observe that $\hat{\Sigma}_S$ consists of a $1:1$ lift of $S$, that we denote $\hat{S}$, together with open annuli attached to each component of $\partial \hat{S}$. Since $S_1$ retracts by deformation onto $S$ we have that $\hat{\Sigma}_S$ also contains a $1:1$ lift of $S_1$, that we call $\hat{S}_1$, with $\hat{S}\subset \hat{S}_1$.

Let $\overline{f}_t$ be the lift of the homotopy $f_t$ to this cover with $\overline{f}_0=Id$.   We denote by $\overline{f}=\overline{f}_1$  the corresponding lift of $f$. Since $\hat{S}_1\setminus \hat{S}\subseteq\pi_S^{-1}(\cup A_{\gamma})$, which does not meet $\pi_S^{-1}(\Omega(f))$, we deduce that $\pi_S^{-1}(\Omega(f))\cap \hat{S}$ is $\overline{f}$-invariant. If $\Lambda\subset S$ is a basic piece and $x\in\Lambda$, we take a lift $\hat{x}\in\hat{S}$ and notice that $\overline{f}^n(\hat{x})\in \hat{S}$ for every $n\in\N$. Since $\hat{\Sigma}_S$ retracts by deformation onto $\hat{S}$ we get that $\{\overline{f}_t(\hat{x})\}_{t\in[0,n]}$ is homotopic with fixed endpoints to a curve contained in $\hat{S}$. Projecting this homotopy by $\pi_S$ we get a homotopy with fixed endpoints of $\{f_t(x)\}_{t\in[0,n]}$ into $S$. As this holds for every $x\in\Lambda$,  we get that   $$\rho_{\Lambda}(f)\subseteq H_1(S)$$ for every basic piece $\Lambda\subseteq S$.

\end{proof}

For a fixed point $x_0$ of $f$ we introduce the notation $\rho(f,x_0)$ for the corresponding rotation vector. In terms of the homotopy $f_t$  of $Id$ to $f$, we may write $\rho(f,x_0)$ as the homology class of the loop $\{f_t(x_0)\}_{t\in [0,1]}$. In terms of the abelian lift $\hat{f}$, we have $\rho(f,x_0)=\hat{f}(\hat{x}_0)-\hat{x}_0 $ for $\hat{x}_0\in \Sigma_{ab}$ lifting $x_0$.

\begin{lemma} \label{l.filling}
Let $f\in \mathcal{A}_0(\Sigma)$ and $S\subset \Sigma$ be a non-trivial connected surface that is quasi-invariant under $f$.
\begin{enumerate}
\item If $S$ is curved, then there is a zero-rotation fixed point in $S$.
\item If $\Lambda$ is a basic piece contained in $\text{Fill}(S)\setminus S$, then there is a fixed point $x_0$ in $S$ with $\rho_{\Lambda}(f)=\rho(f,x_0)$.
\end{enumerate}
\end{lemma}

\begin{proof} Following the notation in the proof of point (4) of Lemma \ref{l.qisurface}, consider $\pi_S:\hat{\Sigma}_S\to \Sigma$ the covering space associated to $\textrm{i}_*\pi_1(S)=\textrm{i}_*\pi_1(\text{Fill}(S))$, and $\hat{S}$ the closed lift of $S$ to this cover. In this case $S$ is not necessarily filled, but the construction is the same as the one for $\text{Fill}(S)$, and $\hat{S}$ is still homeomorphic to $S$. Also consider an homotopy $f_t$  between $f_0=Id$ and $f_1=f$, take $\overline{f}_t$ the lift of $f_t$ to $\hat{\Sigma}_S$ starting at $\overline{f}_0 = Id$, and let $\overline{f}=\overline{f}_1$, which is a lift of $f$.

Define $$S_0:= \text{Fill}\left(\bigcup_{n\in\Z}\overline{f}^n(\hat{S})\right).$$

Notice that $S_0$ is $\overline{f}$-invariant by construction, and by the properties of quasi-invariance for $S$, it consists of $\text{Fill}(\hat{S})$ together with semi-open annuli $C_{\gamma}$  attached to each boundary component $\gamma\subset\partial\text{Fill}(\hat{S})$. Moreover, for each such $\gamma$ we have $C_{\gamma}\cap\pi_S^{-1}(\Omega(f))=\emptyset$, and in particular $C_{\gamma}$ contains no fixed points of $\overline{f}$. We shall consider $\overline{f}$ restricted to $S_0$. It is isotopic to the identity in $S_0$ (which can be seen by a standard construction), and has finitely many fixed points.

The key of the proof is the {\em Lefschetz-Nielsen index theorem} applied to $\overline{f}$ in $S_0$. The first equation given by this theorem relates the Euler characteristic of $S_0$ with the sum of indices of contractible fixed points of $\overline{f}$:
\begin{equation}\label{ecu1}
\sum_{x\in\text{Fix}(\overline{f})\,:\,\rho(\overline{f},x)=0} \text{Ind}(\overline{f},x) = \chi(S_0)
\end{equation}
On the other hand, recall that by Lemma \ref{l.qisurface} (point (2)), if $D$ is a quasi-invariant disk, there exists $[\alpha]\in H_1(S_0)$ (possibly $0$) so that every basic piece $\Lambda\subseteq D$ satisfies $\rho_{\Lambda}(\overline{f})=[\alpha]$. In particular, all fixed points in $D$ have the same rotation vector. Therefore, by the classical Lefschetz index theorem we get:

\begin{equation}\label{ecu2}
\sum_{x\in\text{Fix}(\overline{f})\cap D\,:\,\rho(\overline{f},x)=v} \text{Ind}(\overline{f},x) =  \left\{ \begin{array}{ll}
         1 & \mbox{if $v =[\alpha]$};\\
        0 & \mbox{otherwise}.\end{array} \right.
\end{equation}

To prove point (1) notice that since $S$ is curved we have $\chi(S_0)<0$. Therefore equations \ref{ecu1} and \ref{ecu2} imply the existence of $p\in\text{Fix}(\overline{f})\cap\hat{S}$ with $\rho(\overline{f},p)=0$. Thus $\pi_S(p)$ is the desired fixed point.

The proof of point (2) is very similar. In this case we need to consider the second part of the Lefschetz-Nielsen index theorem, for fixed points in a non-zero rotation class. Namely, if $[\alpha]\neq 0$ then:

\begin{equation}\label{ecu3} \sum_{x\in\text{Fix}(\overline{f})\,:\,\rho(\overline{f},x)=[\alpha]} \text{Ind}(\overline{f},x) = 0
\end{equation}

Consider $\Lambda$ the basic piece in the hypothesis of point (2), and let $\hat{\Lambda}$ be the corresponding lift in $\text{Fill}(\hat{S})$. Notice that $\rho_{\Lambda}(f) = (\pi_S)_*\rho_{\hat{\Lambda}}(\overline{f})$, thus it suffices to work with $\overline{f}$ in $S_0$ . First assume that $\rho_{\hat{\Lambda}}(\overline{f})\neq 0$. Notice that the disk containing $\hat{\Lambda}$ contributes with a $+1$ in the left side of \ref{ecu3} while the other disks in the complement of $\hat{S}$ contribute with non-negative values. Therefore, applying equations \ref{ecu2} and \ref{ecu3} to $v=\rho_{\hat{\Lambda}}(\overline{f})$ we prove point (2) for this case.

If $\rho_{\hat{\Lambda}}(\overline{f})=0$ and $S$ is curved we are in the hypothesis of case (1). It only remains the case where $S$ is annular, in which $\chi(S_0)=0$ and the disk containing $\hat{\Lambda}$ produces a $+1$ contribution in the left side of equation \ref{ecu1}. Thus equations \ref{ecu1} and \ref{ecu2} give the desired result.

\end{proof}

\begin{rem}\label{r.filling} With similar arguments as those in the proof of Lemma \ref{l.filling}, we can show that if $S$ is a quasi-invariant surface that is not homeomorphic to an annulus, then there exists a fixed point $p\in S$. This holds even when $S$ is trivial, which is the only case not covered in Lemma \ref{l.filling}.
\end{rem}

Next we recall Conley's Fundamental Theorem for homeomorphisms in $\mathcal{A}_0(\Sigma)$.

\begin{thm}[\cite{conleyfundthm}]\label{t.conley}

Let $f\in\mathcal{A}_0(\Sigma)$. Then there exists a continuous function
$\mathcal{L}:\Sigma\to\R$ so that:

\begin{enumerate}

\item $\mathcal{L}(x)\geq \mathcal{L}(f(x))$ for all $x\in\Sigma$, and

\item for any pair $x,y\in\Omega(f)$ we have $\mathcal{L}(x)=\mathcal{L}(y)$ iff $x$ and $y$
belong to the same basic piece of $f$.

\end{enumerate}

\end{thm}

A function $\mathcal{L}$ as in Theorem \ref{t.conley} is called a \emph{Lyapunov function}. Lyapunov functions are not unique, and neither are the partitions of $\Sigma$ induced by their
level sets. By the construction in \cite{conleyfundthm}, $\mathcal{L}$ can be considered smooth in the complement of $\Omega(f)$.

Our aim is to give a decomposition of $\Sigma$ into subsurfaces, each containing one basic piece of $f$, that comes from some partition by level sets of a Lyapunov function. That will be useful in order to decompose $\Sigma$ into surfaces with quasi-invariance properties, each containing a single basic piece. The starting idea is to take a partition of $\Sigma$ using level sets of a Lyapunov function $\mathcal{L}$, choosing the levels as to isolate the different basic pieces of $f$. There are technical issues with such a partition by level sets that force us to make some modifications. Namely, we do not want these surfaces to have components contained in the wandering set of $f$. The gist of these modifications is to attach each of these wandering components, if they exist, to a suitable neighbouring subsurface. Next we give the detail, including a precise definition for what we consider as {\em Conley surface} associated to $\mathcal{L}$.

We point out that, for ease of notation, we call a subsurface {\em wandering} if it does not meet $\Omega(f)$.

Given a surface $S$ we define $\text{WFill}(S)$ as the union of $S$ with all the connected components of $\Sigma\setminus S$ which are homeomorphic to either disks or annuli and do not meet $\Omega(f)$.  We say that $\text{WFill}(S)$ is the \emph{wandering filling}, or \emph{W-filling} of $S$, and that $S$ is {\em W-filled} when $S=\text{WFill}(S)$. Notice that disjoint subsurfaces meeting $\Omega(f)$ have disjoint W-fillings. On the other hand, the W-filling of their union may be strictly larger than the union of their respective W-fillings, as there may be wandering annuli bounded between the two subsurfaces.

As motivation for our definition of W-filling, we point out that for a surface of the form  $S=\mathcal{L}^{-1}([a,b])$, where  $\mathcal{L}$ is a Lyapunov function and $a,b\in\R\setminus\mathcal{L}(\Omega(f))$, the W-filling of $S$ agrees with the union of {\em all} the components of $\Sigma\setminus S$ that do not meet $\Omega(f)$. The proof of this fact follows that of Lemma \ref{l.filling}, and we omit it.

In what follows we consider a Lyapunov function $\mathcal{L}$ that is smooth in the complement of $\Omega(f)$. Given a basic piece $\Lambda$, we say that a subsurface $S$ is a \emph{Conley surface} of $\Lambda$ if it verifies the following conditions:

\begin{enumerate}

\item Each component of $\partial S$ is contained in some level set $\mathcal{L}^{-1}(a)$ with $a\notin \mathcal{L}(\Omega(f))$. (Where $a$ depends on the component of $\partial S$).
\item $S\cap\Omega(f)=\Lambda$,
\item every connected component of $S$ meets $\Lambda$,
\item $S$ is W-filled.
\end{enumerate}

Points (3) and (4) are technical, their purpose is to avoid wandering components in $S$ or $\Sigma\setminus S$. The existence of Conley surfaces follows from Theorem \ref{t.conley}.
Moreover, we can decompose $\Sigma$ into Conley surfaces, as we show in the next result.

\begin{lem}\label{l.masconley} Given $f\in\mathcal{A}_0(f)$ with basic pieces $\Lambda_1,\ldots,\Lambda_N$ and a Lyapunov function $\mathcal{L}$ that is smooth in $\Sigma\setminus\Omega(f)$, we can find a family of surfaces $S_{1},\ldots,S_{N}$ associated to $\mathcal{L}$ so that:
\begin{enumerate}

\item $\textrm{int}(S_i)\cap\textrm{int}(S_j)=\emptyset$ whenever $i\neq j$.

\item $\bigcup_{i=1}^N S_i=\Sigma$,

\item For every $i=1,\ldots,N$ we have that $S_i$ is a Conley surface for $\Lambda_i$.

\end{enumerate}

\end{lem}

A family of surfaces that satisfies the statement of Lemma \ref{l.masconley} will be called a \emph{Conley decomposition} of $\Sigma$ for $f$.

\tiny
\begin{figure}[h!]\begin{center}

\centerline{\includegraphics[scale=.5]{CSFig.pdf}}
\caption{The picture illustrates in gray a Conley surface, which can be obtained as follows: starting with $S'_i$, then taking its \textrm{WFill} by adding up the disk $D_1$ and not $D_2$, as this last disk meets the non-wandering set, which is depicted by the black box. Then one can add adjacent
wandering annuli as is the case of $A_1$, and could further add the annuli $A_2$ and $A_3$.}\label{shift}
\end{center}\end{figure}
\normalsize

\begin{proof}

As we outlined earlier, we begin by taking $a_1,\ldots,a_{N+1}\in\R\setminus\mathcal{L}(\Omega(f)) $ as regular values of $\mathcal{L}$ so that $S^0_i=\mathcal{L}^{-1}([a_i,a_{i+1}])$ contains $\Lambda_i$. Let $S'_i$ be the union of all the connected components of $S^0_i$ that meet $\Omega(f)$. These surfaces satisfy (1)-(3) in the definition of Conley surfaces, and are pairwise disjoint except possibly at their boundaries. We also observe that if $\Sigma\setminus \cup_i S'_i$ is non empty, then its connected components are contained in the wandering set and consist on either disks or annuli, as can be proved by the same argument used for Lemma \ref{l.filling}.

We take $S''_i = \textrm{WFill}(S'_i)$. Notice that $S''_i$ is already a Conley surface for the basic piece $\Lambda_i$. Their interiors are also pairwise disjoint, by the properties of the W-filling.

On the other hand, $\cup_iS''_i$ may not yet cover $\Sigma$. Passing to $S''_i = \textrm{WFill}(S'_i)$ adds in the wandering disk components of $\Sigma\setminus \cup_i S'_i$, as well as the wandering annuli that have both boundary components in the same surface $S'_i$. (In fact, these annuli are not present, but the proof is cumbersome and we will not need it). Thus the components of $\Sigma\setminus \cup_i S''_i$ are wandering annuli, with boundary components on different surfaces $S''_i$. For each of these annuli, we choose one neighbouring surface $S''_i$ and attach the annuli to it.

This gives the construction of the surfaces $S_i$ in the statement.

\end{proof}

\begin{rem} \label{r.masconley}
\noindent
\begin{enumerate}
\item The construction in Lemma \ref{l.masconley} gives Conley surfaces $S$ that have the following form:

There are $a,b\in\R\setminus\mathcal{L}(\Omega(f))$ so that $S$ is the W-fill of the union of the components of $\mathcal{L}^{-1}([a,b])$ that meet $\Omega(f)$, together with wandering annuli, each attached to one boundary component.

\item Every Conley decomposition of $\Sigma$ associated to the Lyapunov function $\mathcal{L}$ is obtained from the construction in the proof of Lemma \ref{l.masconley}, except possibly by varying each curve on the decomposition within the same class of the Morse Theory of $\mathcal{L}$.

\item With similar arguments as in the proof of Lemma \ref{l.masconley}, we can obtain that any Conley surface for $f$ is part of a Conley decomposition. Thus Lemma \ref{l.masconley} gives a general way to construct Conley surfaces.

\end{enumerate}

\end{rem}

As we pointed out at the beginning of this section, Conley and quasi-invariant surfaces are closely related. Showing this relationship will be our next goal. We begin with a preliminary result that will be useful in general for proving quasi-invariance of subsurfaces. It is worth to point out that this is specific for hyperbolic surfaces, and is not true for the torus.

\begin{lemma}\label{l.wandering} Consider $f\in\mathcal{A}_0(\Sigma)$ and $\gamma$ an essential simple closed curve satisfying $$f^n(\gamma)\cap\gamma=\emptyset \text{ for every } n>0$$ Then the annulus bounded by $\gamma$ and $f^n(\gamma)$ does not meet $\Omega(f)$ for every $n>0$.
\end{lemma}
\begin{proof} Since $f$ is isotopic to the identity, $f^n(\gamma)$ is homotopic to $\gamma$  for all $n$, and the subsurface of $\Sigma$ with boundary $\gamma\cup f(\gamma)$ is an annulus $A_0$. We consider the semi-open annulus $A=A_0\setminus f(\gamma)$.
Suppose there exists $k>0$ so that $f^k(A)\cap A\neq\emptyset$. We can consider the first such $k$, so that $f^m(A)\cap  A=\emptyset$ for $0<m< k$. Notice that $f^k(\gamma)$ cannot meet $A$, since $f^{k-1}(A)$ is adjacent to $f^k(A)$ at $f^k(\gamma)$. Since $f^{k+1}(\gamma)\cap (\gamma\cup f(\gamma))=\emptyset$, this leaves us with two possibilities: either $A\subset f^k(A)$ or $f^{k+1}(\gamma)$ is contained in the interior of $A$. The first possibility is ruled out by considering the annulus $B$ between $f^k(\gamma)$ and $\gamma$, and observing that $f$ would map $B$ to $f(B)$ reversing orientation. To exclude the second one, observe that $\cup_{i=0}^{i=k}f^i(A)$ would be homeomorphic to a torus, which contradicts the hypothesis that $\Sigma$ is a connected surface of higher genus.

Therefore we have that $f^i(A)\cap A=\emptyset$ for every $i>0$, which implies that $\cup_{i=0}^{i=\infty}f^i(A)$ is a semi-open annulus disjoint from $\Omega(f)$. In particular, the annulus between $\gamma$ and $f^n(\gamma)$ does not meet $\Omega(f)$ for every $n>0$.
\end{proof}

The following result gives the relationship we seek, going from Conley surfaces to quasi-invariant surfaces.

\begin{lemma}\label{l.conleyesquasi} Let $f\in\mathcal{A}_0(\Sigma)$ and $S$ be a Conley surface which has a non-trivial connected component. Then the following hold.
\begin{enumerate}
\item $S$ is connected,
\item there exists $k\in\N$ so that $S$ is quasi-invariant under $f^k$,
\item $\text{Fill}(S)$ is quasi-invariant under $f$.
\end{enumerate}
Moreover, if $S'$ is another connected and non-trivial Conley surface that is adjacent to $S$ in the same Conley decomposition of $\Sigma$, then $\text{Fill}(S)\cup \text{Fill}(S')$ is also quasi-invariant.
\end{lemma}

\begin{proof} Take $S_{\ast}$ a non-trivial connected component of $S$. We shall split the proof into a series of claims, showing first that $\text{Fill}(S_{\ast})$ is quasi-invariant and that $S=S_{\ast}$. This will imply points (1) and (3) immediately, and will be useful for the rest of the statement, that will be left for the end of the proof.

In order to simplify notation we will assume first that $S$ is the W-filling of the connected components that meet $\Omega(f)$ of $\mathcal{L}^{-1}([a,b])$ for some $a,b\in\R$. That is to say that $S$ has the form given in Remark \ref{r.masconley} point (1), with no wandering annuli attached to the boundary. In other words, we are assuming $S$ is of the form $S''_i$ in the proof of Lemma \ref{l.masconley}. We shall prove Claims 1 to 3 below assuming this form for $S$, and afterwards we will generalize for $S$ as in point (1) of Remark \ref{r.masconley}, possibly with wandering annuli attached at the boundary. By points (2) and (3) of the same Remark \ref{r.masconley}, this gives the lemma for $S$ a general Conley surface.

Denote $S_0:=\text{Fill}(S_{\ast})$ and for each boundary component $\gamma\subseteq\partial S_0$ define

\begin{equation}
\gamma':=  \left\{ \begin{array}{ll}
         f(\gamma) & \mbox{if $\mathcal{L}(\gamma)=a$};\\
        f^{-1}(\gamma) & \mbox{if $\mathcal{L}(\gamma)=b$}.\end{array} \right.
\end{equation}

{\bf Claim 1:} $S_0$ is quasi-invariant.

\vspace{.1cm}

Observe that $\gamma'\subseteq\Sigma\setminus S_0$ for every $\gamma\subseteq\partial S_0$, by point (1) in Theorem \ref{t.conley}. First we see that $$\gamma_1'\cap\gamma_2'=\emptyset \text{ whenever } \gamma_1\neq\gamma_2$$
When $\gamma_1$ and $\gamma_2$ have the same image under $\mathcal{L}$ it follows from injectivity of either $f$ or $f^{-1}$. When they have different images under $\mathcal{L}$, it follows from point (1) in Theorem \ref{t.conley}.

Notice that since $S_0$ is filled and non-trivial, each component $\gamma\subseteq\partial S_0$ is essential and so $\gamma$ and $\gamma'$ bound an annulus $A_{\gamma}$, adjacent to $S_0$ by $\gamma$. By point (1) in Theorem \ref{t.conley} we see that $f^n(\gamma)\cap\gamma=\emptyset \text{ for every } n>0$, thus we may apply Lemma \ref{l.wandering} to conclude that $A_{\gamma}\cap\Omega(f)=\emptyset$. Next we see that $$A_{\gamma_1}\cap A_{\gamma_2}=\emptyset \text{ whenever } \gamma_1\neq\gamma_2$$

Suppose this is not the case. Since all the boundary curves are pairwise disjoint and both annuli are disjoint from $S_0$, we get that $\gamma'_1\subset \text{int}(A_{\gamma_2})$ and $C:=A_{\gamma_1}\cup A_{\gamma_2}$ is an annulus. Also notice that $\text{int}(C)$ is a component of $\Sigma\setminus S_0$ and $C\cap \Omega(f)=\emptyset$, but this contradicts the fact that $S_0$ is W-filled.

Define $$S_1:=S_0\cup(\cup_{\gamma} A_{\gamma})$$

We have that $f(S_0)\subseteq S_1$, since  $f(S_0)\cap S_1$ is non-empty and both open and closed in $f(S_0)$. Similarly $f^{-1}(S_0)\subseteq S_1$. Therefore we have shown that $S_0$ is quasi-invariant under $f$, using the family of annuli $A_{\gamma}$. This shows Claim 1.

\vspace{.1cm}
{\bf Claim 2:} $S=S_{\ast}$.

\vspace{.1cm}

Let $\Lambda$ be the basic piece associated to $S$. We obtained from the previous claim that $\Omega(f)\cap S_0$ is $f$-invariant, so $\Lambda\subset S_0$. Thus if there are any components of $S$ other than $S_{\ast}$, they must be contained in $S_0 = \text{Fill}(S_{\ast})$, for they have to meet $\Lambda$. In particular these components must be trivial and contained in $S_0\setminus S_{\ast}$. We shall introduce some notation for the components of $S_0\setminus S_{\ast}$, which are all disks with boundaries in either $\mathcal{L}^{-1}(a)$ or $\mathcal{L}^{-1}(b)$. Let $D^-_1,\ldots,D^-_m$ be the components of $S_0\setminus S_{\ast}$ whose boundaries are contained in $\mathcal{L}^{-1}(a)$, and let $\mathcal{D}^- = \cup_i D_i$. Analogously define $D^+_1,\ldots,D^+_l$ and $\mathcal{D}^+$ for the components with boundary in $\mathcal{L}^{-1}(b)$.

We will show that $\mathcal{D}^-\cap\Omega(f)$ is $f$-invariant. This will imply, by transitivity of the basic pieces, that $\Lambda$ does not meet $\mathcal{D}^-$ and thus $S$ has no components in $\mathcal{D}^-$. The same will hold for $\mathcal{D}^+$, and that will give this Claim.

To prove that $\mathcal{D}^-\cap\Omega(f)$ is $f$-invariant, we look at $f(D_i^-)$ for each $i=1,\ldots,m$. On one hand $D_i^-\subset S_0$, thus $f(D_i^-)\subset S_1 $. On the other, by Theorem \ref{t.conley} and the fact that $\partial D_i^-\subset \mathcal{L}^{-1}(a)$, we see that $f(D_i^-)$ lies in $S_1\cap \mathcal{L}^{-1}((-\infty,a))$ which is contained in $\mathcal{D}^-\cup (\cup_{\gamma}A_{\gamma})$. Since $D_i^-$ is connected, $f(D_i^-)$ must be contained either in some component of $\mathcal{D}^-$, or in $A_{\gamma}$ for some $\gamma\subset\partial S_0$. Since $S_{\ast}$ is W-filled, we have that $D_i^-$ meets $\Omega(f)$, and so does $f(D_i^-)$, thus it cannot be contained in any of the wandering annuli $A_{\gamma}$. We obtain that $f(D_i^-)\subset \mathcal{D}^-$. As this holds for every $i=1,\ldots,m$, we get $f(\mathcal{D}^-)\subset \mathcal{D}^-$ and so $\mathcal{D}^-\cap\Omega(f)$ is $f$-invariant, as we desired. The same proof works for $\mathcal{D}^+$, and this proves Claim 2.

\vspace{.1cm}
{\bf Claim 3:} Points (1), (2) and (3) in the statement.
\vspace{.1cm}

Point (1) comes directly from Claim 2, recalling the definition of $S_{\ast}$, and point (3) from combining Claims 1 and 2, recalling that $S_0 = \text{Fill}(S_{\ast}) = \text{Fill}(S)$.

It remains to show point (2). From the proof of Claim 2 we see that for each $i=1,\ldots,m$ we have $f(D_i^-)\subset D_{\sigma(i)}^-$ for some $\sigma(i)\in\{1,\ldots,m\}$. By transitivity of basic pieces $\sigma$ must be a permutation of $\{1,\ldots,m\} $. Similarly $f^{-1}(D_j^+)\subset D_{\tau(j)}^+$ for a permutation $\tau$ of $\{1,\ldots,l\}$. Pick $k$ so that $\sigma^k=Id$ and $\tau^k=Id$. Thus $f^k(D_i^-)\subset\text{int}(D_i^-)$ and $f^{-k}(D_j^+)\subset\text{int}(D_j^+)$ for every $i=1,\ldots,m$ and $j=1,\ldots,l$, and this defines annuli $A^-_i$ and $A^+_j$ as the respective closures of $D_i^-\setminus f^k(D_i^-)$ and $D_j^+\setminus f^{-k}(D_j^+)$. It is clear that these annuli do not meet $\Omega(f)$. For $\gamma$ in the boundary of $S_0$, which is the essential boundary of $S$, let

$$
A'_{\gamma} :=  \left\{ \begin{array}{ll}
  \cup_{s=1}^k f^s(A_{\gamma})        & \mbox{if $\mathcal{L}(\gamma)=a$};\\
    \cup_{s=1}^k f^{-s}(A_{\gamma}) & \mbox{if $\mathcal{L}(\gamma)=b$}.\end{array} \right.
$$

It is straightforward to verify the definition of quasi-invariance under $f^k$ for the set of annuli we just described, consisting on $A'_{\gamma}$ for $\gamma$ in the essential boundary of $S$, $A^-_i$ for $i=1,\ldots,m$, and $A^+_j$ for $j=1,\ldots,l$.
This concludes Claim 3.

\vspace{.1cm}
We have shown Claims 1 through 3 assuming that $S$ is one of the surfaces $S''_i$ in the construction for Lemma \ref{l.masconley}. Attaching some wandering annuli to $S$ would make the notation more cumbersome, and would actually add difficulty only in proving the disjointness of the boundary annuli for the quasi-invariance in Claim 1. This can be sorted out by observing that if an essential wandering annulus is adjacent to $S$ by $\gamma \subset \partial S$, it is eventually contained in $\cup_{s=0}^m f^{s}(A_{\gamma})$ or $\cup_{s=0}^m f^{-s}(A_{\gamma})$ for large enough $m$. (Where $A_{\gamma}$ was defined in Claim 1.)

\vspace{.1cm}
{\bf Claim 4:} If $S'$ is another connected and non-trivial Conley surface that is adjacent to $S$ in the same Conley decomposition, then $\text{Fill}(S)\cup \text{Fill}(S')$ is quasi-invariant under $f$.
\vspace{.1cm}

To obtain the Conley decomposition under consideration, which includes $S$ and $S'$, we follow the construction for Lemma \ref{l.masconley}. To obtain the statement for a general Conley decomposition we resort to point (2) of Remark \ref{r.masconley}.

By such construction of the Conley decomposition we see that $S\cup S'$ must be W-filled, since any wandering annulus between $S$ and $S'$ should have been already contained in $S$ or $S'$. For any component $\gamma$ in the boundary of $\text{Fill}(S)\cup \text{Fill}(S')$ we can associate an annulus $A_{\gamma}$ that comes from Claim 1 for either $S$ or $S'$. These annuli give the quasi-invariance of $\text{Fill}(S)\cup \text{Fill}(S')$ by similar arguments as those used for Claim 1, so we omit the details. The only delicate point is the disjointness of the annuli $A_{\gamma}$, in which we need to use that $\text{Fill}(S)\cup \text{Fill}(S')$ is W-filled.

\end{proof}

\begin{rem}\label{r.discos} If $S$ is a non-trivial Conley surface then $\text{cl}[\text{Fill}(S)\setminus S]$ can be written as $\mathcal{D}^+\sqcup \mathcal{D}^-$, where $\mathcal{D}^+$ and $\mathcal{D}^-$ are disjoint unions of disks, with $f(\mathcal{D}^+)\subseteq \text{int}(\mathcal{D}^{+})$ and $f^{-1}(\mathcal{D}^-)\subseteq \text{int}(\mathcal{D}^{-})$.
 \end{rem}

Once we choose a fixed Conley decomposition, Lemma \ref{l.conleyesquasi} lets us classify the basic pieces of $f$ into {\em trivial}, {\em annular} or {\em curved}, according to their associated Conley surfaces. Given a basic piece $\Lambda$, let us call $S_{\Lambda}$ to its Conley surface, then

\begin{itemize}
\item $\Lambda$ is {\em trivial} if every component of $S_{\Lambda}$ is trivial. In this case $S_{\Lambda}$ is contained in  a disjoint union of disks,  either attracting or repelling.
\item Otherwise, $S_{\Lambda}$ is connected by Lemma \ref{l.conleyesquasi}. We say that $\Lambda$ is {\em annular/curved} if $S_{\Lambda}$ is annular/curved respectively.
\end{itemize}

\begin{cor} Let $f\in\mathcal{A}_0(\Sigma)$ and $S$ be a Conley surface
of a non-trivial basic piece $\Lambda$. If $\Lambda'\subset \text{Fill}(S)$ is a trivial basic piece, then $\rho_{\Lambda'}(f)\subset\rho_{\Lambda}(f)$.
\end{cor}
\begin{proof} This follows from Lemma \ref{l.conleyesquasi} point (1) and Lemma \ref{l.filling} point (2).
\end{proof}

In general, for $f$ a homeomorphism of $\Sigma$, a subsurface $S\subset\Sigma$, and $n>0$, we denote $$S^{(n)}= \text{Fill}\left(\bigcup_{|j|\leq n} f^j(S)\right)$$
We point out that if $S$ is filled and quasi-invariant then $S^{(1)}=S\cup(\cup A_{\gamma})$ where $A_{\gamma}$ are annuli attached to the components of $\partial S$. The same is true for $S^{(n)}$, by taking $f^n$. Getting this property is the reason why we fill {\em after} taking the union, and not {\em before}, which could yield a strictly smaller set.
Then Lemma \ref{l.conleyesquasi} implies the following.
\begin{lem} \label{l.conleycasesess0} Assume $S$ is the fill of either
\begin{itemize}
\item  a non-trivial Conley surface or
\item the union of two non-trivial Conley surfaces which are adjacent in the same Conley decomposition of $\Sigma$.
\end{itemize} Then the following hold for every $n>0$,
\begin{enumerate}
\item $S^{(n)}\cap\Omega(f)=S\cap\Omega(f)$,
\item $S^{(n)}$ is connected, filled and retracts by deformation onto $S$ rel $\Omega(f)$,
\item $S^{(n)}$ is quasi-invariant under $f^n$.
\end{enumerate}
\end{lem}

Although Conley decompositions will play the central role in our main Theorem \ref{mainthm}, we will also need another construction of quasi-invariant surfaces for sections \ref{s.agujas} and \ref{s.triviales}. This is given in the next lemma, with the extra assumption of having topologically mixing basic pieces, which by Remark \ref{r.piecespower} applies to some power of any $f\in\mathcal{A}_0(\Sigma)$. Recall that for a basic piece $\Lambda$ of an axiom A diffeomoprhism
$f$ we defined $\Lambda^u=\textrm{cl}[W^u(\Lambda,f)]$.

\begin{lemma}\label{l.construquasi} Suppose $f\in\mathcal{A}_0(\Sigma)$ has topologically mixing basic pieces. Assume $U$ is a subsurface such that $\text{cl}[f(U)]\subseteq\text{int}(U)$, and every connected component of $U$ meets $\Omega(f)$. Let $K\subseteq U$ be a continuum and $\Lambda_1,\ldots,\Lambda_N$ basic pieces satisfying:
\begin{itemize}
\item $\Lambda_i\subseteq U$ for $i=1,\ldots,N$
\item $K\cap[\cup_{i=1}^{i=N}\Lambda_i^u]=\emptyset$

\end{itemize}
Then there exists a quasi-invariant surface $S$ such that $K\subseteq S$ and $$S\cap\Omega(f)=(U\cap\Omega(f))\setminus\{\Lambda:\Lambda_i\preceq\Lambda \text{ for some }i=1,\ldots,N\}$$
\end{lemma}

Together with Remark \ref{r.piecespower}, this will let us build quasi-invariant surfaces with some control over the basic pieces contained in them. By choosing $K$ appropriately, we will also be able to join some of those basic pieces with an arc within the surface $S$. These specific requirements make it difficult to obtain $S$ through an arbitrary Conley decomposition, as union of Conley surfaces, so we provide this alternative construction.

\begin{proof}

Notice that we may replace $U$ with $\text{WFill}(U)$ without loss of generality, so we assume $U$ is W-filled. Consider $U_1,\ldots,U_k$ the connected components of $U$. Since they all meet $\Omega(f)$, by recurrence we have   $f(U_i)\subseteq U_{\sigma(i)}$ where $\sigma$ is a permutation of $\{1,\ldots,k\}$. Since $f$ is topologically mixing on its basic pieces, we get that $\sigma=Id$.

First we shall prove that $U$ is quasi-invariant: For each component $\gamma$ of $\partial U$ define $\gamma':=f^{-1}(\gamma)$. We will show that $\gamma$ and $\gamma'$ bound an annulus $A_{\gamma}$ adjacent to $U$ by $\gamma$. When $\gamma$ is essential this is clear, since $f$ is isotopic to $Id$. When $\gamma$ is inessential there are two cases to distinguish:
\begin{enumerate}
\item $\gamma$ bounds a disk that contains a trivial component $U_j$ of $U$.
\item $\gamma$ bounds a disk in $\text{Fill}(U_j)\setminus U_j$, for $U_j$ any component of $U$.
\end{enumerate}
The first case follows from the fact that $f(U_j)\subset U_j$. For the second one, let $D_1,\ldots,D_m$ be the components of $\text{Fill}(U_j)\setminus U_j$, and $\gamma_l = \partial D_l$ for $l=1,\ldots,m$. Since $U$ is W-filled we have $D_l\cap\Omega(f)\neq \emptyset$ for all $l=1,\ldots,m$, thus $f(D_l)\supset D_{\tau(l)}$ where $\tau$ is a permutation of $\{1,\ldots,m\}$. We get that $\tau=Id$ by the topologically mixing property of the basic pieces, therefore $\gamma_l$ and $\gamma'_l$ bound an annulus for every $l=1,\ldots,m$. The disjointness of the annuli $A_{\gamma}$ comes from the injectivity of $f$ and the fact that $U$ is W-filled, as in the proof of Lemma \ref{l.conleyesquasi}. Thus the quasi-invariance of $U$ follows.

We will arrive at our desired surface $S$ by removing a subsurface $Z$ from $U$, that we shall construct next.

Since $\Lambda_i\subseteq \text{int}(U)$ for $i=1,\ldots,N$ and $U$ is forward-invariant under $f$, we get that $\Lambda_i^u\subseteq U$ for every $i=1,\ldots,N$. Recall that by Proposition \ref{p.propfaxa} point (3) and Lemma \ref{l.entornos1} we have that for every $i=1,\ldots,N$:
\begin{itemize}
\item $\Lambda_i^u$ is an attractor for $f$, and
\item $\Lambda_i^u\cap\Omega(f)=\cup_{\Lambda\succeq\Lambda_i}\Lambda$.
\end{itemize}

Thus for every $i=1,\ldots,N$ there is a subsurface $Z_i\subseteq U$, obtained as a neighbourhood of $\Lambda_i^u$, that satisfies:
\begin{itemize}
\item $\text{cl}[f(Z_i)]\subseteq \text{int}(Z_i)$, and
\item $Z_i\cap\Omega(f)=\Lambda_i^u \cap\Omega(f) =\cup_{\Lambda\succeq\Lambda_i}\Lambda$.
\end{itemize}

Define $Z$ as the union of the connected components of $\text{WFill}(\cup_{i=1}^{i=N}Z_i)$ that meet $\Omega(f)$. We can choose the surfaces $Z_i$ so that their boundaries intersect transversally, thus making $Z$ a subsurface. Since $\text{cl}[f(Z)]\subset \text{int}(Z) $, we get that $Z$ is quasi-invariant, by the same argument we used for $U$. Taking the surfaces $Z_i$ as small enough neighbourhoods of $\Lambda_i^u$ we may assume that  $\text{cl}[Z]\subset \text{int}(U)$ and $Z\cap K=\emptyset$. Then $S = U\setminus Z$ is the desired surface. Checking the details is straightforward.

\end{proof}

As the final point of this section, we turn back to Conley surfaces and look at the dynamic behaviour of their boundaries.

\begin{lem}\label{sl.borde} Let $S$ be a non-trivial Conley surface and $\gamma\subseteq \partial S$ be an essential boundary component. Then the following holds.

\begin{enumerate}
\item If $f(\gamma)\cap S=\emptyset$ then the annulus between $\gamma$ and $f^n(\gamma)$  meets $S$ only at $\gamma$ for every $n>0$,
\item If $f(\gamma)\cap S \neq\emptyset$ then $f^n(\gamma)\cap S\neq\emptyset$ for every $n>0$.
\end{enumerate}

\end{lem}
\begin{proof} Since $\gamma$ is essential, the surface between $\gamma$ and $f^n(\gamma)$ is an annulus $A_n$. If for some $n$ we have that $f^n(\gamma)\cap S=\emptyset$, then $(S\setminus\gamma)\cap A_n$ is both open and closed in $S\setminus\gamma$, thus either $(S\setminus\gamma)\cap A_n=\emptyset$ or $S\setminus\gamma\subseteq A_n$. Lemma \ref{l.wandering} excludes the second possibility, so $(S\setminus\gamma)\cap A_n=\emptyset$ whenever $f^n(\gamma)\cap S=\emptyset$.

For point (1) assume that $f(\gamma)$ does not intersect $S$. By Lemma \ref{l.conleyesquasi} and the fact that $(S\setminus\gamma)\cap A_1=\emptyset$ we have that $f^n(\gamma)\cap S=\emptyset$ for every $n>0$. Therefore we must have $(S\setminus\gamma)\cap A_n=\emptyset$, which gives point (1).

We prove point (2) by contradiction, supposing that $f^n(\gamma)\cap S=\emptyset$ for some $n>0$. Then we get that $(S\setminus\gamma)\cap A_n=\emptyset$, which contradicts the fact that $f(\gamma)\cap S\neq\emptyset$.
\end{proof}

The following relates the essential boundary of a Conley surface with the stable and unstable manifolds of its basic piece.

\begin{lemma}\label{l.conleycasesess}

Let $f\in\mathcal{A}_0(\Sigma)$ and $S$ be a non-trivial Conley surface
associated to a basic piece $\Lambda$. Then for any essential component $b$ of $\partial S$ we have that either $W^s(\Lambda,f)\cap b\neq\emptyset$ or $W^u(\Lambda,f)\cap b\neq\emptyset$.

\end{lemma}

We remark that
\begin{itemize}
\item if $f(b)\cap S =\emptyset$ then $b$ meets $W^u(\Lambda,f)$, while

\item if $f(b)\cap S \neq\emptyset$ (which happens iff $f^{-1}(b)\cap S =\emptyset$), then $b$ meets $W^s(\Lambda,f)$.

\end{itemize}

\begin{proof}
We shall assume that $f(b)\cap S\neq\emptyset$ and prove that $W^s(\Lambda,f)\cap b\neq\emptyset$. The other case is analogous.  Consider the set
$$A=\{x\in b\,:\, \mbox{there is some }k>0 \mbox{ such that } f^{k}(x)\in\Sigma \setminus S \}\subseteq b$$

We wish to show that $A\neq b$, for if there is a point $z \in b\setminus A$ then all the future iterates of $z$ are contained in $S$, which implies that $z\in W^{s}(\Lambda,f)$ as every point in $\Sigma$ must belong to the stable manifold of some basic piece. Thus we would find $z\in b\cap W^{s}(\Lambda,f)$ as desired.

For $x\in A$ define $$k_0(x)=\min \{k>0:f^k(x)\in\Sigma\setminus S\} $$

\vspace{.1cm}
{\bf Claim:} If $x\in A$ then $f^n(x)\in\Sigma\setminus S $ for every $n\geq k_0(x)$.
\vspace{.1cm}

Recall from Remark \ref{r.discos} that we can write $\text{Fill}(S)\setminus S = \mathcal{D}^+\sqcup \mathcal{D}^- $ where $\mathcal{D}^+$ and $\mathcal{D}^-$ are disjoint unions of disks with $f(\mathcal{D}^+)\subseteq \text{int}(\mathcal{D}^+)$ and $f^{-1}(\mathcal{D}^-)\subseteq \text{int}(\mathcal{D}^-)$. On the other hand we consider the annuli $A_{\gamma}$ associated to the essential components of $\partial S$ as in Lemma \ref{l.conleyesquasi}, namely:
\begin{enumerate}
\item If $f(\gamma)\cap S = \emptyset$ then $A_{\gamma}$ is the annulus between $\gamma$ and $f(\gamma)$,
\item if $f(\gamma)\cap S \neq \emptyset$ then $A_{\gamma}$ is bounded by $\gamma$ and $f^{-1}(\gamma)$
\end{enumerate}

Let $\partial S^+$ be the essential boundary of $S$ that belongs in case (1) above. Then we have that $$f(S)\subseteq S_1:= S\cup \mathcal{D}^+ \cup \left(\bigcup_{\gamma\subset \partial S^+} A_{\gamma}\right) $$
Thus if $x\in A$ and $k_0=k_0(x)$, we have either $f^{k_0}(x)\in \mathcal{D}^+$ or $f^{k_0}(x)\in A_{\gamma}$ for some component $\gamma$ of $\partial S^+$. In the first case, since $\mathcal{D}^+$ is an attractor, we have $f^n(x)\in\mathcal{D}^+$ for all $n>k_0$, so we get the Claim. In the second one, if $n>k_0$ then $f^n(x)\in f^{n-k_0}(A_{\gamma})$, which lies in the annulus between $\gamma$ and $f^{n-k_0}(\gamma)$, and does not meet $\gamma$. Then Lemma \ref{sl.borde} implies $f^n(x)\notin S$, concluding the proof of the Claim.

\vspace{.1cm}

We are ready to prove that $A\neq b$. Suppose they are equal. Then by compactness of $b$ and continuity of $f$ (upper-semicontinuity of $k_0$) there exists $k_1=\max\{k_0(x):x\in b\}$, and by the  Claim we have $f^{k_1}(b)\cap S =\emptyset$. This contradicts point (2) of Lemma \ref{sl.borde}, since $f(b)\cap S \neq\emptyset$ by the hypothesis of the case we are proving. This concludes the proof, by the remarks at the beginning of it.

\end{proof}


text:
\begin{document}

\maketitle

{\bf Abstract.}
A topological structure of the solution set to differential inclusions with impulses at variable times is investigated. In order to do that an appropriate Banach space is defined. It is shown that the solution set is an $R_{\delta}$-set. Results are new also in the case of~differential equations with impulses at variable times.

{\bf Mathematics Subject Classification (2010):} Primary 34A37; Secondary 34A60, 34K45.

{\bf Keywords and phrases:} solution set, impulsive differential inclusions, variable times, $R_\delta$-set, topological structure.

\section{Introduction} \label{section:introduction}
Impulsive differential equations and inclusions have a~lot of applications in~the~various fields. The moments of impulses can be chosen in various ways: randomly, fixed beforehand, determined by the state of a system. The problems with fixed time of impulses were recently investigated \cite{BeRu, CarRub, DjebaliGor, DjebaliGorOua, GabGru, ObuYao}. The problems with impulses with variable times bring much more difficulties and up to now there were only existence theorems \cite{BeBeOu,BeOu,BeOu2}. Our results develop this research area and show that the solution set is an $R_{\delta}$-set.

There are many motivations to study the structure of solution sets of~differential equations and inclusions. One of them is considering the Poincar\'{e} translation operator and discussing the problem of the existence of periodic solutions \cite{KryPla,CS,HX}.

We have to have the space of functions which contains solutions of~given problem to study the structure of solution set. Obviously standard Banach space with norm $sup$ is insuficient to impulsive problem with the times of~jumps that depend on the state. B-topology on spaces of solutions of impulsive differential inclusions is introduced in \cite{AK}, however, it is only Hausdorff topology. We use this concept to create Banach space that have the same topology on common functions, and is sufficient to the considered problem.

In the Section \ref{section:preliminaries} we describe the problem and recall useful theorems. In~the~main Section \ref{section:main} we present the main results of the paper. The main idea is included in Theorem \ref{twr:dokladniejedenskokINKLUZJA} in which we show that the solution set for the problem with exactly one jump is an $R_{\delta}$-set. In the Theorem \ref{twr:drugieskonWym} we are using the result from previous Theorem proving by induction analogical statement for any fixed number of jumps. We also provide the reader with a transparent example.

\section{Preliminaries} \label{section:preliminaries}

\par The problem we deal with is
\begin{equation}\label{eq:zagadnienieINKLUZJA}
\left\{ \begin{array}{ll}
    \dot{y}(t)\in{F(t,y(t))}, & \hbox{for $t\in[0,a]$, $t\neq{\tau_j(y(t))},$ $j=1,\dots,m,$} \\
    y(0)=y_0, \\
    y(t^{+})=y(t)+I_j(y{(t)}), & \hbox{for $t={\tau_j}(y(t))$, $j=1,\dots,m,$}\\
\end{array}
\right.
\end{equation}
where $F:[0,a]\times\R^N\multimap\R^N$, $I_j:\R^N\to\R^N,$ $j=1,\dots,m,$ are given impulse functions, $\tau_j\in{C^1(\R^N,\R)}$ with $0<\tau_j(y)<a$, and $t_y=\{t~~|~~t=\tau_k(y(t))\}$.
The hypersurface $t-\tau_j(y)=0$ is called the $j$-th pulse hypersurface and we denote it by $\Sigma_j$.
If for each $j=1,\dots,m,$ $\tau_j$ is a different constant function, then impulses are in the fixed times.

Our goal is to find the structure of the solution set of the previous problem, but to do that we need a space of functions with $m$ jumps. We introduce considering space as $CJ_m([0,a]):=C([0,a])\times(\R\times\R^N)^m$ with following interpretation:
the element $(\varphi,(l_j,v_j)_{j=1}^m),$ where $l_j\in[0,a]$ we will interpret as the function with $m$ jumps in the times $j_k$ defined as follows:
$$\hat{\varphi}(t):=\left\{
\begin{array}{ll}
\varphi(t), &  0\leq{t}\leq l_{\sigma(1)}, \\
\varphi(t)+\sum\limits_{i=1}^{j}v_{\sigma(i)}, &  l_{\sigma(j)}<t\leq l_{\sigma(j+1)},\\
\varphi(t)+\sum\limits_{i=1}^{m}v_{\sigma(i)}, &  l_{\sigma(m)}<t\leq{a},
\end{array}
\right.$$
where $\sigma$ is a permutation of $\{1, 2, \ldots, m\}$ such that $l_{\sigma(i)}\leq l_{\sigma(i+1)}\mbox{.}$

There is a mutual correspondence between the functions on interval $[0,a]$ with $m$ jumps and the sets  $\{(\varphi,(l_j,v_j)_{j=1}^m)\in CJ_m([0,a]) \; | \; l_j<l_{j+1}\}$, with $\zeta \mapsto (\check{\zeta},(l_j,I_j(\check{\zeta}(l_j)))_{j=1}^m)$, where the function $\check{\zeta}$ is $\zeta$ with reduced jumps, $l_j$ is $j$-th time of jump and the function $I_j$ is an impulse functions.

The space $CJ_m([0,a])$ with the norm
$$\|(\varphi,(l_j,v_j)_{j=1}^m)\|:=\sup_{t\in[0,a]}\|\varphi(t)\|+\sum_{j=1}^{m}(|l_j|+\|v_j\|)$$
is a Banach space.

We will find out that the wanted structure is $R_{\delta}$-type. In order to show that we will use the following well-known theorems:

\begin{twr}[see \cite{Hyman}] \label{twr:Hyman}
Let $X$ be an absolute neighbourhood retract and \mbox{$A\subset{X}$}  be a compact nonempty subset.\par
Then the following statements are equivalent:
\begin{enumerate}
    \item [(a)] $A$ is an $R_\delta$-set,
\item [(b)] for every $\epsilon>0$ the set $A$ is contractible in $O_{\epsilon}(A)=\{x\in{X} ~|~ dist(x,A)<{\epsilon}\}$,
\item [(c)] $A$ is an intersection of a decreasing sequence $\{A_n\}$ of compact contractible spaces,
\item [(d)] $A$ is an intersection of a decreasing sequence $\{A_n\}$ of closed contractible spaces, such that $\beta(A_n)\to 0$, where $\beta$ is the Hausdorff measure of noncompactness.
\end{enumerate}
\end{twr}
 \begin{twr} {\em(Convergence theorem) (see \cite{AE}\label{twr:convergenceinkluzje})}
Let $E$ and $E'$ be Banach spaces, $(T,\Omega,\mu)$ be a measurable space, and the multivalued map $F\! : \! T\times E\multimap E'$ has closed and convex values and for a.e. $t\in T$ the map $F(t,\cdot)\! : \! E\multimap E'$ is uhc. Let $(u_n\! :\! T\to E)$ be a sequence of functions such that $u_n\to u$ in $L^p(T,E)$ and let sequence $(w_n)\subset L^p(T,E')$, $1\leq p<\infty$ be such that $w_n\rightharpoonup{w}$ in $L^p(T,E')$. If for a.e. $t\in T$ and for arbitrary $\varepsilon>0,$ there exists $N\in\N$ such that
$$w_n(t)\in \mbox{cl}\,\mbox{conv}\, B(F(t,B(u_n(t),\varepsilon)),\varepsilon)$$ for $n>N$, then $w(t)\in F(t,u(t))$ for a.e. $t\in T$.
\end{twr}

We recall Arzela- Ascoli Theorem:
\begin{twr}\label{twr:Arzela-Ascoliklasyczne}
If the family $\mathcal{F}\subset{C([a,b],\R^N)}$ of continuous functions is equicontinuous and uniformly bounded, then  there exists a subsequence that converges uniformly.
\end{twr}

A piecewise absolutely continuous function $y \! :\! [0,a]\to{\R^N}$ is a solution of the problem with impulses (\ref{eq:zagadnienieINKLUZJA}) if:
\begin{enumerate}
\item[(a)] $y(0)=y_0,$
\item[(b)] there exists a function $f\in{L^1([0,a],\R^N)}$ such that $f(t)\in{F(t,y(t))}$ for a.e. $t\in[0,a]$ and $y(t)=y_0+\sum_{j=1}^{m}I_j(y(t_j))+\int_{0}^{t}f(s)ds,$ where $t_j=\tau_{j}(y(t_j)),$
\item[(c)] the function $y$ is left continuous at $t={\tau}_j(y(t))\in[0,a]$ and the limit $y(t^+)$ exists and  $y(t^+)=y(t)+I_j(y(t))$ for $t={\tau}_j(y(t)),$ $j=1,\dots,m.$
\end{enumerate}

\section{The structure of the solution set}\label{section:main}
We assume the following conditions on multivalued perturbation $F:[0,a]\times{\R^N}\multimap{\R^N}$:

   \begin{enumerate}
    \item[$(F0)$] $F$ has compact and convex values,
    \item[$(F1)$] $F(\cdot,y):[0,a]\multimap{\R^N}$ has a measurable selection for every $y\in\R^N$,
    \item[$(F2)$] is almost uniformly with respect to $t$ H-usc, i.e. for every $y\in\R^N$ and $\varepsilon>0$ there exists $\delta>0$ such that for a.e. $t\in[0,a]$ and for all $x\in\R^N$ if $\|y-x\|<\delta$, then $\sup_{\varphi\in{F(t,x)}} d(\varphi,F(t,y))<\varepsilon,$
    \item[$(F3)$] has a sublinear growth, i.e., there exists $\alpha\in{L^1([0,a])}$ such that
 $$\sup_{\varphi\in{F(t,y)}}\|\varphi\|\leq\alpha(t)(1+\|y\|) \textrm{ for a.e. }
 t\in[0,a] \mbox{ and } y\in{\R^N}\mbox{.}$$

Moreover, we assume the following hypotheses about impulse functions:
    \item[$(H1)_m$] $I_j\in{C(\R^N,\R^N)}$, $j=1,\dots,m$,
    \item[$(H2)_m$] $\tau_j\in{C^1(\R^N,\R)},$ $j=1,\dots,m$, \\
for $j=1,\dots,m-1$ we have:
$$0<\tau_j(y)<\tau_{j+1}(y)<a,$$
$$\tau_j(y+I_j(y))\leq\tau_j(y)<\tau_{j+1}(y+I_j(y)),$$
$$\tau_m(y+I_m(y))\leq\tau_{m}(y),$$ and there exists a constant $M\geq{0}$ such that $\|{\tau_j}'(y)\|\leq{M}$ for all $y\in{\R^N}$, $j=1,\dots,m$,
    \item[$(H3)_m$] there exists a constant $p>0$ such that for a.e. $t\in[0,a]$  $$\sup_{\varphi\in{F(t,y)}}{\tau}_j'(y)\cdot \varphi-1\leq{-p}<0 \mbox{ for all } y\in{\R^N}, j=1,\dots,m \mbox{.}$$
 \end{enumerate}

Note taht if ${\tau}'_j(y)=0$ for $j=1,\dots,m$, then the problem is reduced to a problem with a fixed impulse time.

The assumption $(F2)$ and compact values of the multivalued map $F$ implies that $F(t,\cdot)$ is usc.

\begin{twr}\label{twr:dokladniejedenskokINKLUZJA} Let the assumptions $(F0)-(F3)$ and $(H1)_1-(H3)_1$ hold. Then every solution of the problem (\ref{eq:zagadnienieINKLUZJA}), where $m=1$, meets $\Sigma_1$ exactly once and the solution set $S$ of this problem is an $R_\delta$-set in the space $CJ_1([0,a])$.
\end{twr}

\begin{proof}
To simplify notation we write $I$ and $\tau$ instead of $I_1$ and $\tau_1$.
We will proceed in several steps.

 \vspace{0,3cm}
\underline{Step 1.}
A Lipschitz selection and the uniqueness of jump.

For each $n$ let $\{B(y,r_n(y))\}_{y\in{\R^N}}$ be an open covering (open balls, such that $r_n(y)\leq{\frac{1}{n}}\to{0}$) of the space $\R^N,$ such that for every $x\in B(y,r_n(y))$ we have

\be\label{eq:szacowanie}
\sup_{\varphi\in{F(t,x)}}d(\varphi,F(t,y))<\frac{1}{n}.\ee
There exists locally finite open point-star refinement $\mathcal{U}_n=\{U_{n,s}\}_{s\in{S}}$ of the cover
$\{B(y,r_n(y))\}_{y\in{\R^N}},$ i.e. for every  $y\in{\R^N}$ there exists $x_{y,n}\in\R^N$ such that $\mbox{st}(y,\mathcal{U}{_n})\subset{B(x_{y,n},r_n(x_{y,n}))}.$

We can choose it in a way that $\mathcal{U}_{n+1}$ is a~refinement of the cover $\mathcal{U}_{n}$. Let $\{\lambda_s\}_{s\in{S}}$ be a locally Lipschitz partition of unity subordinated to the cover $\mathcal{U}_n$ i.e. for every $s\in{S}$ the function $\lambda_s:[0,a]\to\R^N$ satisfies the locally Lipschitz condition.
For every $y_s\in\R^N$, $s\in{S}$ let a~function $q_s$ be a~measurable selection of ${F}(\cdot,y_s).$ We define
the function $g_n:[0,a]\times{\R^N}\to{\R^N}$ in the following way
$$g_n(t,y):=\sum_{s\in{S}}\lambda_s(y)\cdot{q_s(t)}.$$

The set $S(y):=\{s\in{S}~|~{\lambda}_s(y)\neq{0}\}$ is finite. If $\lambda_s(y)>0,$ so $s\in{S(y)},$ then $y\in\mbox{supp\,}\lambda_s\subset U_{n,s}\subset{\mbox{st\,}(y,\mathcal{U}_n)}.$ There exists ${x_{y,n}}\in\R^N$ such that $\mbox{st\,}(y,\mathcal{U}{_n})\subset B(x_{y,n},r_n(x_{y,n})).$ We know that $y_s\in{\mbox{st\,}(y,\mathcal{U}_n)}$.
We get
\begin{align*}\nonumber
g_n(t,y)&=\sum_{s\in{S}}\lambda_s(y)\cdot{q_s(t)}\in\mbox{conv\,}F(t,\mbox{st\,}(y,\mathcal{U}{_n}))\\
&\subset G_n(t,y):=\mbox{cl conv\,}F(t,\mbox{st\,}(y,\mathcal{U}{_n}))\mbox{.}\\
\end{align*}
Moreover,
$${G_n(t,y)}\subset\mbox{cl conv\,}F(t,B(x_{y,n},r_n(x_{y,n}))),$$
so from the inequality $(\ref{eq:szacowanie})$  we have
\be\label{eq:GnsiedziwOtoczceF}
G_n(t,y)\subset \mbox{ cl\,}O_{\frac{1}{n}}(F(t,x_{y,n})).\ee
We have
$$F(t,y)\subset\bigcap_{n\geq{1}}G_n(t,y).$$

From  usc (the map $F$ has compact values) we get that for every $y\in\R^N$ and for every $\varepsilon>0$ there exists $\delta>0$ such that
$$F(t,B(y,\delta))\subset{O_{\varepsilon}}(F(t,y)).$$

We have
\begin{align*}\nonumber
\bigcap_{n\geq{1}}G_n(t,y)&\subset \bigcap_{n\geq{1}}\mbox{ cl\,}O_{\frac{1}{n}}(F(t,x_{y,n}))\\
&\subset\bigcap_{n\geq{1}}\mbox{ cl\,}O_{\frac{1}{n}}\left(F\left(t,B(y,2r_n(x_{y,n}))\right)\right)\\
&\subset\bigcap_{n\geq{1}}\mbox{ cl\,}O_{\frac{1}{n}+\varepsilon}\left(F(t,y)\right)=F(t,y)\mbox{.}\\
\end{align*}
We obtain that
$$F(t,y)=\bigcap_{n\geq{1}}G_n(t,y).$$
We have
$$G_{n+1}(t,y)\subset G_n(t,y).$$

Let us introduce the Nemitski\u{i} (substitution) operator $P_{G_n} : CJ_1([0,a])\multimap L^1(J,\R^N)$
as follows
$$P_{G_n}(y):=\{\phi\in{L^1([0,a],\R^N)}~|~ \phi(t)\in{G_n(t,y(t))} \textrm{ for a.e. } t\in[0,a] \}.$$
Let $S_n$ denote the set of solutions of the problem
\begin{equation}\label{eq:zagadnieniezGnvariabletimes}
\left\{ \begin{array}{ll}
    \dot{y}(t)\in{G_n(t,y(t))}, & \hbox{for $t\in[0,a]$, $t\neq{\tau(y(t))},$} \\
    y(0)=y_0, \\
    y(t^{+})=y(t)+I(y{(t)}), & \hbox{for $t={\tau}(y(t))$. }\\
\end{array}
\right.
\end{equation}
It is obvious that the sets $S_n$ are nonempty, because the problem
\begin{equation}\label{eq:zagadnieniegnvariabletimes}
\left\{ \begin{array}{ll}
    \dot{y}(t)=g_n(t,y(t)), & \hbox{for $t\in[0,a]$, $t\neq{\tau(y(t))},$} \\
    y(0)=y_0, \\
    y(t^{+})=y(t)+I(y{(t)}), & \hbox{for $t={\tau}(y(t))$, }\\
\end{array}
\right.
\end{equation}
for every $n\in\N$ has exactly one solution.

 \vspace{0,3cm}
\underline{Step 1a.}

We denote by $t^j_{y_n}$ the time of $j$-th jump for the function $y_n$ and if the function $y_n$ has less that $j$ jumps we take $t^j_{y_n}=a$.
Let $y_n$ be an arbitrary solution of the system (\ref{eq:zagadnieniezGnvariabletimes}) for $0\leq t\leq t^2_{y_n}$. For $t\leq{t^1_{y_n}}$ we get the following form of the solution:
$$y_n(t)=y_0+\int_{0}^{t}\phi_n(s)ds,$$
where $\phi_n\in P_{G_n}(y_n)$.

There exists selection $f_n$ (not necessarily measurable) of the multivalued map $F(\cdot,x_{y,n}(\cdot))$ such that for a.e.  $t$ we get $\|\phi_n(t)-f_n(t,x_{y,n}(t))\| \leq \frac{1}{n}$.
We have
$$\|\phi_n(t)\|\leq \frac{1}{n}+\|f_n(t,x_{y,n}(t))\|.$$
From the assumption $(F3)$ we obtain
$$\|f_n(t,x_{y,n}(t))\|\leq \alpha(t)\left(1+\|x_{y,n}(t)\|\right)\leq \alpha(t)\left(1+\|y(t)\|+\frac{1}{n}\right).$$
So
$$
\|y_n(t)\|\leq \|y_0\|+\int_{0}^{t}\left(\alpha(s)\left(1+\|y_n(s)\|+\frac{1}{n}\right)+\frac{1}{n}\right)ds\mbox{.}
$$
From Gronwall inequality, we have:
\be\label{eq:wspOgr}\sup_{t\in[0,t^1_{y_{n}}]}\|y_n(t)\|\leq{\left(\|y_0\|+
\int_{0}^{a}2\alpha(s)ds+\frac{a}{n}\right)e^{\int_0^a\alpha(s)ds}}:=K.\ee
By continuity of the function $I$ there exists a constant $c>0$ such that $\|I(y_n(t^1_{y_n}))\|\leq c$ for all $n$.
Next for $t^1_{y_n}<t\leq{t^2_{y_n}}$ we obtain
\begin{align*}\nonumber
\|y_n(t)\|&\leq\|y_0\|+\|I(y_n(t^1_{y_n}))\|+\int_{0}^{t}\|\phi_n(s)\|ds\\
&\leq\|y_0\|+c+\int_{0}^{t}\left(\alpha(s)\left(1+\frac{1}{n}+\|{y_n(s)}\|\right)+\frac{1}{n}\right)ds\mbox{.}\\
\end{align*}
Again, from Gronwall inequality we get:
\be\label{eq:wspOgrPosokou}\sup_{t\in[0,t^2_{y_n}]}\|y_n(t)\|\leq{Ce^{\int_0^t\alpha(s)ds}}<{Ce^{\int_0^a\alpha(s)ds}}=:\bar{K},\ee
where $C:=\|y_0\|+c+\int_{0}^{a}2\alpha(s)ds+\frac{a}{n}.$\\
If the solution $y_n$ does not have jumps, then of course $\sup_{t\in[0,a]}\|y_n(t)\|\leq K.$ \\

\vspace{0,3cm}
\underline{Step 1b.}

Let $\bar{y}$ be a fixed function with values in $\mbox{cl\,}B(0,\bar{K}).$ Let $\phi_n\in{G_n(t,\bar{y}(t))}$, where $t$ is such that $(H3)_1$ is satisfied. We denote $\bar{y}(t)=:y\mbox{.}$
From the assumptions $(H2)_1$ and $(H3)_1$ for some $v\in\mbox{cl }{B(0,1)}$ we get:
\begin{align*}\nonumber
{\tau}'(y)\cdot{\phi}_n-1
&=\left({{\tau}'(x_{y,n})+\tau}'(y)-{\tau}'(x_{y,n})\right)\cdot{\phi}_n-1\\
&={\tau}'(x_{y,n})\cdot\left(\bar{\varphi}+\frac{1}{n}v\right)-1+\left({\tau}'(y)-{\tau}'(x_{y,n})\right)\cdot{\phi}_n\\
&={\tau}'(x_{y,n})\cdot\bar{\varphi}-1+\frac{1}{n}{\tau}'(x_{y,n})\cdot{v}+({\tau}'(y)-{\tau}'(x_{y,n}))-{\phi}_n,\\
\end{align*}
where $\bar{\varphi}\in{F(t,x_{y,n})}.$
Hence
\begin{align*}\nonumber
{\tau}'(y)\cdot{\phi}_n-1&\leq{-p+\frac{\|{\tau}'(x_{y,n})\|}{n}}+\left({\tau}'(y)-{\tau}'(x_{y,n})\right)\cdot{\phi}_n\\
&\leq{-p+\frac{M}{n}}+\|{\tau}'(y)-{\tau}'(x_{y,n})\|~~\|{\phi}_n\|.\\
\end{align*}

The function ${\tau}'$ is continuous on the compact set $\mbox{ cl\,}B(0,\bar{K}+1)$, therefore it is uniformly continuous. Hence, for $y\in\mbox{ cl\,}B(0,\bar{K})$ ${\tau}'(y)-{\tau}'(x_{y,n})\to{0}$ (we have $\|x_{y,n}\|\leq{\bar{K}}+\frac{1}{n}$). Moreover, the set $\{{\phi}_n\}$ is bounded (from the sublinear growth of $F$). We have $\|{\tau}'(y)-{\tau}'(x_{y,n}))\|~~\|{\phi}_n\|\to{0}.$
We take $N_0\in{\N}$ such that for every $n\geq{N_0}$ we have $-p+\frac{M}{n}+\|{\tau}'(y)-{\tau}'(x_{y,n})\|~~\|{\phi}_n\|<\frac{-p}{2}.$  There exists a constant $p'>0$ such that
$${\tau}'(y)\cdot \phi_n-1<-p'\mbox{.}$$

\vspace{0,3cm}
\underline{Step 1c.}

Let us fix $y_n$, where $n>N_0,$ the solution of the problem (\ref{eq:zagadnieniezGnvariabletimes}).

We define the function $w_n:[0,a]\to\R$ by:
$$w_n(t):=\tau(y_n(t))-t.$$
The function $w_n$ has value $0$ in any time, in which the function $y_n$ has a~jump.
By the condition $(H2)_1$ we get $w_n(0)=\tau(y_0)>{0}$ and $w_n(a)=\tau(y_n(a))-a<a-a=0$.
If $w_n(t)\neq{0}$ on $[0,a]$, then there would not be any impulse effect, therefore there would not be any jump time, so $w_n$ would be continuous, which would contradict with the earlier inequalities. Hence every solution of~the~problem (\ref{eq:zagadnieniezGnvariabletimes}) has at least one jump.

Suppose that $0<t^1_{y_n}<a$ is the first time in which the solution $y_n$ hits the hypersurface $\Sigma_1$. Then
$$w_n(t^1_{y_n})=0\;\mbox{ and } w_n(t)>0,\; \mbox{ for } t\in[0,t^1_{y_n}).$$
By  assumption $(H2)_1$ we get that
$$w_n({t^1_{y_n}}^+)=\tau(y_n({t^1_{y_n}}^+))-{t^1_{y_n}}= \tau\left(y_n(t^1_{y_n})+I(y_n(t^1_{y_n}))\right)-t^1_{y_n}\leq{0}\mbox{.}$$
For a.e. $t\geq{t^1_{y_n}}$ we have
$$w'_n(t)={\tau}'(y_n(t))\cdot{y_n'(t)}-1={\tau}'(y_n(t))\cdot{\phi}_n(t)-1<-p'<0,$$
where $\phi_n\in{P_{G_n}(y_n)}$.
The function $w_n$ in $[t^1_{y_n},a]$ is decreasing, hence $y_n$ hits the hypersurface $\Sigma_1$ exactly once and the time of this jump we denote $t_{y_n}\mbox{.}$

\vspace{0,3cm}
\underline{Step 2.}

Now we will show that each sequence $(y_n)$, where $y_n\in{S_n},$ has a~convergent subsequence to the solution $\tilde{y}$ of the problem (\ref{eq:zagadnienieINKLUZJA}).\\

There exists exactly one jump, so from the previous estimations we have

$$\|y_n(t)\|\leq\bar{K}\mbox{.}$$
Consequently, the values of solutions of the problem (\ref{eq:zagadnienieINKLUZJA}) are contained in a ball $\mbox{cl\,}B(0,\bar{K})$, which is convex, so in particular we know that function $g_n|_{[0,a]\times\mbox{cl}\,B(0,\bar{K})}$ has integrable Lipschitz constant $\Lambda$.

For $t<t'\leq t_{y_{n}}$ we have
\begin{equation}\begin{split} \label{eq:JednakowaCG}
\|y_n(t)-y_n(t')\|&=\left\|\int_{t}^{t'}\phi_n(s)ds\right\|\leq \frac{1}{n}|t-t'|+\int_{t}^{t'}\alpha(s)\left(1+\frac{1}{n}+\|y_n(s)\|\right)ds\\
&\leq |t-t'|+(2+K)\int_{t}^{t'}\alpha(s)ds,
\end{split}\end{equation}
and for $t_{y_{n}}<t<t'$ we have
\begin{equation}\begin{split} \label{eq:JednakowaCG2}
\|y_n(t)-y_n(t')\|&=\left\|\int_{t}^{t'}\phi_n(s)ds\right\|\leq \frac{1}{n}|t-t'|+\int_{t}^{t'}\alpha(s)\left(1+\frac{1}{n}+\|y_n(s)\|\right)ds\\
&\leq |t-t'|+(2+\bar{K})\int_{t}^{t'}\alpha(s)ds.
\end{split}\end{equation}

\underline{Step 2.a}

Let us consider convergence to the time $t_*$, where $t_*$ the limit of~a~convergent subsequence $(t_{y_{n_k}})$ of the sequence $(t_{y_n})$ (with $y_n\in{S_n}$), which exists due to the compactness of $[0,a]$.
For every $\varepsilon>0$ there exists $N_0$ such that for $n_k>N_0$ we have $t_*-\varepsilon<t_{y_{n_k}}$.

Note that
$$\sup_{t\in[0,t_*-\varepsilon-\delta]}\int_{t}^{t+\delta}\alpha(s)ds\to{0}$$
with $\delta\to 0\mbox{.}$

From inequality (\ref{eq:JednakowaCG}) we obtain that for all $\xi>0$ and $0<t<t_*-\varepsilon$ there exists $\delta>0$ such that for all $n_k>N_0$ and $t<t'<t+\delta\leq t_*-\varepsilon$ we have $\|y_{n_k}(t)-y_{n_k}(t')\|\leq |t'-t|+(2+K)\int_{t}^{t'}\alpha(s)ds<\xi$.
Therefore family $\{y_{n_k}\}_{n_k>N_0}$ is equicontinuous and by the inequality (\ref{eq:wspOgr}) uniformly bounded.
By Arzela-Ascoli Theorem \ref{twr:Arzela-Ascoliklasyczne} (possibly going to the subsequences) we can assume that $y_{n_{k}}\to{y_{\varepsilon}}$ on $[0,t_*-\varepsilon]$, where $y_{\varepsilon}$ is continuous function. This can be done in such a way that for any ${\varepsilon}_1,{\varepsilon}_2>0$ such that ${\varepsilon}_1>{\varepsilon}_2$ functions $y_{{\varepsilon}_1}$, $y_{{\varepsilon}_2}$ fulfil condition $y_{\varepsilon_2}|_{[0,t_*-{\varepsilon}_1]}=y_{{\varepsilon}_1}\mbox{.}$
For ${\varepsilon}\searrow{0}$ we obtain an extension of the function $y_{\varepsilon}$, ie the function $y_*\! :\! [0,t_*) \to \R^N$, where $y_{n_k}$ converges pointwise to $y_{*}\mbox{.}$

Moreover
\begin{equation}
\label{eq:calkOgranphi_n} \|\phi_{n_k}(t)\|\leq{\alpha(t)\left(1+\|x_{y_{n_k},n_k}(t)\|\right)}+\frac{1}{n_k}\leq \alpha{(t)}(2+\bar{K})+1. \end{equation}
We know that:
\begin{itemize}
 \item ${\phi}_{n_k}(t)\in \mbox{cl conv}\; F\left(t,\mbox{st\,}(y_{n_k}(t),\mathcal{U}_{n_k})\right)
\subset{\mbox{cl}}\,O_{\frac{1}{n_k}}\left(F\left(t,B(y_{n_k}(t),\frac{1}{n_k})\right)\right)$, by inclusion (\ref{eq:GnsiedziwOtoczceF}),
 \item $y_{n_k}(t)\to y_{\varepsilon}(t)$ a.e. on $[0,t_*-\varepsilon]$,
 \item ${\phi}_{n_k}\in{L^1([0,t_*-\varepsilon],\R^N)}$,
 \item by estimation (\ref{eq:calkOgranphi_n}) and weak compactness of the closed ball we get ${\phi}_{n_{k_l}}\rightharpoonup{\phi}$ on $[0, t_*-\varepsilon]$.
\end{itemize}
Thus by Theorem \ref{twr:convergenceinkluzje} we obtain ${\phi}(t)\in F(t,y_{\varepsilon}(t))$ for a.e. $t\in[0,t_*-\varepsilon]$.
By analogy, we conclude that ${\phi}(t)\in F(t,y_*(t))$ a.e. on $[0,t_*)$.
By weak convergence
${\phi}_{n_{k_l}}\rightharpoonup{\phi}$ on $[0, t_*-\varepsilon]$, for $\Psi(\phi_n):=\int_{0}^{t}\phi_n(s)ds$ we have
\begin{align*}
\int_{0}^{t}\phi(s)ds&=\Psi(\phi)=\lim_{k\to\infty}\Psi(\phi_{n_k})=\lim_{k\to\infty}\int_{0}^{t}{\phi}_{n_k}(s)ds\\
&=\lim_{k\to\infty}{y_{n_k}(t)}-y_0=y_*(t)-y_0\mbox{.}\end{align*}
For an increasing sequence $(s_n)$ convergent to $t_*$ with $n<n'$ we obtain
$$\|y_*(s_n)-y_*(s_{n'})\|=\left\|\int_{s_n}^{s_{n'}}\phi(s)ds\right\|\leq \int_{s_n}^{t_*}(\alpha(s)(2+K)+1)ds\mbox{.}$$
We have convergence of the right hand side of the inequality to $0$ with $n\to \infty$,
consequently $(y_*(s_n))$ is Cauchy sequence. It is convergent ($\R^N$ is complete) and we denote its limit $y_*(s_n)\to y_1$. We define $y_*(t_*):=y_1$ and we obtain continuous extension $y_*$ on $[0,t_*]$.
We will show, that $\tau(y_*(t_*))-t_*=0$, which means that $t_*$ is time of jump for $y_*$.

Let $\epsilon>0$. The function $\alpha(s)(2+\bar{K})+1$ is integrable, so we can choose $t_{\epsilon}<t_*$ so that  $$\int_{t_{\epsilon}}^{t_*}2[\alpha(s)(2+\bar{K})+1]ds<\frac{\epsilon}{2}\mbox{.}$$
There exists $K_0\in\N$ such that for $k>K_0$ we have $\|y_*(t_{\epsilon})-y_{n_k}(t_{\epsilon})\|<\frac{\epsilon}{2}$.
We can estimate
\begin{align*}
\|y_{n_k}(t_{y_{n_k}})-y_*(t_{*})\|&\leq \left\|y_{n_k}(t_{\epsilon})+\int_{t_{\epsilon}}^{t_{y_{n_k}}}\phi_{n_k}(s)ds-y_*(t_{\epsilon})-\int_{t_{\epsilon}}^{t_*}\phi(s)ds\right\|\\
\leq&\|y_{n_k}(t_{\epsilon})-y_*(t_{\epsilon})\|+\int_{t_{\epsilon}}^{t_{y_{n_k}}}\|\phi_{n_k}(s)\|ds+\int_{t_{\epsilon}}^{t_*}\|\phi(s)\|ds\\
\leq&\|y_{n_k}(t_{\epsilon})-y_*(t_{\epsilon})\|+\int_{t_{\epsilon}}^{t_{y_{n_k}}}\left(\alpha(s)(2+\bar{K})+1\right)ds\\
+&\int_{t_{\epsilon}}^{t_*}\left(\alpha(s)(2+\bar{K})+1\right)ds\\
\leq& \frac{\epsilon}{2}+\int_{t_{\epsilon}}^{t_*}2(\alpha(s)(2+\bar{K})+1)ds\leq \epsilon\mbox{.}
\end{align*}
For $t_{y_{n_k}}>t_*$ we get
\begin{align*}\|y_*(t_*)-y_{n_k}(t_{y_{n_k}})\|\leq \|y_*(t_*)-y_*(t_*-\varepsilon)\|&+\|y_*(t_*-\varepsilon)-y_{n_k}(t_*-\varepsilon)\|\\
&+\|y_{n_k}(t_*-\varepsilon)-y_{n_k}(t_{y_{n_k}})\|\mbox{,}
\end{align*}
but it is easy to see that:
\begin{align*}
\|y_{n_k}(t_*-\varepsilon)-y_{n_k}(t_{y_{n_k}})\|&=\left\|\int_{t_*-\varepsilon}^{t_{y_{n_k}}}\phi_{n_k}(s)ds\right\|
\leq\int_{t_*-\varepsilon}^{t_{y_{n_k}}}\|\phi_{n_k}(s)\|ds\\
&\leq  \int_{t_*-\varepsilon}^{t_{y_{n_k}}} \big(\alpha(s)(2+K)+1\big)ds\\
&= (2+K)\int_{t_*-\varepsilon}^{t_{y_{n_k}}} \alpha(s)ds +t_{y_{n_k}}-t_*+\varepsilon \mbox{,}
\end{align*}
so
\begin{align*}
\|y_*(t_*)-y_{n_k}(t_{y_{n_k}})\|&\leq \|y_*(t_*)-y_*(t_*-\varepsilon)\|+\|y_*(t_*-\varepsilon)-y_{n_k}(t_*-\varepsilon)\|\\
&+(2+K)\int_{t_*-\varepsilon}^{t_{y_{n_k}}} \alpha(s)ds +t_{y_{n_k}}-t_*+\varepsilon\\
&\to_{k\to \infty}
\|y_*(t_*)-y_*(t_*-\varepsilon)\|+(2+K)\int_{t_*-\varepsilon}^{t_*} \alpha(s)ds +\varepsilon\mbox{.}
\end{align*}
From the arbirary $\epsilon$ and $\varepsilon$ we get
$\|y_*(t_*)-y_{n_k}(t_{y_{n_k}})\|\to 0$.

Summarising, we have that
$y_{n_k}(t_{y_{n_k}})\to y_*(t_*)$.
By the continuity of $\tau$ we obtain $\tau(y_*(t_*))-t_*=\lim_{{n_k}\to\infty}(\tau(y_{n_k}(t_{y_{n_k}}))-t_{y_{n_k}})=0$ which means that $t_*$ is the time of jump for $y_*$.

 \vspace{0,3cm}
\underline{Step 2.b}

We will make the similar reasoning with the part of segment $[0,a]$ after jump.

Form inequalities (\ref{eq:wspOgrPosokou}) and (\ref{eq:JednakowaCG2}) we conclude that the family $\{y_{n_k}\}$ is equicontinuous and equibounded on $[t_*+\varepsilon,a]$.
Therefore, from Arzeli-Ascolego \ref{twr:Arzela-Ascoliklasyczne} theorem (passing to a subsequence if it is
needed) we can assume that $y_{n_k}\to{y^{\varepsilon}}$, where $y^{\varepsilon}$ is a continuous function and we extend
it to a continuous function $y^*\! :\! (t_*,a] \to \R^N$ with $y_{n_k}$ convergent pointwise to $y^*$.

We know that:
\begin{itemize}
 \item ${\phi}_{n_k}(t)\in \mbox{cl conv}\; F\left(t,\mbox{st\,}(y_{n_k}(t),\mathcal{U}_{n_k})\right)
\subset{\mbox{cl}}\,O_{\frac{1}{n_k}}\left(F\left(t,B(y_{n_k}(t),\frac{1}{n_k})\right)\right)$.
 \item $y_{n_k}(t)\to y^{\varepsilon}(t)$ a.e. on $[t_*+\varepsilon,a]$,
 \item ${\phi}_{n_k}\in{L^1([t_*+\varepsilon,a],\R^N)}$,
 \item by estimation (\ref{eq:calkOgranphi_n}) and the weak compactness of closed ball we get ${\phi}_{n_{k_l}}\rightharpoonup{\phi}$ on $[t_*+\varepsilon,a]$.
\end{itemize}
Again, by theorem \ref{twr:convergenceinkluzje} we obtain ${\phi}(t)\in F(t,y^{\varepsilon}(t)),$ for a.e. $t\in{[t_*+\varepsilon,a]}$, so we get an information that ${\phi}(t)\in F(t,y^*(t))$ a.e. on $(t_*,a]$.
By weak convergence
${\phi}_{n_{k_l}}\rightharpoonup{\phi}$ on $[t_*+\varepsilon,a]$ for $\Psi(\phi_n):=\int_{t}^{a}\phi_n(s)ds$ we have
\begin{align*}\nonumber
\int_{t}^{a}\phi(s)ds&=\Psi(\phi)=\lim_{k\to\infty}\Psi(\phi_{n_k})=\lim_{k\to\infty}\int_{t}^{a}{\phi}_{n_k}(s)ds\\
&=\lim_{k\to\infty}{y_{n_k}(a)}-y_*(t)=y_*(a)-y_*(t)\mbox{.}\\
\end{align*}
For decreasing sequence $(s_n)$ convergent to $t_*$ we get for $n<n'$:
$$\|y^*(s_n)-y^*(s_{n'})\|=\left\|\int_{s_{n'}}^{s_n}\phi(s)ds\right\|\leq \int^{s_n}_{t_*}\alpha(s)(1+\bar{K})ds\mbox{.}$$
By analogy to Step 2a, we get that $(y^*(s_n))$ is the Cauchy sequence, which is convergent in Banach space, so $y^*(s_n)\to y_2$ for some $y_2\in\R^N$.

By continuity of $I$ we have that $I(y_{n_k}(t_{y_{n_k}}))\to I(y_*(t_*))$, therefore $y_2=y_*(t_*)+I(y_*(t_*))$.

We can define a function $\tilde{y} \! : \! [0,a] \to \R^N$ by concatenation $y_*$ on $[0,t_*]$ with the function $y^*$ on $(t_*,a]$.
Obviously, for $t\leq t_*$ the function $\tilde{y}$ is the solution of the problem~(\ref{eq:zagadnienieINKLUZJA}). For $t>t_*$ we have:
$$\tilde{y}(t)=\tilde{y}(t_*)+I(\tilde{y}(t_*))+\int_{t_*}^t \phi(s)ds=y_0+\int_0^{t_*}\phi(s)ds+ I(\tilde{y}(t_*))+\int_{t_*}^t \phi(s)ds\mbox{,}$$
where $\phi\in P_{G_n}(\tilde{y})$, so $\tilde{y}$ is the solution of the problem~(\ref{eq:zagadnienieINKLUZJA}) for all $t\in[0,a]$, hence $\tilde{y}\in{S}$. The function $\tilde{y}$ is
the limit of the sequence $(y_{n_k})$ in the space $CJ_1([0,a])$.
 \vspace{0,3cm}

\underline{Step 3.}
We will show, for every $n\in\N$,  the contractibility of the set $\mbox{cl}\, S_n$.

Fix $n$ such that we can define $p'$ (see Step 1.) and take $\bar{y}\in\mbox{cl}\,S_{n}$. We divide the interval $[0,1]$ into two halfs.

Let $r\in[0,\frac{1}{2}]$. We consider the problem
\begin{equation}\left\{
\begin{array}{ll}
    \dot{y}(t)=g_n(t,y(t)), & \mbox{for } t\in[a-2r(a-t_{\bar{y}}),a] \mbox{, } t\neq{\tau(y(t)}) \mbox{,}  \\
    y(t)=\bar{y}(t), & \mbox{for } t\in[0,a-2r(a-t_{\bar{y}})] \mbox{,}\\
    y(t^+)=y(t)+I({y}(t)), & \mbox{for } t=\tau(y(t)) {.}\\
\end{array}
\right.\label{eq:impulDlaselekcjigSKwym}\end{equation}
In the previous problem we denote by $g_n$ selection of the map $G_n.$
There exists exactly one solution of this problem; we denote it by $y^{2}_{\bar{y},r}.$ Then
$y^2_{\bar{y},r}\in\mbox{cl}\,S_n.$\\
Next for $r\in(\frac{1}{2},1]$ we consider the problem
\begin{equation}\left\{
\begin{array}{ll}
    \dot{y}(t)=g_n(t,y(t))\mbox{,} & \mbox{for } t\in[t^{\bar{y},r},a] \mbox{, } t\ne\tau(y(t)) \mbox{,} \\
    y(t)=\bar{y}(t)\mbox{,} & \mbox{for } t\in[0,t^{\bar{y},r}] \mbox{,}\\
y(t^+)=y(t)+I({y}(t))\mbox{,} & \mbox{for } t=\tau(y(t)) \mbox{.} \\
\end{array}
\right.\label{eq:mimpulDlaselekcjigSKwym}\end{equation}
where $t^{\bar{y},r}:=t_{\bar{y}}-2(r-\frac{1}{2})t_{\bar{y}}\mbox{.}$
There exists exactly one solution of this problem, denoted by $y^{1}_{\bar{y},r},$ which also belongs to $\mbox{cl}\,S_n.$\\
Finally we consider the following function $h:[0,1]\times{\mbox{cl\,}S_n}\rightarrow \mbox{cl\,}S_n$ given by:
\begin{equation}h(r,\bar{y}):=\left\{
\begin{array}{ll}
y^2_{\bar{y},r}, & \hbox{ $r\in[0,\frac{1}{2}]$},\\
y^1_{\bar{y},r}, & \hbox{ $r\in(\frac{1}{2},1]$}.\\
\end{array}
\right.\end{equation}

\vspace{0,5cm}
Now, we will show that the function h is continuous.

\vspace{0,5cm}
Due to the continuous dependence of solutions on initial conditions \cite{Hale} we know that the function $h$ is continuous on $[0,\frac{1}{2})\times{\mbox{cl}\,S_n}$ and left continuous on $\{\frac{1}{2}\}\times\mbox{cl}\,S_n$.

Let $((r_k,\bar{y}_k))_k$  be a sequence convergent to $(\frac{1}{2}^+,\bar{y})$.

We know that if $t_{\bar{y}}<t_{h(r_k,\bar{y}_k)}$ then $\tau(h(r_k,\bar{y}_k)(t_{\bar{y}}))-t_{\bar{y}}>0$ and we have
$$ t_{\bar{y}}-\tau(h(r_k,\bar{y}_k)(t_{\bar{y}}))= \int_{t_{\bar{y}}}^{t_{h(r_k,\bar{y}_k)}}(\tau(h(r_k,\bar{y}_k)(\cdot))-\cdot)'(\theta)d\theta < -p'(t_{h(r_k,\bar{y}_k)}-t_{\bar{y}}),$$
so
$$ t_{h(r_k,\bar{y}_k)}<t_{\bar{y}}+(\tau(h(r_k,\bar{y}_k)(t_{\bar{y}}))-t_{\bar{y}})/p'\mbox{,}$$
but for $k$ such that $t_{\bar{y}}\leq t_{h(r_k,\bar{y}_k)}$ one sees that
\begin{align*}
\Bigg\|\bar{y}(t^{\bar{y}_k,r_k})&+\int_{t^{\bar{y}_k,r_k}}^{t_{\bar{y}}}\bar{\phi}_n(s)ds- \left(y^1_{\bar{y}_k,r_k}(t^{\bar{y}_k,r_k})
+\int_{t^{\bar{y}_k,r_k}}^{t_{\bar{y}}}g_n(s,y^1_{\bar{y}_k,r_k}(s))ds\right)\Bigg\|\\
&=\left\|\int_{t^{\bar{y}_k,r_k}}^{t_{\bar{y}}}\bar{\phi}_n(s)-g_n(s,x^1_{\bar{y}_k,r_k}(s))ds\right\|\\
&\leq\int_{t^{\bar{y}_k,r_k}}^{t_{\bar{y}}}\|\bar{\phi}_n(s)\|+\|g_n(s,y^1_{\bar{y}_k,r_k}(s))\|ds\\
&\leq\int_{t^{\bar{y}_k,r_k}}^{t_{\bar{y}}}2\left(\alpha(s)(1+K)+\frac{1}{n}\right)ds\mbox{.}
\end{align*}
Hence, if $k\to\infty$ we get
$$ h(r_k,\bar{y}_k)(t_{\bar{y}})\to \bar{y}(t_{\bar{y}})\mbox{,}$$
so
$$ \tau( h(r_k,\bar{y}_k)(t_{\bar{y}}))-t_{\bar{y}}\to 0\mbox{.}$$
For every $k$ we have
$$t^{\bar{y}_k,r_k}<t_{h(r_k,\bar{y}_k)}\leq t_{\bar{y}}+{1{\hskip -2.5 pt}\hbox{l}}_{t_{\bar{y}}<t_{h(r_k,\bar{y}_k)}}(\tau(h(r_k,\bar{y}_k)(t_{\bar{y}}))-t_{\bar{y}})/p'$$
and
$$t^{\bar{y}_k,r_k}\rightarrow t_{\bar{y}}\leftarrow t_{\bar{y}}+{1{\hskip -2.5 pt}\hbox{l}}_{t_{\bar{y}}<t_{h(r_k,\bar{y}_k)}}(\tau(h(r_k,\bar{y}_k)(t_{\bar{y}}))-t_{\bar{y}})/p'\mbox{,}$$
therefore by squeeze theorem
\be\label{eq:t}
t_{h(r_k,\bar{y}_k)}\rightarrow t_{\bar{y}}\mbox{.}
\ee

\vspace{0,5cm}
Let us fix k for a while. By $1{\hskip -2.5 pt}\hbox{l}$ we denote the function:
\begin{equation}\nonumber
1{\hskip -2.5 pt}\hbox{l}_{t>t_0}(y(t))=
\left\{ \begin{array}{ll}
    y(t), & \hbox{for $t>t_0,$} \\
    0, & \hbox{for $t\leq{t_0}.$} \\
   \end{array}
\right.
\end{equation}
We define function $\varrho_{\bar{y}_k,r_k}\! :\! [0,a]\to\R^N$

\begin{align*}
\varrho_{\bar{y}_k,r_k}(t):=&y^1_{\bar{y}_k,r_k}(t)- {1{\hskip -2.5 pt}\hbox{l}}_{t>t_{y^1_{\bar{y}_k,r_k}}}(I(y^1_{\bar{y}_k,r_k}(t_{y^1_{\bar{y}_k,r_k}})))\\
-&(y^2_{\bar{y},1/2}(t)-{1{\hskip -2.5 pt}\hbox{l}}_{t>t_{\bar{y}}} (I(y^2_{\bar{y},1/2}(t_{\bar{y}}))),
\end{align*}
which is function of differences between $y^1_{\bar{y}_k,r_k}\mbox{,}$ and $y^2_{\bar{y},1/2}$ with deleted changes caused
by jumps. It is easy to see that $t_{\bar{y}}=t_{y^2_{\bar{y},1/2}}\mbox{.}$
For $t^{\bar{y}_k,r_k}\leq t\leq t_{\bar{y}}$ we get that:
\begin{align*}
\|\varrho_{\bar{y}_k,r_k}(t)\|&\leq \|\bar{y}-\bar{y}_k\|+\int_{t^{\bar{y}_k,r_k}}^t \|g_n(s,{y^1_{\bar{y}_k,r_k}}(s))-\bar{\phi}_n(s)\|ds\\
&\leq\|\bar{y}-\bar{y}_k\|+\int_{t^{\bar{y}_k,r_k}}^{t}\left(\|\bar{\phi}_n(s)\|+\|g_n(s,y^1_{\bar{y}_k,r_k}(s))\|\right)ds\\
&\leq\|\bar{y}-\bar{y}_k\|+\int_{t^{\bar{y}_k,r_k}}^{t}2\left(\alpha(s)(1+\bar{K})+\frac{1}{n}\right)ds=:z_{\bar{y}_k,r_k}(t)
\end{align*}
and for $t> t_{\bar{y}}$ we have
\begin{align*}
\|\varrho_{\bar{y}_k,r_k}(t)\|&\leq z_{\bar{y}_k,r_k}(t_{\bar{y}})+ \int_{t_{\bar{y}}}^{t}\|g_n(s,{y^1_{\bar{y}_k,r_k}}(s))-g_n(s,{y^2_{\bar{y},1/2}}(s))\|ds
\\
\leq z_{\bar{y}_k,r_k}(t_{\bar{y}})&+\int_{t_{\bar{y}}}^{t} \Lambda(s)\|{y^1_{\bar{y}_k,r_k}}(s)-{y^2_{\bar{y},1/2}}(s)\|ds
\\
=z_{\bar{y}_k,r_k}(t_{\bar{y}})&+\int_{t_{\bar{y}}}^{t}\Lambda(s)\|{\check{y}^1_{\bar{y}_k,r_k}}(s)
+{1{\hskip -2.5 pt}\hbox{l}}_{s>t_{y^1_{\bar{y}_k,r_k}}}I(y^1_{\bar{y}_k,r_k}(t_{y^1_{\bar{y}_k,r_k}}))
-{\check{y}^2_{\bar{y},1/2}}(s)
\\
&-I({y}^2_{\bar{y},1/2}(t_{\bar{y}}))\|ds
\\
\leq z_{\bar{y}_k,r_k}(t_{\bar{y}})&+ \left|\int_{t_{\bar{y}}}^{t_{y^1_{\bar{y}_k,r_k}}}\Lambda(s)\|I({y}^2_{\bar{y},1/2}(t_{\bar{y}})\|ds\right|
\\
&+\int_{\max\{t_{\bar{y}},t_{y^1_{\bar{y}_k,r_k}}\}}^t \Lambda(s)\|I(y^1_{\bar{y}_k,r_k}(t_{y^1_{\bar{y}_k,r_k}}))-I({y}^2_{\bar{y},1/2}(t_{\bar{y}}))\|ds
\\
&+\int_{t_{\bar{y}}}^{t}\Lambda(s)\|{\check{y}^1_{\bar{y}_k,r_k}}(s)-{\check{y}^2_{\bar{y},1/2}}(s)\|ds,
\end{align*}
hence by the Gronwall inequality we obtain
\begin{align*}
\|\varrho_{\bar{y}_k,r_k}(t)\|&\leq \Bigg(z_{\bar{y}_k,r_k}(t_{\bar{y}})+\left|\int_{t_{\bar{y}}}^{t_{y^1_{\bar{y}_k,r_k}}} \Lambda(s)\|I({y}^2_{\bar{y},1/2}(t_{\bar{y}}))\|ds\right|\\
&+\int_{\max\{t_{\bar{y}},t_{y^1_{\bar{y}_k,r_k}}\}}^t \Lambda(s)\|I(y^1_{\bar{y}_k,r_k}(t_{y^1_{\bar{y}_k,r_k}}))- I({y}^2_{\bar{y},1/2}(t_{\bar{y}}))\|ds\Bigg)\\
&*\exp{\int_{t_{\bar{y}}}^{t}\Lambda(s)ds}=:z_{\bar{y}_k,r_k}(t)\mbox{.}
\end{align*}
By previous convergences and continuity of I we obtain

\begin{equation}\begin{split}\label{eq:jumps}
\Big\| &I(h(r_k,\bar{y}_k)(t_{h(r_k,\bar{y}_k)})) -I(h(1/2,\bar{y})(t_{h(1/2,\bar{y})}))\Big\|\to 0
\end{split}\end{equation}
and
\be\label{eq:supy}
\sup_{t\in[0,a]}\|\varrho_{\bar{y}_k,r_k}(t)\|\leq z_{\bar{y}_k,r_k}(a)\to 0.
\ee

Summing up, by (\ref{eq:supy}), (\ref{eq:t}) and (\ref{eq:jumps}), if $(r_k,\bar{y}_k)$ converges to $(\frac{1}{2}^+,\bar{y}),$ then $y^1_{\bar{y}_k,r_k}$ converges to $y^2_{\bar{y},\frac{1}{2}}$ in norm in the space $CJ_1([0,a])$.

The function h, as continuous on
$[0,1]\times{\mbox{cl}\,S_n},$ is a homotopy. By definition of the function $h$ we have
$h(0,\bar{y})=\bar{y}$ and $h(1,\bar{y})=y^1_{\bar{y},1}$, so
${\mbox{cl}\,S_n}$ is a~contractible set.
 \vspace{0,3cm}

\underline {Step 4.}

We will show that properties needed to theorem \ref{twr:Hyman} are fulfilled.

The sets $\mbox{cl}\,S_n$ are contractible in the power of Step 3.

If $x\in \bigcap_{n\in\N}\; \mbox{cl}\,S_n$, then $x\in \mbox{cl}\,S_n$ for every $n$. Therefore, there exists sequence $(d_n)\subset\R_+$ converges to~$0$ such that $B(x,d_n)$ (in $CJ_1([0,a])$) contains $y_n$ ($y_n\in{S_n}$). Hence $y_n\to x$ in the space $CJ_1([0,a])$. Moreover, we know that subsequence $y_{n_k}$ converges to a solution of problem (\ref{eq:zagadnienieINKLUZJA}), where $m=1$, so $x\in S$. We get
$$S\subset\bigcap_{n\in\N}\, S_n\subset\bigcap_{n\in\N}\,\mbox{cl}\, S_n\subset S\mbox{,}$$
so $S=\bigcap_{n\in\N}\,\mbox{cl}\, S_n$.

We will show that $\sup\{d(z,S)| z\in{S_n}\}\to_{n\to\infty}{0}$. Assume that there exists $\varepsilon>0$ and a sequence $(y_n)$ such that $y_n\in{S_n}$ and $d(y_n,S)\geq\varepsilon$. From the Step 3 we know taht this sequence has subsequence $(y_{n_k})$ such that $y_{n_k}\to{\tilde{y}}\in{S},$
so $d(y_{n_k},S)\to{0}$. It is contrary to the choice of the sequence $(y_n),$ hence $\sup\{d(z,S)|
z\in{S_n}\}\to{0}$. Therefore
$\sup\{d(z,S)| z\in{\mbox{cl}\,{S_n}}\}\to{0}$.

We get
$S_n\subset S + B(0,p_n)$, where $p_n:=\sup_{z\in S_n}d(z,S)\to 0$ with $n\to\infty$.

Compactness of $S$ implies
$$\beta(\mbox{cl}\, S_n)=\beta(S_n)\leq \beta(S)+p_n=p_n,$$
so $\beta(\mbox{cl}\, S_n)\to 0$.

Summing up, we can use theorem \ref{twr:Hyman}, which implies that the set $S$ is an $R_{\delta}$-set.
\end{proof}

We will use this theorem to prove more general case.

\begin{twr} \label{twr:drugieskonWym} Let the assumptions $(F0),$-$(F3)$ and $(H1)_m,$-$(H3)_m$ hold. Then every
solution of the problem (\ref{eq:zagadnienieINKLUZJA}) for every $j=1,\dots,m$ meets $\Sigma_j$ exactly once and the solution set $S$ of this problem is an $R_\delta$-set in the space $CJ_m([0,a]).$
\end{twr}
\begin{proof}
We will show that we can divide the interval $[0, a]$ into $m$ disjoint parts and any
of them will have exactly one jump effect. Then we will be able to use the reasoning
of theorem \ref{twr:dokladniejedenskokINKLUZJA} on every such part, which will end the proof. By analogy to Step 1. in
the proof of Theorem \ref{twr:dokladniejedenskokINKLUZJA} we define a~multivalued map $G_n:[0,a]\times\R^N\multimap\R^N$ and consider the following problem
\begin{equation}\label{eq:zagadnieniezGnvariabletimesk}
\left\{ \begin{array}{ll}
    \dot{y}(t)\in{G_n(t,y(t))}, & \hbox{for $t\in[0,a]$, $t\neq{\tau_j(y(t))},$ $j=1,\dots,m,$} \\
    y(0)=y_0, \\
    y(t^{+})=y(t)+I_j(y{(t)}), & \hbox{for $t={\tau}_j(y(t)),$ $j=1,\dots,m$. }\\
\end{array}
\right.
\end{equation}

We denote by $t^j_{y_n}$ the time of $j$-th jump for the function \mbox{$y_n:[0,a]\to\R^N$}. If the function $y_n$ has less than $j$ jumps we take $t^j_{y_n}:=a$.
Let $y_n$ be an~arbitrary solution of the problem (\ref{eq:zagadnieniezGnvariabletimesk}) for $0\leq{t}\leq{t^2_{y_n}}\mbox{.}$
By analogy to theorem \ref{twr:dokladniejedenskokINKLUZJA} in Step 1a. we show that there exists a constant $\bar{K}$ such that
$$\sup_{t\in[0,t^2_{y_n}]}\|y_n(t)\|\leq{\bar{K}}\mbox{.}$$
Next we will proceed similary to the proof of theorem \ref{twr:dokladniejedenskokINKLUZJA} (Step 1b), we show that there exists a constant $p'>0$ such that
$${\tau}'_j(y)\cdot\phi_n-1<-p'$$
for all $j=1,\dots,m,$ and for enough big $n,$
where $\bar{y}$ is fixed function with values in $\mbox{cl}\,B(0,\bar{K})\mbox{,}$ $t$ is such that the assumption $(H3)_m$ is satysfied, $\bar{y}(t)=y,$ $\phi_n\in{G_n(t,\bar{y}(t))}.$
We define the function $w_{n,j}:[0,a]\to\R$ in the following way:
$$w_{n,j}(t):=\tau_j(y_n(t))-t,~~~~ j=1,\dots,m\mbox{.}$$
Let us fix a solution $y_n$ of problem (\ref{eq:zagadnieniezGnvariabletimesk}).

Now we will prove by induction that the $j$-th time of
jump is zero of~the~function $w_{n,j}$.\\
{\underline {Basis.}}  By the assumption $(H2)_m$ we have:\\
a) $w_{n,j}(0)=\tau_j(y_0)>0,$\\
b) $w_{n,j}(a)={\tau}_j(y_n(a))-a<0,$\\
c) $w_{n,j}(t)=\tau_j(y_n(t))-t<\tau_{j+1}(y_n(t))-t=w_{n,j+1}(t)$ for all $t\in[0,a]\mbox{.}$

If there are no impulses then $w_{n,j}(t)\neq{0}$ on $[0,a],$ but by the
definition of $w_{n,j}$ for every $j$ the function $w_{n,j}$ is continuous, which contradicts with a) and b). Hence there is at least one jump time.

Let $0<t^1_{y_n}<a$ be the first time of jump for the solution $y_n\mbox{.}$ Then
$$w_{n,j}(t^1_{y_n})=0,\quad w_{n,j}(t)>0,\; \mbox{ dla } t\in[0,t^1_{y_n})\mbox{.}$$
By the assumption $(H2)_m$ we get that:
$$w_{n,j}({t^1_{y_n}}^+)=\tau_j(y_n({t^1_{y_n}}^+))-{t^1_{y_n}}= \tau_j(y_n(t^1_{y_n})+I_j(y_n(t^1_{y_n})))-t^1_{y_n}\leq{0}\mbox{.}$$
For a.e. $t\geq{t^1_{y_n}}$ we have
$$w'_{n,j}(t)={\tau}'_j(y_n(t))\cdot{y_n'(t)}-1={\tau}_j'(y_n(t))\cdot{\phi}_n(t)-1<-p'<0\mbox{.}$$
Hence $w_{n,j}$ is decreasing function at $[t^1_{y_n},a]$, so $y_n$ for every $j$ meets $\Sigma_j$ exactly once.
By c) we know that the time $t^1_{y_n}$ of the first jump is zero of the function $w_{n,1}$.\\
{\underline {Inductive step.}}
We assume that the $j$-th time of jump is zero of $w_{n,j},$ $j<m$.\\
Denote by $t_{y_n}^j$ the time for which we have $t=\tau_j(y_n(t))$.
We know that $w_{n,l}({t^j_{y_n}}^+)=\tau_l(y_n(t^j_{y_n}))-{t^j_{y_n}}>\tau_j(y_n(t^j_{y_n}))-{t^j_{y_n}}=0$ for $j>1\mbox{,}$ $l>j$.
We consider interval $J:=(t^j_{y_n},a]$. By analogy to Step 1 in proof of theorem \ref{twr:dokladniejedenskokINKLUZJA} we have $w_{n,i}(t)\neq{0},$ $t\in{J},$ $i\leq{j}$, as long as there is
no jump caused by $w_{n,l}$ for $l>j$.
There have to be at least one jump after $t^j_{y_n}$ and we denote it by~$\tilde{t}$. By c) there are no jumps before the jump caused by $w_{n,j+1}$, so $\tilde{t}=t^{j+1}_{y_n}$.

Since both the basis and the inductive step have been proved, it has been proved
by mathematical induction that there are exactly $m$ jumps one for each $\Sigma_j$, $j=1,\dots,m$.

Next we will proceed similary to the proof of theorem \ref{twr:dokladniejedenskokINKLUZJA}.\\
Let $(t^1_*, t^2_*, \dots, t^m_*)$ be the limit of the sequence $((t^1_{y_n}, t^2_{y_n}, \dots, t^m_{y_n}))$.\\
We can show that on $[0, t^1_*]$ there exists the subsequence of $(y_{n_k})$ such that $I_1(y_{n_{k_l}}(t_{y_{n_{k_l}}}))\to{I_1(y_*(t^1_*))}$ and $\tau_1(y_*(t_*^1))=t^1_*\mbox{,}$ so
$$\tau_2(y_*((t_*^1)^+))=\tau_2(y_*(t_*^1)+I_1(y_*(t_*^1)))>\tau_1(y_*(t_*^1))=t_*^1\mbox{.}$$
By the assumptions $(H1)_m$ and $(H2)_m$ there exists $\varepsilon_1>0$ such that there is only
one jump time of $y_*$ on $[0,t^1_*+2\varepsilon_1]$ and we get that for all sufficiently big $l$ there is only
one jump time of $y_{n_{k_l}}$ on $[0,t^1_*+\varepsilon_1]$. By analogy we find subsequence and $\varepsilon_2>0$ such that there is exactly one jump on $[t_*^1+\varepsilon_1, t_*^2+\varepsilon_2]$, and so on. On every such interval we
proceed with reasoning from theorem \ref{twr:dokladniejedenskokINKLUZJA}.
\end{proof}

The following example shows the case of inclusion with exactly one jump, but can easly be rearranged to a multijump case.

\bex
There are two trust funds with interest rates (dependent on time and amount of money) $\alpha(t,y_1)\in A(t,y_1)$ and $\beta(t,y_2)\in B(t,y_2)$ respectively where $A$ and $B$ are (multivalued) investment plans.
It is available to transfer money once (in or out) without loosing interest.

We decided to start both trust funds with the same amount of money and transfer money from worse deposit to better one after clarifying which one is better.
The amount of money that we wish to transfer would be proportional to difference in incomes.

 This situation can be represented in the form of following differential inclusion:
$$
\left\{ \begin{array}{ll}
    \dot{(y_1,y_2)}(t)\in F(t,(y_1,y_2)(t)),&\hbox{for $t\in[0,1]$, $t\neq{\tau((y_1,y_2)(t))}$,} \\
    (y_1,y_2)(0)=(y_0,y_0), \\
    (y_1,y_2)(t^{+})=(y_1,y_2)(t)+I((y_1,y_2)(t)),&\hbox{for $t={\tau}((y_1,y_2)(t))$}\\
\end{array}
\right.
$$

with
$$ F(t,y_1,y_2)=A(t,y_1)\times B(t,y_2),$$
where $A$ and $B$ fulfill assumptions $(F1)$ and $(F2)$ (for example continuous) and have interval values with $A(t,y), \; B(t,y) \in [-6y,6y]$
$$I(y_1,y_2)=\left\{
\begin{array}{ll}
(-y_1,y_1), &  0\leq (1+\frac{1}{\rho}) y_1<y_2, \\
(y_2,-y_2), &  0\leq (1+\frac{1}{\rho}) y_2<y_1, \\
(\rho (y_1-y_2), \rho (y_2-y_1)), &  0\leq y_1\leq (1+\frac{1}{\rho}) y_2\leq (1+\frac{1}{\rho})^2 y_1, \\
(0,0), & y_1<0 \mbox{ or } y_2<0
\end{array}
\right.$$
and
$$\tau(y_1,y_2):=\mbox{arccot}(y_1+y_2)/\pi.$$
\vspace{0.1cm}

For any $\alpha\in A(t,y)$ we have that
$$\alpha \geq -3(1/2+2y^2)=(\pi(p-1))(1/2+2y^2)$$
where $1>p=1-3/\pi>0$. We have alnalogous inequality for $\beta$ therefore we get
\begin{align*}
\sup_{\varphi\in{F(t,y_1,y_2)}}{\tau}_j'(y_1,y_2)\cdot \varphi &= \sup_{\alpha\in{A(t,y_1)}}\sup_{\beta\in{B(t,y_2)}} \frac{\alpha+\beta}{-\pi(1+(y_1+y_2)^2)}\\
&\leq (1-p)\frac{1+2y_1^2+2y_2^2}{1+(y_1+y_2)^2}\leq{1-p}<1
\end{align*}

All assumptions of Theorem \ref{twr:dokladniejedenskokINKLUZJA} are satisfied, therefore the solution set of~this~problem is an~$R_\delta$-set.
\eex