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{\bf Abstract}

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{\bf AMS Classification numbers}\quad Primary:\quad \theprimaryclass

Secondary:\quad \thesecondaryclass

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{\bf Keywords:}\quad \thekeywords

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\title{The Cone Length and Category of Maps:  Pushouts, Products and Fibrations}
\authors{Martin Arkowitz, Donald Stanley and Jeffrey Strom}
\address{}

\abstract
For any collection of spaces $\A$, we investigate
two non-negative integer homotopy invariants of maps:  $\LL_\A(f)$,
the $\A$-cone length of $f$, and
$\L_\A(f)$, the $\A$-category of $f$.  When $\A$ is the collection of
all spaces, these are the cone length and category of $f$, respectively,
both of which have been studied previously.
The following results are obtained:

(1)  For a map of one homotopy pushout diagram into another, we derive
an upper bound for $\LL_\A$ and $\L_\A$ of the induced map of homotopy
pushouts in terms of $\LL_\A$ and $\L_\A$ of the other maps.  This has
many applications, including an inequality for $\LL_\A$ and $\L_\A$ of the
maps in a mapping of one mapping cone sequence into another.

(2)  We establish an upper bound for $\LL_\A$ and $\L_\A$ of the product
of two maps in terms of $\LL_\A$ and $\L_\A$
of the given maps and the $\A$-cone length of their domains.

(3) We study our invariants in a pullback square and obtain as a
consequence an upper bound for the $\A$-cone length and $\A$-category
of the total space of a fibration in terms of the
$\A$-cone length and $\A$-category of the base and fiber.

We conclude with several remarks, examples and open questions.

\endabstract

\primaryclass{55M30}
\secondaryclass{55P99, 55R05}
\keywords{Lusternik-Schnirelmann category, cone length,
homotopy pushouts, products of maps, fibrations}

\maketitlepage

\begin{changemargin}{-1.1cm}{-1.1cm}

\section{Introduction}

In this paper we continue our investigation, begun in \cite{ASS}, of the cone length
and category of   maps relative to a fixed collection of spaces.  For a
collection $\A$ of spaces we consider
two non-negative integer homotopy invariants of maps: the $\A$-category,
denoted  $\L_\A$,
and the $\A$-cone length, denoted $\LL_\A$.   When $\A$ is the collection of
all spaces, $\L_\A(f)$ is the category of the map $f$ as defined and studied in
\cite{Fa-Hu}  and \cite{Cornea2} and $\LL_\A(f)$ is the
cone length of the map $f$ as defined and studied in \cite{Marcum} and \cite{Cornea1}.
If, in this special case, $f$ is the inclusion of the base point
into $Y$, then $\L_\A(f)$ is just the category of the space $Y$,
which was introduced in 1934 by Lusternik and Schnirelmann in their work
on the number of critical points of
smooth functions on a manifold \cite{L-S}.
In addition, $\LL_\A(f)$ is the cone length of $Y$ which
has been studied by several people
\cite{Cornea1,Cornea4,Cornea5,Ganea,Marcum,Stanley,Takens}
in the context of homotopy theory.  For an arbitrary collection
$\A$, the $\A$-category and the $\A$-cone length
of the inclusion of the basepoint into $Y$ coincide  with
the $\A$-category and $\A$-cone length of $Y$.
Variants of this concept have been studied previously
\cite{S-T}.

Thus our invariants are common generalizations of the category and cone length of a
map and the $\A$-category and $\A$-cone length of a space.  In addition to
providing a general framework for many
existing notions and retrieving known results as special cases, they have
led to several new concepts and
results.  To discuss this, we first briefly summarize that part of our
previous work which is relevant to this
paper.  More details are given in \sec 2.

In \cite{ASS} we introduced, for a fixed collection $\A$, five simple axioms which an
integer valued function of based maps may satisfy.
Then $\L_\A$ was defined as the maximum of
all such functions.  Similarly, $\LL_\A$ was defined as the maximum of all
functions which satisfy an analogous
set of five axioms.  We then gave alternate characterizations of these
invariants in terms of
certain decompositions of maps.  For instance, $\LL_\A(f)$ is essentially the smallest
integer $n$ such that $f$ admits a decomposition up to homotopy as
$$
X = X_0\maprt{j_0}X_1\maprt{j_1}\cdots \maprt{j_{n-1}} X_n = Y,
$$
where $L_i\maprt{}X_i\maprt{j_i}X_{i+1}$ is a mapping cone sequence with $L_i \in \A$.
When $f$ is the inclusion of the base point into $Y$, we obtain $\cat_\A(Y)$ and $\cl_\A(Y)$,
as noted above, and when $g$ is the map of $X$ to a  one point space, we obtain two new
invariants of spaces, the $\A$-kitegory of $X$, $\kit_\A(X) = \L_\A(g)$, and the
$\A$-killing length of $X$, $\kl_\A(X) = \LL_\A(g)$.   In \cite{ASS}
we made a preliminary study of these invariants and their interrelations.

From the time the concept of category was first introduced to the present,
many people have been interested in the following questions:
What is the relationship between the categories of the spaces
which appear in a homotopy pushout \cite{Marcum,Hardie}?
What is the category of the product of two spaces in terms of the categories
of the factors \cite{A-Sta,Bassi,ClP2,Ganea2,Iwase1,Roitberg,Stanley2,Takens,Van}?
What is the relationship of the categories of the spaces which appear
in a fiber sequence \cite{Hardie2,James-Singhof,V}?
Similar questions have also been considered for cone length.
In this paper we study these questions for the $\A$-category and $\A$-cone
length of maps, and provide reasonably complete answers.
This gives both new and known results for the $\A$-category and $\A$-cone length
of spaces as well as new results for the $\A$-kitegory and the $\A$-killing length.

We now summarize the contents of the paper.  In \sec 2 we give our
terminology and notation and discuss our earlier work in more detail.
In \sec 3 we prove one of our main results, the Homotopy Pushout Mapping Theorem.
This theorem gives an inequality for the $\A$-categories and
the $\A$-cone lengths of the four maps which constitute a map of one
homotopy pushout square into another.
Many applications  are given in \sec 4.  In particular,
we obtain results about the
$\A$-categories and $\A$-cone lengths of the maps which appear in a mapping
of one mapping cone sequence into
another.  In \sec 5 we establish an upper bound for the $\A$-category
(resp., $\A$-cone length) of the product of
two maps in terms of the $\A$-categories (resp., $\A$-cone lengths) of the original
maps and the $\A$-cone lengths of
their domains.  By specializing to spaces and letting $\A$ be the collection of
all spaces, we retrieve classical
results on the category and cone length of the product of two spaces.
We study pullbacks in \sec 6.  As a consequence of our main result on pullbacks
we obtain an inequality
for the $\A$-category of the total space of a fiber sequence in terms of the
$\A$-category of the base and fiber, and a similar result for $\A$-cone length.
Section 7 contains a potpouri of
results, examples and questions.   We begin by presenting a few simple,
but useful, results about $\A$-category and
$\A$-cone length.  We then give some examples to illustrate the difference
between these invariants for different
collections $\A$.  In particular, we show that some results that are known
for the collection of all spaces  do
not hold for arbitrary collections.  Finally, we state and discuss some
open problems.

We conclude this section by emphasizing two important points.
First of all, there are several different notions of the category of a map
in the literature.   The one that we generalize here to the $\A$-category
of a map has been studied in \cite{Fa-Hu} and \cite{Cornea2}.  It is not the
same as the one considered in \cite{Fox,B-G}.    In addition, Clapp and Puppe have
considered the category of a map with respect to a collection of
spaces \cite{ClP,ClP2}.  However, their notion is completely different from ours.
Secondly, although we state and prove our results in the category of well-pointed
spaces and based maps, it should be clear that  nearly all our results hold in a
(closed) model category \cite{Quillen} and that all of our results hold in a
J-category \cite{Do,H-L}.

\bigskip

\section{Preliminaries}

In this section we give our notation and terminology and also recall some
results from \cite{ASS} which will be needed later.

All topological spaces are based and have the based homotopy type of
CW-complexes, though we could more generally consider well-pointed based spaces.
All maps and homotopies are to preserve base points.  We {\it do} distinguish
between a map and a homotopy class.  By a commutative diagram we mean one which is
{\it strictly} commutative.

We next give some notation which is standard for homotopy theory:
$*$ denotes the base point of a space or the space consisting of a single point,
$\simeq$ denotes homotopy of maps and $\equiv$ denotes same homotopy type of spaces.
We let $0:X\maprt{} Y$ stand for the constant map and id$:X\maprt{} X$ for the
identity map.   We use $\Sigma $ for (reduced) suspension, $*$
for (reduced) join, $\vee$  for wedge sum and $\smsh$ for smash product.

We call a sequence $A\maprt{f} X\maprt{j} C$ of spaces and maps
a  {\it mapping cone sequence} if
$C$  is the mapping cone of $f$ and $j$ is the standard inclusion.
Then $j$ is a cofibration with cofiber $\Sigma A$.
Using the mapping cylinder construction, we see that the
concept of a cofiber sequence  and the concept of a mapping cone sequence are equivalent
\cite[Ch.\thinspace 3]{Hi}.  For maps $C\maplft{g}A\maprt{f}B$ we can form the homotopy pushout $Q$
$$
\xymatrix{
A\ar[rr]^f \ar[d]_g && B\ar[d] \\
C\ar[rr] && Q \\ }
$$
by defining $Q$ to be the quotient of $B\vee (A\times I) \vee C$
under the equivalence relation $(*,t)\sim *$, $(a,0) \sim f(a)$
and $(a,1) \sim g(a)$ for   $t\in I$ and $a \in A$.
Note that $A \maprt{f} X \maprt{j} C$ is a mapping
cone sequence if and only if
$$
\xymatrix{
A\ar[rr]^f \ar[d]  && X\ar[d]^j \\
{*}\ar[rr] && C \\ }
$$
is a homotopy pushout square.
The pullback $P$ of $C\maprt{g}A\maplft{f}B$ is defined by
$$
P = \{ (b,c)\, |\, b\in B, c\in C, f(b) = g(c)\} \subseteq B\times C.
$$
We only use this construction when $f$ is a fibration.
Thus all our pullbacks are homotopy pullbacks as well.
Given a map $f:X\maprt{} Y$ we say that a map $g:X'\maprt{} Y'$ {\it homotopy dominates}
$f$ (or $f$ is a {\it homotopy retract} of $g$) if there is a homotopy-commutative
diagram
$$
\xymatrix{
X\ar[rr]^i\ar[d]_f  &&   X'\ar[rr]^r\ar[d]^g   &&  X\ar[d]^f
\\
Y\ar[rr]^j        &&   Y'\ar[rr]^s         &&  Y
}
$$
such that $ri \simeq \id$ and $sj \simeq \id$.
If the diagram is strictly commutative and both homotopies
are equality, we delete the word `homotopy' from the definition.
If $g$ homotopy dominates $f$ as above and in addition
$ir\simeq \id$ and $js \simeq \id$ (i.e., $r$ and $s$ are homotopy equivalences
with homotopy
inverses $i$ and $j$), we say that $f$ and $g$ are {\it homotopy equivalent}.

Next we recall some definitions and results from \cite{ASS} which will be used in the
sequel.   By a {\it collection} $\A$ we mean a class of spaces containing $*$ such that if
$A \in \A$ and $A\equiv A'$, then $ A' \in \A$.  We say that (1) $\A$ is {\it closed under
suspension} if
$A \in \A $ implies $\Sigma A \in \A $, (2) $\A $ is {\it closed under wedges}
if $A, A' \in \A $
implies $A\vee A' \in \A $ and (3) $\A $ is {\it closed under joins} if
$A, A' \in \A $
implies $A *  A' \in \A $.  Examples of collections that we consider are
(1) the collection $\A = \{{\mathrm{all\ spaces}}\}$ of all spaces,
(2) the collection $\Sigma$ of all suspensions, and
(3) the collection $\S$ of all wedges of spheres (including $S^0$).

Let $\A$ be a collection and $\l_{\A }$ an function which assigns to
each map $f$  an integer
$0\le \l_{\A }(f) \le \infty $.  We say that $\l_{\A }$ satisfies the
$\A$-{\it cone axioms} if
\begin{enumerate}
\item[(1)] (Homotopy Axiom) If
$f \simeq g$, then $\l_{\A }(f)  = \l_{\A }(g)$.
\item[(2)]  (Normalization Axiom)  If
$f$ is a homotopy equivalence, then $\l_{\A }(f) =0$.
\item[(3)]
(Composition Axiom)  $\l_{\A }(fg) \le  \l_{\A }(f) +  \l_{\A }(g)$.
\item[(4)]
(Mapping Cone Axiom)  If $A\maprt{} X   \maprt{f} Y$ is a mapping cone sequence
with $A \in \A$, then $\l_{\A }(f) \le 1$.
\item[(5)]  (Equivalence Axiom)
If $f$ and $g$ are homotopy equivalent, then $\l_{\A }(f) = \l_{\A }(g) $.
\end{enumerate}
We say that $\l_{\A }$ satisfies the $\A $-{\it category axioms} if $\l_{\A}$ satifies
(1) -- (4) and
\begin{enumerate}
\item[(5$'$)] (Domination Axiom) If $f$ is dominated by $g$, then $\l_{\A }(f) \le \l_{\A }(g) $.
\end{enumerate}

\bdfn{\rm
We denote by $\LL_{\A}(f)$ the maximum of all $\l_{\A }(f)$ where $\l_{\A }$ satisfies
(1)--(5) and by $\L_{\A}(f)$ the maximum of all $\l_{\A }(f)$ where $\l_{\A }$ satisfies
(1)--(4) and (5$'$).  We call $\LL_{\A}(f)$ the $\A $-{\it cone length} of $f$ and
$\L_{\A}(f)$ the $\A $-{\it category} of $f$.  }
\edfn

Since (5) is weaker than (5$'$),
$\L_{\A}(f) \leq \LL_{\A}(f)$.   In  \cite{ASS} it is proved that when
$\A = \{ {\mathrm{all\ spaces}} \}$, $\LL_{\A}(f)$ is
the cone length of $f$ as defined in \cite{Cornea2,Marcum},
and $\L_{\A}(f)$ is the category of $f$
as defined in \cite{Fa-Hu,Cornea2}.

One of the main results of \cite{ASS} gives alternate characterizations
of $\LL_{\A}(f)$ and
$\L_{\A}(f)$ in terms of decompositions of the map $f$.
If $f:X\maprt{} Y$ is a map, then an {\it $\A $-cone decomposition of length $n$}
of $f$  is
a homotopy-commutative diagram
$$
\shortlabeldecomp{X}{f}{Y}{n}{K}{i}{j}{s}
$$
in which $f_n$ is a homotopy equivalence with homotopy inverse $s$
and each map $j_i$ is part of a mapping cone sequence
$
\xymatrix@1{
A_i \ar[r] &  X_i \ar[r]^-{j_i}& X_{i+1}}
$
with $A_i\in \A$.  Thus $f_nj_{n-1}\cdots j_0
\simeq f$, $sf \simeq j_{n-1}\cdots j_0$,
$f_n s \simeq \id$ and $s f_n\simeq \id$.
The
homotopy-commutative diagram above is an
{\it $\A$-category  decomposition of $f$ of length $n$}
if $s$ is simply a homotopy section of $f_n$, i.e., if
$f_nj_{n-1}\cdots j_0 \simeq f$, $sf \simeq j_{n-1}\cdots j_0$ and
$f_n s \simeq \id$, but $sf_n$ need not be homotopic to the identity.
We prove in \cite[Thm.\thinspace 3.7]{ASS} that
$$ \LL_{\A}(f) =\left\{
\begin{array}{ll}
0  & {\mathrm{if}}\ f\ {\mathrm {is\ a\ homotopy\ equivalence}}
\\
\infty & {\mathrm {if\ there\ is\ no}}\ \A{\mathrm{-cone\ decomposition\ of}}\ f
\\
n & {\mathrm {if}}\  n\
{\mathrm{is\ the\ smallest\ integer\ such\ that\ there\ exists\ an}}\\ &
\A{\mathrm{-cone\ decomposition\ of\ length}}\  n\ {\mathrm{of}}\ f .\\
\end{array} \right.
$$
We similarly characterize $\L_{\A}(f)$ using $\A$-category decompositions instead of
$\A$-cone decompositions.  Observe that if  the induced map $\pi_0(f):\pi_0(X) \to \pi_0(Y)$ on
path components is not surjective, then $L_\A(f)$ and $\L_\A(f)$ are
infinite for {\it every} collection $\A$.

We have also studied four numerical invariants of
spaces, defined in terms of the invariants $\LL_\A$ and $\L_\A$ as follows:
$$
\begin{array}{rclcrcl}
\cl_{\A}(X) & = & \LL_\A(*\maprt{}X) &\quad &
\cat_{\A}(X) & = & \L_\A(*\maprt{}X)\\
      \kl_{\A}(X) & = & \LL_\A(X\maprt{}*) &\quad &
      \kit_{\A}(X) & = & \L_\A(X\maprt{}*) . \\
\end{array}
$$
When $\A =\{ {\mathrm{all\ spaces}}\}$,
$\cat_\A(X) = \cat(X)$, the reduced Lusternik-Schnirelmann category of $X$
\cite[Prop.\thinspace 4.1]{ASS}, and
$\cl_\A(X)= \cl(X)$, the cone length of $X$.
Moreover, $\kit_\A(X)\leq 1$ and $\kl_\A(X)\leq 1$ for every space $X$ in this case.

\bigskip

\section{The Homotopy Pushout Mapping Theorem}

In this section we prove the first main result of this paper.
This consists of two inequalities, one for the $\A$-cone length
and one for the $\A$-category of the maps from one homotopy pushout
square to another.   In Section \sec 4
we will derive numerous consequences of this result.

We begin with a technical result that plays a key role in the proof of
the main theorem.

\blem\label{lem:strict}
Let $f: X\maprt{}Y$ be a  map with $\L_\A(f) = n$.  Then there exists
a map $g:X\maprt{} Z$ and maps $i:Y\maprt{}Z$ and $r:Z\maprt{}Y$
such that the diagram
$$
\xymatrix{
           && X\ar[lld]_f\ar[rrd]^f \ar[d]_g         \\
Y\ar[rr]_i && Z\ar[rr]_r                        && Y,\\
}
$$
commutes, $ri= \id$ and $\LL_\A(g) = n$.
\elem

\begin{prf}
By \cite[Cor.\thinspace 4.4]{ASS} there is a homotopy-commutative diagram
$$
\xymatrix{
&&X \ar[lld]_f   \ar[d]_h \ar[rrd]^f\\
Y\ar[rr]_j && W \ar[rr]_s && Y \\
}
$$
with $sj\simeq\id$ and $\LL_\A(h) = n$.  We factor $s$ as
$W\maprt{s^0} E\maprt{s^1}Y$, where $s^0$ is a homotopy equivalence
and $s^1$ is fibration \cite[Ch.\thinspace 3]{Hi}.  Then $s^0j :Y\maprt{}E$ and $s^1s^0j\simeq \id$.
Thus there is a map $t:Y\maprt{}E$ such that
$t\simeq s^0 j$ and $s^1t =\id$.  Now we have a homotopy-commutative diagram
$$
\xymatrix{
&&X  \ar[lld]_f  \ar[d]_{s^0h} \ar[rrd]^f   \\
Y\ar[rr]_t            && E \ar[rr]_{s^1}             && Y.          \\
}
$$
We factor $s^0$ as $W\maprt{k'}E'\maprt{k''}E$
where $k'$ is a homotopy equivalence and $k''$ is a fibration.
Then $k'h:X\maprt{}E'$ and $k''k'h =s^0h\simeq tf$, and so
there is a map $l:X\maprt{}E'$ with $l\simeq k'h$ and $k''l = tf$.
Then we have a diagram
$$
\xymatrix{
&& X  \ar[lld]_f   \ar[d]_{k''l} \ar[rrd]^f\\
Y\ar[rr]_t && E \ar[rr]_{s^1} && Y \\
}
$$
with $s^1t =\id$ and the left hand triangle strictly commutative.
It then follows that the right hand triangle is strictly commutative.
Note that $k''l\simeq k''k'h =s^0 h$.  Since $s^0$ is a homotopy
equivalence, $\LL_\A(k''l) = \LL_\A(h)$.  This proves the lemma
with $Z= E$, $g= k''l$, $i=t$ and $r=s^1$.
\end{prf}

\bthm\label{thrm:MPT}
Let $\A$ be a collection of spaces that is closed under wedges and
suspension and let
$$
\xymatrix{
C\ar[d]^{c} && A \ar[ll]_{g} \ar[rr]^{f} \ar[d]^{a} && B\ar[d]^{b}
\\
C' && A' \ar[ll]_{g'}\ar[rr]^{f'} && B'
\\  }
$$
be a commutative diagram.  Let $D$ be the homotopy pushout of
the top row, $D'$ be the homotopy pushout of the bottom row,
and $d:D\maprt{}D'$ the induced map.  Then
\begin{enumerate}
\item  $\LL_\A(d) \leq \LL_\A(a) +
\max ( \LL_\A(b),\LL_\A(c))$;
\item  $\L_\A(d) \leq \L_\A(a) +
\max ( \L_\A(b),\L_\A(c))$.
\end{enumerate}
\ethm

\brmk{\rm
Our proof will show that
if $\A$ is only assumed to be closed under wedges, then (1) and (2) hold when
$A=A'$ and $a=\id$.
In Corollary \ref{cor:nowedge} below we derive some slightly weaker formulas
that require only closure under suspension.
In particular, if $\A$ is only known to be closed under
suspensions, then (1) and (2) hold with $B= B'$, $C= C'$, $b=\id$ and $c=\id$.}
\ermk

\begin{prf}
First we prove (1).  We factor the given diagram as
$$
\xymatrix{
C\ar[d]_c && A \ar[ll]_g\ar[rr]^f\ar@{=}[d]  && B\ar[d]^b   \\
C' \ar@{=}[d] && A \ar[ll]_{cg}\ar[rr]^{bf}\ar[d]^{a} && B'\ar@{=}[d]   \\
C' && A' \ar[ll]_{g'}\ar[rr]^{f'} && B'    \\  }
$$
and let $\overline D$ denote the homotopy pushout of the middle row.
Then we have a factorization
$$
\xymatrix{
D\ar[rd]_{d'} \ar[rr]^d && D' \\
&\overline D \ar[ru]_{d''}\\ }
$$
of $d$.
Since $\LL_\A(d) \leq \LL_\A(d') + \LL_\A(d'')$ by the Composition Axiom,
it suffices to prove the result in the two special cases
\begin{enumerate}
\item[(a)]  $A=A'$ and $a=\id$,
\item[(b)] $B=B'$, $C=C'$, $b=\id$ and $c=\id$.
\end{enumerate}

We begin with (a) and let $m = \max ( \LL_\A(b),\LL_\A(c))$.
We consider an $\A$-cone decomposition of $b$ of length
$m$.  This yields a homotopy factorization of
$b\simeq hi_{m-1}\cdots i_1i_0$:
$$
B = X_0\maprt{i_0}X_1\maprt{i_1}\cdots \maprt{i_{m-1}}X_m\maprt{h}B',
$$
where $A_l\maprt{}X_l\maprt{i_l}X_{l+1}$ is a mapping cone sequence
with $A_l\in \A$ for each $l$ and $h$ is a homotopy equivalence.
Since $i_{m-1}\cdots i_1i_0$ is a cofibration, $h$ is homotopic
to a map (also called $h$) such that $b = hi_{m-1}\cdots i_1i_0$.
Similarly, we have an $\A$-cone decomposition of $c$ of length
$m$ which gives a factorization $c = k j_{m-1}\cdots j_1 j_0$:
$$
C = Y_0\maprt{j_0}Y_1\maprt{j_1}\cdots \maprt{j_{m-1}}Y_m\maprt{k}C',
$$
where $B_l\maprt{} Y_l\maprt{j_{l}}Y_{l+1}$ is a mapping cone sequence
with $B_l\in\A$ for each $l$ and $k$ is a homotopy equivalence.  Thus we
have a commutative diagram
$$
\xymatrix{
Y_0=C \ar[d]_-{j_0} && A \ar[ll]_-g\ar[rr]^-{f}\ar@{=}[d] && B=X_0 \ar[d]^-{i_0}
\\
Y_1\ar[d]_-{j_1}   && A \ar[ll]_-{j_0g}\ar[rr]^-{i_0f}\ar@{=}[d] && X_1 \ar[d]^-{i_1}
\\
\vdots \ar[d] && \vdots \ar@{=}[d] && \vdots \ar[d]
\\
Y_{m-1} \ar[d]_-{j_{m-1}} && A
\ar[ll]_-{j_{m-2} \cdots j_0 g}
\ar[rr]^-{i_{m-2} \cdots i_0 f}
\ar@{=}[d] && X_{m-1} \ar[d]^-{i_{m-1}}
\\
Y_{m} \ar[d]_k   && A
\ar[ll]_-{j_{m-1} \cdots j_0 g}
\ar[rr]^{i_{m-1} \cdots i_0 f}
\ar@{=}[d] && X_{m} \ar[d]^-{h}
\\
C' && A  \ar[ll]_-{cg}\ar[rr]^-{bf} &&   B' .
\\  }
$$
We number the rows $0, 1, \ldots, m+1$ and let $D_l$ be the homotopy
pushout of the $l^{th}$ row, with induced maps $d_l:D_l\maprt{} D_{l+1}$.
Then $D_0 = D$, $D_{m+1} = \overline D$ and $d_m\cdots d_0 = d':D\maprt{} \overline D$.
Thus it suffices to prove
\begin{enumerate}
\item[(i)]  $\LL_\A(d_l) \leq 1$ for $l= 0, \ldots, m-1$,
\item[(ii)]  $\LL_\A(d_m) =0$.
\end{enumerate}
We first establish (i).  Consider the commutative diagram
$$
\xymatrix{
B_l\ar[d] && {*}\ar[ll]\ar[rr]\ar[d]  && A_l\ar[d]  &&&& A_l\wdg B_l\ar[d] \\
Y_l\ar[d]_{j_l} && A \ar[ll]\ar[rr]\ar@{=}[d] && X_l\ar[d]^{i_l} &
{}\ar@{.>}[rr]^{ {\mathrm{homotopy}}\atop {\mathrm{pushout}} } &&&
  D_l\ar[d] \\
Y_{l+1}      && A\ar[ll]\ar[rr]         && X_{l+1}  &&&& D_{l+1},      \\   }
$$
where the columns are regarded as mapping cone sequences.
The homotopy pushouts of the rows form a sequence
$A_l\wdg B_l \maprt{} D_l\maprt{} D_{l+1}$.  By the Four
Cofibrations Theorem, this is a cofiber sequence (see \cite[p.\thinspace 21]{Do2}).
Since $\A$ is closed under wedges, $A_l\wdg B_l\in\A$.
Therefore $\LL_\A(d_l)\leq 1$ by the Mapping Cone Axiom.  For (ii) we note that
$d_m: D_m\maprt{}\overline D$ is a homotopy equivalence since
$h$ and $k$ are homotopy equivalences \cite[Ch.\thinspace XII,\thinspace \sec\thinspace 4.2]{B-K}.  Thus
$\LL_\A(d_m) =0$, which completes the proof of (a).

For (b) we proceed similarly by assuming that $\LL_\A(a) = m$ and taking
an $\A$-cone length decomposition of $a$ of length $m$:
$$
A = X_0\maprt{i_0} X_1\maprt{} \cdots\maprt{} X_{m-1}\maprt{i_{m-1}} X_m \maprt{h} A',
$$
where $A_l \maprt{} X_l\maprt{i_l} X_{l+1}$ is a mapping cone sequence
with $A_l\in \A$, $h$ a homotopy equivalence and $a= hi_{m-1}\cdots i_0$.
This yields a commutative diagram
$$
\xymatrix{
C'                                         \ar@{=}[d]
&&
A=X_0
\ar[ll]_-{cg}\ar[rr]^-{bf} \ar[d]^-{i_{0}}
&& B'                                      \ar@{=}[d]
\\
C'                                         \ar@{=}[d]
&&
X_1
\ar[ll]_-{g'hi_{m-1}\cdots i_1 }
\ar[rr]^-{f'hi_{m-1}\cdots i_1 }
\ar[d]^-{i_{1}}
&&
B' \ar@{=}[d]
\\
\vdots \ar@{=}[d]  && \vdots  \ar[d]  && \vdots    \ar@{=}[d] \\
C'                                         \ar@{=}[d]
&&
X_{m-1}
\ar[ll]_-{g'hi_{m-1}}
\ar[rr]^-{f'hi_{m-1}}
\ar[d]^-{i_{m-1}}
&&
B'                                          \ar@{=}[d]
\\
C'                                          \ar@{=}[d]
&&X_m
\ar[ll]_-{g'h}
\ar[rr]^-{f'h}
\ar[d]^-{h}
&& B'                                       \ar@{=}[d]    \\
C'
&&
A'
\ar[ll]_-{g'}\ar[rr]^-{f'}
&&   B' .  \\  }
$$
We number the rows $0,1,\ldots, m+1$ and
let $\twdl D_l$ be the homotopy pushout of the $l^{th}$ row
with induced maps $\twdl d_l:\twdl D_l\maprt{} \twdl D_{l+1}$.  Then
$\twdl D_0 = \overline D$, $\twdl D_{m+1}= D'$ and
$d'' = \twdl d_m\cdots \twdl d_1\twdl d_0$.  It suffices to show (i)
$\LL_\A(\twdl d_l) \leq 1$ for $l= 0, \ldots , m-1$ and (ii) $\LL_\A(\twdl d_m) =0$.
The argument is similar to (a), and so we content ourselves with noting
that
$$
\xymatrix{
{*}\ar[d]    &&   A_l\ar[ll]\ar[rr]\ar[d]       &&   {*}\ar[d]          &&&& \s A_l\ar[d]
\\
C'\ar@{=}[d] &&   X_l\ar[ll]\ar[rr]\ar[d]^{i_l} && B'\ar@{=}[d]
& {}\ar@{.>}[rr]^{  {\mathrm{homotopy}}\atop {\mathrm{pushout}} }        &&& \twdl D_l\ar[d]^{\twdl d_l}
\\
C'           && X_{l+1} \ar[ll]\ar[rr]          && B'                   &&&& \twdl D_{l+1}
\\  }
$$
determines a sequence $\s A_l \maprt{} \twdl D_l \maprt{\twdl d_l}D_{l+1}$
since $\s A_l$ is the homotopy pushout of the top row.  By the
Four Cofibrations Theorem this is a cofiber sequence.  Since $\A$
is closed under suspension, $\s A_l\in\A$, and so $\LL_\A(\twdl d_l)\leq 1$.
This completes the proof of (1).

To prove (2), we apply Lemma \ref{lem:strict}.  Thus, there are
commutative diagrams
$$
\xymatrix{
& A\ar[rd]^a\ar[ld]_a \ar[d]_\alpha  \\
A'\ar[r]^i & X \ar[r] ^r & A' \\
}\quad
\xymatrix{
& B\ar[rd]^b\ar[ld]_b \ar[d]_\beta  \\
B'\ar[r]^j & Y \ar[r] ^s & B' \\
}\quad \xymatrix{{}
\ar@{}[d]^{\txt{and}}\\{}}\quad
\xymatrix{
& C\ar[rd]^c\ar[ld]_c \ar[d]_\gamma  \\
C'\ar[r]^k & Z \ar[r] ^t & C' \\
}
$$
with $ri=\id$, $sj=\id$, and $tk =\id$, and $\LL_\A(\alpha) = \L_\A(a)$,
$\LL_\A(\beta) = \L_\A(b)$, and $\LL_\A(\gamma) =\L_\A(c)$.  Thus we have
a diagram
$$
\xymatrix{
& C \ar[rdd]^(.35)c  \ar[dd]^(.35)\gamma  \ar[ldd]_(.35)c
\\
&&&& A \ar[lllu]_g\ar[rrrd]^f\ar[rdd]^(.35)a\ar[ldd]_(.35)a\ar[dd]^(.35)\alpha
\\
C'\ar[r]^k &Z\ar[r]^t & C'  &&&&& B\ar[rdd]^(.35)b\ar[ldd]_(.35)b\ar[dd]^(.35)\beta
\\
&&&A'\ar[r]^(.3)i \ar[lllu]^(.65){g'}\ar[rrrd]_(.3){f'} & X
\ar[r]^(.35)r\ar[lllu]_(.6)v\ar[rrrd]^(.4)u & A'\ar[lllu]_(.65){g'}\ar[rrrd]^(.35){f'}
\\
&&&&&&B'\ar[r]_j & Y\ar[r]_s & B'
\\
}
$$
where $u=jf'r:X\maprt{}Y$ and $v=kg'r:X\maprt{}Z$.
All triangles and rectangles in the above diagram are commutative.
If we denote by $E$ the homotopy pushout of $Z\maplft{v} X\maprt{u}Y$
and the induced maps of homotopy pushouts by $e:D\maprt{}E$,
$l:D'\maprt{} E$ and $m:E\maprt{}D'$, then we have a commutative
diagram
$$
\xymatrix{
&& D\ar[rrd]^d\ar[lld]_d \ar[d]_e \\
D'\ar[rr]^l && E \ar[rr] ^m && D' \\
}
$$
with $ml = \id$.  Therefore
$$
\begin{array}{rclcl}
\L_\A(d)  & \leq  & \L_\A(e)    &\quad &  {\mathrm{since}}\ e\ {\mathrm{dominates}}\ d \\
          & \leq  & \LL_\A(e) \\
          & \leq & \LL_\A(\alpha) + \max(L_\A(\beta),L_\A(\gamma) ) && {\mathrm{by\ part}}\ (1) \\
          & =    & \L_\A(a) + \max(\L_\A(b),\L_\A(c)).\\
\end{array}
$$
\end{prf}

We next show that
some of Theorem \ref{thrm:MPT} remains true with weaker hypotheses on the
collection $\A$.

\bcor\label{cor:nowedge}

Assume the hypotheses of Theorem \ref{thrm:MPT},
except that $\A$ is not necessarily closed under wedges.
Then
\begin{enumerate}
\item  $\LL_\A(d) \leq \LL_\A(a) + \LL_\A(b) + \LL_\A(c)$;

\item  $\L_\A(d) \leq \L_\A(a) + \L_\A(b) + \L_\A(c)$.
\end{enumerate}
The result remains true without assuming that
$\A$ is closed under suspensions if $A = A'$ and
$a =\id$.
\ecor

\begin{prf}
We simply decompose the given map of homotopy pushouts into
a composition of three maps:
$$
\xymatrix{
C  \ar[d]_c   && A \ar[ll]_g\ar[rr]^f\ar@{=}[d]       && B\ar@{=}[d]         \\
C' \ar@{=}[d] && A \ar[ll]_{cg}\ar[rr]^f\ar@{=}[d]       && B\ar[d]^b           \\
C' \ar@{=}[d] && A \ar[ll]_{cg}\ar[rr]^{bf}\ar[d]^{a} && B'\ar@{=}[d]        \\
C'            && A' \ar[ll]_{g'}\ar[rr]^{f'}          && B'  .                \\  }
$$
The method of the proof of Theorem \ref{thrm:MPT} is then applied to each
factor.
\end{prf}

\section{Applications of the Homotopy Pushout Mapping Theorem}

In this section we illustrate the power of the homotopy pushout mapping theorem by
obtaining as a consequence a large number of results, some known (in the case
$\A = \{ {\mathrm {all\ spaces}} \}$), and some new.

\subsection{Homotopy Pushouts}

\bcor\label{cor:pushouts}
Let $\A$ be any collection of spaces.  Let
$$
\xymatrix{ A\ar[rr] \ar[d] && B\ar[d]\\ C\ar[rr] && D\\ }
$$
be a homotopy pushout square.  Then
\begin{enumerate}

\item
\begin{enumerate}
\item  $\LL_\A(B\maprt{}D) \leq \LL_\A(A\maprt{}C)$,
\item  $\L_\A(B\maprt{}D) \leq \L_\A(A\maprt{}C)$;
\end{enumerate}

\item
\begin{enumerate}
\item  $\cl_\A( D ) \leq \cl_\A(B) + \LL_\A(A\maprt{}C)$,
\item  $\cat_\A( D ) \leq \cat_\A(B) + \L_\A(A\maprt{}C)$;
\end{enumerate}

\item
\begin{enumerate}
\item $\kl_\A(B) \leq \LL_\A(A\maprt{}C) + \kl_\A(D)$,
\item $\kit_\A(B) \leq \L_\A(A\maprt{}C) + \kit_\A(D)$.
\end{enumerate}

\item  If $A$ is closed under wedges, then
\begin{enumerate}
\item  $\LL_\A( A\maprt{}D)\leq \max( \LL_\A(A\maprt{}B) , \LL_\A(A\maprt{}C) )$,
\item  $\L_\A( A\maprt{}D)\leq \max( \L_\A(A\maprt{}B) , \L_\A(A\maprt{}C) )$.
\end{enumerate}

\end{enumerate}
\ecor

\begin{prf}
The proof of each part amounts to constructing the
correct diagram.

\noindent{\sc Proof of 1}\quad   Apply Corollary \ref{cor:nowedge}
to the diagram
$$
\xymatrix{
& A \ar[rr]\ar@{=}[ld]\ar@{=}'[d][dd]      && B \ar@{=}[dd] \ar@{=}[ld]
\\
A \ar[rr] \ar[dd]                  && B \ar[dd]
\\
& A\ar'[r][rr]\ar[ld]             && B. \ar[ld]
\\
C\ar[rr]                          && D
}
$$

\noindent{\sc Proof of 2 and 3} \quad Apply (1) to the
diagram
$$
\xymatrix{ && {*}\ar[d] \\A\ar[rr] \ar[d] && B\ar[d]\\ C\ar[rr] && D\ar[d] \\
&& {*}.\\  }
$$

\noindent{\sc Proof of 4}\quad
Map the trivial homotopy pushout diagram
$$
\xymatrix{ A\ar@{=}[rr] \ar@{=}[d] &&A\ar@{=}[d]\\ A\ar@{=}[rr] && A\\ }
$$
into the given one, and apply Theorem \ref{thrm:MPT}.
\end{prf}

\bcor\label{cor:susppushouts}
Let $\A$ be a collection of spaces that is closed under
wedges and suspension and let
$$
\xymatrix{ A\ar[rr] \ar[d] && B\ar[d]\\ C\ar[rr] && D\\ }
$$
be a homotopy pushout square.  Then
\begin{enumerate}
\item
\begin{enumerate}
\item  $\cl_\A(D) \leq \cl_\A(A) + \max (\cl_\A(B), \cl_\A(C) )$,
\item  $\cat_\A(D) \leq \cat_\A(A) + \max (\cat_\A(B), \cat_\A(C) )$;
\end{enumerate}
\item
\begin{enumerate}
\item  $\kl_\A(D) \leq \kl_\A(A) + \max (\kl_\A(B), \kl_\A(C) )$,
\item  $\kit_\A(D) \leq \kit_\A(A) + \max (\kit_\A(B), \kit_\A(C) )$.
\end{enumerate}
\end{enumerate}
\ecor

\begin{prf}
For (1), apply Theorem \ref{thrm:MPT} to the map of the trivial homotopy
pushout diagram
$$
\xymatrix{ {*}\ar@{=}[rr] \ar@{=}[d] &&{*}\ar@{=}[d]\\ {*}\ar@{=}[rr] && {*}\\ }
$$
into the given homotopy pushout; for (2), map the given homotopy pushout into the
trivial one.
\end{prf}

\brmk
{\rm In the special case $\A=\{ {\mathrm{all\ spaces}} \}$,
Marcum \cite{Marcum} has proved  Corollary \ref{cor:pushouts}(1a)
and Hardie \cite{Hardie} has proved Corollary \ref{cor:susppushouts}(1b)
(see also \cite{Cornea4}).}
\ermk

\medskip

\subsection{Mapping Cone Sequences}

As noted in \sec 2, a mapping cone sequence $A\maprt{}B\maprt{}C$
can be regarded as a homotopy pushout square.
Therefore the results of 4.1 apply to mapping cone sequences.

\bcor\label{cor:mappingcones}
Let $\A$ be any collection of spaces.
Let $A\maprt{}B\maprt{}C$ be a mapping cone sequence.
Then
\begin{enumerate}

\item
\begin{enumerate}
\item  $\cl_\A(C) \leq \LL_\A(A\maprt{} B )$,
\item  $\cat_\A(C ) \leq \L_\A(A\maprt{} B)$;
\end{enumerate}

\item
\begin{enumerate}
\item   $\LL_\A (B\maprt{}  C) \leq \kl_\A(A)$,
\item   $\L_\A(B\maprt{}  C) \leq \kit_\A(A)$;
\end{enumerate}
\item
\begin{enumerate}
\item  $\cl_\A(C) \leq \kl_\A(A) + \cl_\A(B)$,
\item  $\cat_\A(C) \leq \kit_\A(A) + \cat_\A(B)$;
\end{enumerate}

\item
\begin{enumerate}
\item  $\kl_\A(B)\leq \kl_\A(A) + \kl_\A(C)$,
\item  $\kit_\A(B)\leq \kit_\A(A) + \kit_\A(C)$.
\end{enumerate}

\end{enumerate}
\ecor

\begin{prf}
\noindent{\sc Proof of 1 and 2}\quad  Immediate from Corollary \ref{cor:pushouts}(1).

\noindent{\sc Proof of 3 and 4} \quad Immediate from (2) and (3)
of Corollary \ref{cor:pushouts}.
\end{prf}

\brmk{\rm
Corollary \ref{cor:mappingcones}(4)
shows that $\kl_\A$ and $\kit_\A$ are subadditive on cofibrations in the
following sense (we only state this for $\kl_\A$): If
$A\maprt{}X\maprt{}Q$ is a cofiber sequence, then $\kl_\A(X)
\leq \kl_\A(A) + \kl_\A(Q)$ (see \cite[Thm.\thinspace 3.4]{AS}.
This follows (when $\A$ is closed under wedges) since
every cofiber sequence is equivalent to a mapping cone sequence.  This inequality is not
generally true for $\cl_\A$ or $\cat_\A$ as the cofiber sequence
$$
S^2\maprt{}\cp^3\maprt{}S^4 \wdg S^6
$$
shows for the collections $\A = \S$, $\Sigma$ and $\{ {\mathrm{all\ spaces}} \}$.
}
\ermk

\bcor\label{cor:suspcof} Let $\A$ be a collection of
spaces that is closed under suspension.
Consider the map of one mapping cone sequence into another given
by the commutative diagram
$$
\xymatrix{
A \ar[rr]\ar[d] && B \ar[rr]\ar[d] && C\ar[d]\\
A'\ar[rr] && B'\ar[rr] && C'. \\ }
$$
Then
\begin{enumerate}
\item  $\LL_\A(C\maprt{} C') \leq \LL_\A(A\maprt{} A') + \LL_\A(B\maprt{}B')$,
\item  $\L_\A(C\maprt{} C') \leq \L_\A(A\maprt{} A') + \L_\A(B\maprt{}B')$.
\end{enumerate}
\ecor

\begin{prf}
Apply Corollary \ref{cor:nowedge}
to the homotopy pushouts obtained from the mapping cone sequences.
\end{prf}

\medskip

\subsection{Other Consequences}

\bcor\label{cor:klcl}
Let $\A$ be any collection of spaces.  Then
for any space $B$,
\begin{enumerate}
\item  $\cl_\A(\s B) \leq \kl_\A(B)$;
\item  $\cat_\A(\s B) \leq \kit_\A(B)$.
\end{enumerate}
\ecor

\begin{prf}  Apply  Corollary
\ref{cor:pushouts}(1) to the homotopy pushout
square
$$
\xymatrix{B\ar[rr] \ar[d] &&{*}\ar[d]\\ {*}\ar[rr] && \s B.\\ }
$$
\end{prf}

\bcor\label{cor:4.10}
Let $\A$ be a  collection of spaces that is closed under suspension.
\begin{enumerate}
\item  For any map $f:A\maprt{}B$,
\begin{enumerate}
\item  $\LL_\A(f) \leq \cl_\A(A) + \cl_\A(B) $,
\item  $\L_\A(f) \leq \cat_\A(A) + \cat_\A(B)$;
\end{enumerate}

\item  For any space $A$,
\begin{enumerate}
\item  $\kl_\A( A) \leq \cl_\A(A)$,
\item  $\kit_\A( A) \leq \cat_\A(A)$;
\end{enumerate}

\item  If $f:A\maprt{} B$ and $g:B\maprt{}C$, then
\begin{enumerate}
\item  $\LL_\A(g) \leq \LL_\A(f) + \LL_\A(g  f) $,
\item  $\L_\A(g) \leq \L_\A(f) + \L_\A(g  f) $;
\end{enumerate}

\item  If $f:A\maprt{}B$ and $g: B\maprt{}A$ with $gf =\id$, then
\begin{enumerate}
\item[] $\L_\A(g) \leq \cat_\A(B)$;
\end{enumerate}

\item  If $f:A\maprt{} B$ and $g:B\maprt{}A$ with
$g  f = \id$, then
\begin{enumerate}
\item  $\LL_\A(g) \leq \LL_\A(f)  $,
\item  $\L_\A(g) \leq \L_\A(f) $.
\end{enumerate}

\end{enumerate}
\ecor

\begin{prf}
Again, the proofs depend on finding the appropriate diagram.

\noindent{\sc Proof of 1}\quad   Apply Corollary \ref{cor:nowedge}
to the diagram
$$
\xymatrix{
& {*} \ar@{=}[rr]\ar[ld]\ar'[d][dd]      && {*} \ar[dd] \ar[ld]
\\
A \ar@{=}[rr] \ar@{=}[dd]                  && A \ar[dd]
\\
& A\ar'[r][rr]\ar@{=}[ld]             && B. \ar@{=}[ld]
\\
A\ar[rr]                          && B
}
$$

\noindent{\sc Proof of 2}\quad   Apply (1) to the map $A\maprt{} *$.

\noindent{\sc Proof of 3}\quad   Apply Corollary \ref{cor:nowedge}
to the diagram
$$
\xymatrix{
& A \ar[rr]\ar@{=}[ld]\ar'[d][dd]      && B \ar@{=}[dd] \ar@{=}[ld]
\\
A \ar[rr] \ar[dd]                  && B \ar[dd]
\\
& B\ar@{=}'[r][rr]\ar[ld]             && B. \ar[ld]
\\
C\ar@{=}[rr]                          && C
}
$$

\noindent{\sc Proof of 4}\quad   We consider the following mapping of
homotopy pushout squares
$$
\xymatrix{
& {*} \ar[rr]\ar[ld] \ar'[d][dd]      && B \ar@{=}[dd] \ar[ld]
\\
A \ar[rr] \ar@{=}[dd]                   && A\wdg B \ar[dd]^(.3){(\id,g)}
\\
& B\ar@{=}'[r][rr]\ar[ld]^(.4)g             && B. \ar[ld]^(.4)g
\\
A\ar@{=}[rr]                          && A
}
$$
By Corollary \ref{cor:nowedge}, we immediately conclude that
$\L(\id,g) \leq \cat (B)$.  Now the commutative diagram
$$
\xymatrix{
B\ar[d]_g  \ar[rr]    && A\wdg B \ar[d]^{(\id,g)} \ar[rr]^-{(f,\id)}&& B \ar[d]^g
\\
A\ar@{=}[rr]                   && A\ar@{=}[rr]                     && A
\\
}
$$
shows that $g$ is dominated by $(\id, g)$.  Thus
$\L(g) \leq \L(\id ,g) \leq \cat(B)$.

\noindent{\sc Proof of 5}\quad   Apply (3),
using the fact that $L(\id) = \L(\id ) =0$.

\end{prf}

\bcor\label{cor:lowerbound}
Let $\A$ be a collection that is closed under
suspension and let $f:A\maprt{}B$.  Then
\begin{enumerate}
\item
\begin{enumerate}
\item   $\LL_\A(f) \geq | \kl_\A(B) - \kl_\A(A) |$,
\item   $\L_\A(f) \geq | \kit_\A(B) - \kit_\A(A) |$;
\end{enumerate}
\item
\begin{enumerate}
\item   $\LL_\A(f) \geq   \cl_\A(B) - \cl_\A(A)  $,
\item   $\L_\A(f) \geq   \cat_\A(B) - \cat_\A(A)  $.
\end{enumerate}
\end{enumerate}
\ecor

\begin{prf}
We only prove (1a); the other parts are similar.
The Composition Axiom, applied to $A\maprt{f} B\maprt{} *$, implies that
$$
\kl_\A(A) =    \LL_\A(A\maprt{}*)
          \leq \LL_\A(B\maprt{}*) + \LL_\A(f)
          =    \kl_\A(B) + \LL_\A(f),
$$
so $\LL_\A(f) \geq \kl_\A(A) - \kl_\A(B)$.  On the other hand,
Corollary \ref{cor:4.10}(3) shows that
$$
\kl_\A(B) = \LL_\A(B\maprt{}*) \leq  \LL_\A(A\maprt{}*) + \LL_\A(f)  = \kl_\A(A) + \LL_\A(f),
$$
so $\LL_\A(f) \geq \kl_\A(B) - \kl_\A(A)$. This proves (1a)
\end{prf}

\brmk  {\em  Assume $\A$ is closed under suspension.  Then
by Corollary \ref{cor:4.10}, $\kl_\A(A) \leq \cl_\A(A)$ and $\kit_\A(A)\leq \cat_\A(A)$
(the first inequality was also proved in \cite[Thm.\thinspace 3.3]{AS}).
Corollary \ref{cor:klcl} then shows that $\cl_\A$ and $\kl_\A$
agree stably (this was proved by Christensen in \cite{Christensen}),
and similarly for $\cat_\A$ and $\kit_\A$.  Additionally, Cornea \cite{Cornea2}
has given a completely different proof of Corollary \ref{cor:4.10}(4) in the case
$\A = \{ {\mathrm{all\ spaces}} \}$. }
\ermk

\medskip

\subsection{Partial Converse to Theorem \ref{thrm:MPT}}

In this section we show that the formulas of
Theorem \ref{thrm:MPT} very nearly characterize
those collections $\A$ which are closed under
wedges or under suspensions.

We introduce the following new
construction:  for any collection $\A$,  The collection
$\overline \A$ is defined to be
$$
\overline \A = \{ X \, |\, \kl_\A(X) \leq 1\}.
$$
Our first result shows that passing from $\A$ to $\overline \A$
has no effect on the corresponding cone length and
category invariants.  Note that if every space in
$\A$ is simply-connected, then
$\A = \overline \A $.

\bprp
For any map $f:X\maprt{}Y$,
\begin{enumerate}
\item  $\LL_{\overline \A} (f) = \LL_\A(f)$,
\item  $\L_{\overline \A} (f) = \L_\A(f)$.
\end{enumerate}
\eprp

\begin{prf}
If suffices to prove (1), because for any collection $\A$,
$\L_\A(f)$ is the least $n$ for which $f$ is a retract of a map
$g$ with $\LL_\A(g) \leq n$ \cite[Prop.\thinspace 4.3]{ASS}.

Since $\A\sseq \overline \A$, we have $\LL_{\overline \A} (f) \leq \LL_\A(f)$
for any map $f$, so it remains to prove the reverse inequality.
Suppose $\LL_{\overline \A} (f) = n$, and that

$$
\xymatrix{
A_0  \ar[d]           & A_1\ar[d]  &                & A_{n-1}\ar[d]
\\
X_0 \ar[r]^-{j_0}\ar@{=}[d]  & X_1\ar[r]^-{j_1}  & \cdots\ar[r]^-{j_{n-2}}   & X_{n-1} \ar[r]^-{j_{n-1}}  &
X_n\dequivarrow
\\
X\ar[rrrr]^f &&&& Y\\
}
$$
is a minimal $\overline \A$-cone decomposition for $f$.  Thus
each $A_i\in \overline \A$ and each $A_i\maprt{} X_i\maprt{j_i}X_{i+1}$
is a mapping cone sequence.   Since $A_i\in\overline \A$, $\kl_\A(A_i)\leq 1$,
and hence $\LL_\A(j_i) \leq 1$ by Corollary \ref{cor:mappingcones}(2a).
By the Composition Axiom, $\LL_\A(f) \leq n = \LL_{\overline \A}(f)$.
\end{prf}

We next show that the collection $\overline \A$ satisfies the inequality
of Theorem \ref{thrm:MPT}(1) if and only if $\overline \A$
is closed under both wedges and suspension.
For this it suffices to prove the following corollary.

\bcor
Let $\A$ be any collection and consider commutative diagrams of the form
$$
\xymatrix{
C\ar[d]^{c} && A \ar[ll]_{g} \ar[rr]^{f} \ar[d]^{a} && B\ar[d]^{b}
\\
C' && A' \ar[ll]_{g'}\ar[rr]^{f'} && B'  .
\\  }
$$
If the inequality
\begin{enumerate}
\item $\LL_\A(d) \leq \LL_\A(a) +  \max ( \LL_\A(b),\LL_\A(c))$
\end{enumerate}
of Theorem \ref{thrm:MPT} holds for any such diagram, then $\overline \A$ is
closed under both wedges and suspension.
\ecor

\begin{prf}
We show that $\overline \A$ is closed under suspension; the proof of the other
assertion is similar.  Let $A\in \overline \A$ and  consider the commutative diagram
$$
\xymatrix{
{*} \ar[d] && A   \ar[ll]\ar[d]\ar[rr] && {*} \ar[d]\\
{*}        && {*} \ar[ll]      \ar[rr] && {*}.\\
}
$$
By (1), we have
$$
\begin{array}{rclcl}
\kl_\A(\s A) & = & \LL_{\A}(\s A\maprt{} *) \\
& \leq & \LL_{\A}(A\maprt{} *) \\
& = & \kl_\A(A)  & \leq & 1, \\
\end{array}
$$
so $\s A\in \overline \A$ by definition.
\end{prf}

\brmk  {\rm To conclude that $\overline \A$ is closed under suspension,
it suffices to consider only diagrams in which $b = \id $ and $c=\id$,
and to conclude that $\overline \A$ is closed under wedges,
we only need to consider diagrams with $a=\id$. }
\ermk

\bigskip

\section{Products}

The following is our main result on products of maps.

\bthm\label{thrm:5.1}
Let $\A$ be a collection that is closed under wedges and joins
and let $f:A\maprt{}X$ and $g:B\maprt{}Y$ be maps.  Then

\begin{enumerate}
\item
$\LL_\A(f\cross g) \leq \LL_\A(f) + \LL_\A(g)
+\max( \cl_\A(A) , \cl_\A(B) )$,
\item
$\L_\A(f\cross g) \leq  \L_\A(f) + \L_\A(g)
+\max( \cl_\A(A) , \cl_\A(B) )$.
\end{enumerate}
\ethm

\begin{prf}
In the proof of (1) we write
$a=\cl_\A(A)$, $b=\cl_\A(B)$, $m=L_\A(f)$ and $n=L_\A(g)$ and assume that $a\geq b$.

Now consider the $\A$-cone decompositions of $*\maprt{}A$ and $f$

$$
\xymatrix{
K_0  \ar[d]           & K_1\ar[d]  &                & K_{a-1}\ar[d]   & K_a\ar[d] && K_{m+a-1}\ar[d]
\\
A_0 \ar[r]\ar@{=}[d]  & A_1\ar[r]  & \cdots\ar[r]   & A_{a-1} \ar[r]  & A_a\dequivarrow \ar[r] & \cdots \ar[r]
& A_{m+a-1}\ar[r] & A_{m+a}\dequivarrow
\\
{*}\ar[rrrr] &&&& A \ar[rrr]_f &&& X\\
}
$$
and of $*\maprt{}B$ and $g$
$$
\xymatrix{
L_0  \ar[d]           & L_1\ar[d]  &                & L_{b-1}\ar[d]   & L_b\ar[d] && L_{m+a-1}\ar[d]
\\
B_0 \ar[r]\ar@{=}[d]  & B_1\ar[r]  & \cdots\ar[r]   & B_{b-1} \ar[r]  & B_b\dequivarrow \ar[r] & \cdots \ar[r]
& B_{n+b-1}\ar[r] & B_{n+b}\dequivarrow
\\
{*}\ar[rrrr] &&&& B \ar[rrr]_g &&& Y,\\
}
$$
where we identify $A$ with $A_a$ and $B$ with $B_b$.
Since $A_i\sseq A_{i+1}$ and $B_j\sseq B_{j+1}$,
we may define, for $0\leq k \leq n+m+a + b$,
$$
C_k = A\cross B \ \cup\bigcup_{i+j=k} A_i\cross B_j \ \ \sseq \ \ A_{m+a} \cross B_{n+b} .
$$
Observe that $C_b = A\cross B$, $C_{n+m+a + b} = X\cross Y$
and up to homotopy the composite
$$
C_b \maprt{} C_{b+1} \maprt{} \cdots \maprt{} C_{n+m+a + b}
$$
is $f\cross g$.  From this, we see that it suffices to show that
$\LL_\A(C_k \maprt{} C_{k+1}) \leq 1$ for each $k\geq b$.

For $0\leq i \leq a + m $ and $0\leq j\leq b+ n$, define
$$
P_{ij} = A_i \cross B_j,\ \   T_{ij} = A_i \cross B_{j-1} \cup
A_{i-1}\cross B_j, \ \ {\mathrm{and}}\ \
Q_{ij} = C_{i+j-1}\cup P_{ij}.
$$
Then $C_{k+1}$ is obtained as
the pushout of all the maps $C_k \maprt{} Q_{ij}$
with $i+j = k+1$.
By an induction based on Corollary \ref{cor:pushouts}(4) it follows that
$$ \LL_\A(C_k\maprt{}C_{k+1})\leq
\max( \LL_\A(C_k \maprt{} Q_{ij})).
$$
Thus, it suffices to show that for $i+j = k+1$,
$\LL_\A(C_k \maprt{} Q_{ij})\leq 1$.  Applying
Corollary \ref{cor:pushouts}(1) to the pushout
diagram
$$
\xymatrix{
T_{ij}\ar[rr]\ar[d] && C_k \ar[d]
\\
P_{ij}\ar[rr] && Q_{ij},
}
$$
we have
$\LL_\A(C_k \maprt{} Q_{ij})\leq \LL_\A( T_{ij}\maprt{}P_{ij})$.
According to a result of Baues \cite{Baues1} (see also
\cite{Stanley2}),
there is a mapping cone sequence
$$
K_{i-1} \ast L_{j-1} \maprt{} T_{ij}\maprt{} P_{ij}
$$
when $i,j>0$,   a mapping cone sequence
$$ \xymatrix{
L_{j-1} \ar[r] & T_{0j} \ar@{=}[d]\ar[r] & P_{0j}\ar@{=}[d]
\\
                & B_{j-1}                  & B_j
\\ }
$$
when $i=0$ and a mapping cone sequence
$$ \xymatrix{
K_{i-1} \ar[r] & T_{i0} \ar@{=}[d]\ar[r] & P_{i0}\ar@{=}[d]
\\
  & A_{i-1}                  & A_i
\\ }
$$
when $j=0$.
Since $\A$ is closed under joins,
$\LL_\A(T_{ij}\maprt{} P_{ij})\leq 1$, and this completes the proof of (1).

For (2) we take $f'$ to be a map which dominates $f$,
has the same domain and such that $\L_\A(f) = \LL_\A(f')$, and $g'$
is similarly chosen for $g$ (Lemma \ref{lem:strict}).  Then (2) is a
consequence of (1) since $f'\cross g'$ dominates $f\cross g$.
\end{prf}

\bcor\label{cor:5.2}
If $\A$ is closed under wedges and joins, then
\begin{enumerate}
\item \begin{enumerate}
\item  $\cl_\A (X\cross Y) \leq \cl_\A(X) + \cl_\A(Y)$,
\item  $\kl_\A(X \cross Y) \le  \kl_\A(X) + \kl_\A(Y) + \max(\cl_\A(X),\cl_\A(Y))$;
\end{enumerate}
\item \begin{enumerate}
\item $\cat_\A(X\cross Y) \le \cat_\A(X) + \cat_\A(Y)$,
\item $\kit_\A(X\cross Y) \le \kit_\A(X) + \kit_\A(Y) + \max(\cl_\A(X),\cl_\A(Y))$.
\end{enumerate}\end{enumerate}
\ecor

\brmk
{\em In the case $\A=\{ \mathrm{ all\ spaces} \}$,  Corollary \ref{cor:5.2}(2a)
is a classical result due to Bassi \cite{Bassi}.
Part (1a) has been obtained by Takens \cite{Takens},
Clapp and Puppe \cite{ClP2}, and Cornea \cite{Cornea5}.}
\ermk

It is possible to improve the inequalities in Corollary \ref{cor:5.2} by imposing
stronger conditions on the collection $\A$.  To illustrate this, we state
and sketch a proof of Proposition \ref{prop:below} below.  We say that a collection
$\A$ is a \term{$\smsh$-ideal} if for any $A\in\A$ and any space $B$,
the smash product $A\smsh B\in \A$.

\bprp\label{prop:below}
If $\A$ is a $\smsh$-ideal and is closed under wedges and suspensions, then
\begin{enumerate}
\item  $\kl_\A(X\times Y)\leq \kl_\A(X) + \kl_\A(Y)$ and
\item  $\kit_\A(X\times Y)\leq \kit_\A(X) + \kit_\A(Y)$.
\end{enumerate}
\eprp

\begin{prf}
We only prove (1) since the proof of (2) is similar.
By applying Corollary \ref{cor:mappingcones}(4)
to the sequence $X\wdg Y\maprt{}X\times
Y\maprt{}X\smsh Y$ we conclude that
$\kl_\A(X\times Y)\leq \kl_\A(X\wdg Y) +  \kl_\A(X\smsh Y)$.
By Corollary \ref{cor:susppushouts}(2),
$\kl_\A(X\wdg Y)\leq \max(\kl_\A(X),\kl_\A(Y))$.
Furthermore, a simple argument using the
fact that $\A$ is a $\smsh$-ideal shows that $\kl_\A(X\smsh Y)\leq
\min(\kl_\A(X),\kl_\A(Y))$.  This completes the sketch of the proof.
\end{prf}

\bigskip

\section{Pullbacks and Fibrations}

We prove a result on pullbacks which yields inequalities for the
$\A$-cone length and $\A$-category of the spaces  which appear in a fiber sequence.

We begin with a lemma which may be of independent interest.
In the proof we denote the half-smash $(X \cross Y)/X$ by $X \SDP Y$
and the quotient map by $q:X \cross Y \maprt{} X\SDP Y$.

\blem\label{lem:halfsmash} {\rm (Cf. \cite[Ex.\thinspace 5.4]{Marcum})}
Let $\A$ be a collection which is closed under joins and let $A\in \A$.
If $p_2:A\cross B\maprt{}B$ is the projection, then
\begin{enumerate}
\item[(1)] $\LL_\A(p_2) \leq \cl_\A(B) + 1$
\item[(2)] $\L_\A(p_2) \leq \cat_\A(B) + 1$.
\end{enumerate}
\elem

\begin{prf}
Consider the map $p:A\SDP B\maprt{}B$ induced by $p_2$.
The main step in the proof is to show
$$ \LL_\A(p) \leq \cl_\A(B)\qquad {\mathrm{and}}\qquad   \L_\A(p) \leq \cat_\A(B).
$$
Suppose we have a diagram:
$$
\xymatrix{
L_0  \ar[d]   & L_1\ar[d] & &L_{n-1}\ar[d]
\\
{*}=B_0 \ar[r]^-{j_0} & B_1\ar[r]^-{j_1} & \cdots\ar[r]   & B_{n-1} \ar[r]^-{j_{n-1}} & B_n \\
}
$$
and a map $f_n:B_n\maprt{}B$ with $\LL_i \in A$.
Define $D_i$ as the homotopy pushout in the diagram
$$
\xymatrix{
A\SDP B_i\ar[rr]^{q_i}\ar[d]_{\id\sdp  f_nj_{n-1}...j_i }  && B_i\ar[d]^{r_i}
\\
A\SDP B\ar[rr]^{s_i} && D_i, \\
}
$$
where $q_i$ is the projection.
Then there are maps $k_i:D_i\maprt{}D_{i+1}$ with $k_i s_i =s_{i+1}$.
When $i = 0$ we have $D_0 = A\SDP B$ and when $i =n$ we have
$$
\xymatrix{
A\SDP B_n \ar[rr]^{q_n} \ar[d]_{\id\sdp f_n}&& B_n \ar[d]^{r_n}\\
A\SDP B   \ar[rr]^{s_n}                        && D_n. \\
}
$$
From the above diagram and the maps
$$
 f_n:B_n\maprt{}B  \quad  {\mathrm{and}} \quad  p:A\SDP B\maprt{}B ,
$$
we obtain a map $g_n:D_n\maprt{}B$ such that $g_ns_n = p$ and $g_nr_n = f_n$.
It then follows that
$$
g_nk_{n-1}\cdots k_0 = p.
$$
Now we prove (1).  Suppose $f_n$ is a homotopy equivalence
so our given decomposition is an $\A$-cone
decomposition of $B$ of length $n$.  By the previous
homotopy pushout diagram, $r_n$ is a homotopy equivalence
and from $g_nr_n = f_n$ we obtain that $g_n$ is a
homotopy equivalence.   Since $g_nk_{n-1}\ldots k_0 = p$, we get $\LL_A(p) \leq \LL_\A(k_0) +
\cdots +\LL_\A(k_{n-1})$.  To  complete the proof that $\LL_\A(p)\leq n =\cl_\A(B)$,
it suffices to show  that $\LL_\A(k_i) \leq 1$.  But $L_i\maprt{}B_i\maprt{j_i}B_{i+1}$
is a mapping cone sequence and so
$$
A\SDP \LL_i\maprt{}A\SDP B_i\maprt{}A\SDP B_{i+1}
$$
is also a mapping cone sequence.  Thus we have a commutative  diagram
$$
\xymatrix{
{*}\ar[d]&& A\SDP L_i\ar[ll]\ar[rr]\ar[d] && \LL_i\ar[d]
\\
A\SDP B\ar[d]&& A\SDP B_i\ar[ll]\ar[rr]^{q_i}\ar[d]&& B_i\ar[d]
\\
A\SDP B&& A\SDP B_{i+1}\ar[ll]\ar[rr]^{q_{i+1}} && B_{i+1}.\\
}
$$
Since each column is a cofiber sequence, $P\maprt{}D_i\maprt{k_i}D_{i+1}$ is a cofiber
sequence by the Four Cofibrations Theorem, where $P$ is the homotopy pushout of the top
line.  However it is easily seen that  $P = A*L_i$, the join of $A$ and $\LL_i$.  But $P\in \A$
since $\A$ is closed under joins.  Thus $\LL_\A(k_i)\leq 1$ and so $\LL_\A(p)\leq n =\cl_\A(B)$.
Part 1 of the lemma now follows by factoring $p_2:A\cross B\maprt{}B$ as
$$
A\cross B\maprt{q}A\SDP B\maprt{p}B.
$$
Since $A\maprt{}A\cross
B\maprt{q}A\SDP B$ is mapping cone sequence with $A \in \A$,
$\LL_\A(q)\le 1$.  Thus $\LL_\A(p_2)\leq \LL_\A(p)+\LL_\A(q)\leq \cl_\A(B) +1$
by the Composition Axiom.

The proof of (2) is similar.  Instead of taking an $\A$-cone decomposition
of $B$, we take an $\A$-category decomposition of $B$ of length $n$.
Thus instead of having
$f_n:B_n\maprt{}B$ a homotopy equivalence, we have a map $s: B\maprt{}B_n$ with
$f_n s \simeq\id$.   We define $\sigma:B\maprt{}D_n$ by $\sigma = r_n s$.  Then
the following are easily checked:
$$
\begin{array}{rlrlrl}
{\mathrm (a)} & g_n \sigma \simeq \id  \qquad               &
{\mathrm (b)} & \sigma p \simeq  k_{n-1}\cdots k_0  \qquad   &
{\mathrm (c)} & g_n k_{n-1}\cdots k_0 \simeq p  .     \\
\end{array}
$$
Using the maps $(\id, \sigma)$ and $(\id,g_n)$ we see that $p$ is homotopy dominated by
$k_{n-1}\cdots k_0$. Therefore $\L_\A(p) \leq  \L_\A(k_0) + \cdots +\L_\A(k_{n-1})$.
The rest of the proof is the same as the proof of (1),
using $\L$ for $\LL$.
\end{prf}

Now we prove our pullback theorem.

\bthm\label{thrm:5.3}
Let $\A$ be a collection that is closed under wedges and joins and let
$$
\xymatrix{
A\ar[rr]\ar[d] && B\ar[d]\\
C\ar[rr] && D\\  }
$$
be a pullback diagram.  Let $B\maprt{}D$ be a fibration with fiber
$F$.  Then
\begin{enumerate}
\item $\LL_\A(A\maprt{} B) \leq \LL_{\A} (C\maprt{}D) ( \cl_\A (F) +1)$;
\item $\L_\A(A\maprt{} B) \leq
\L_{\A} (C\maprt{}D) ( \cat_\A (F) +1)$.
\end{enumerate}
\ethm

\begin{prf}
We prove (1).  Let
$$
\xymatrix{
K_0  \ar[d]           & K_1\ar[d]  &                & K_{n-1}\ar[d]
\\
C_0 \ar[r]\ar@{=}[d]  & C_1\ar[r]  & \cdots\ar[r]   & C_{n-1} \ar[r]  & C_n\dequivarrow
\\
C\ar[rrrr] &&&& D\\
}
$$
be a minimal $\A$-cone decomposition for $C\maprt{}D$.
For $0\leq i\leq n$, define $B_i$
to be the pullback indicated by the square
$$
\xymatrix{
B_i\ar[rr]\ar[d] &&  B\ar[d] \\
C_i\ar[rr] && D.\\  }
$$
Thus $B_0 = A$, $B_n\equiv B$ and we obtain maps $B_i\maprt{}B_{i+1}$.
With these identifications, the composition
$B_0\maprt{}B_1 \maprt{}\cdots \maprt{} B_n$ is simply $A\maprt{}B$.
Hence, it suffices to show that $\LL(B_i\maprt{} B_{i+1})\leq
\cl_{\A}(F) +1$.  Consider the cube diagram
$$
\xymatrix{
& K_i \cross F \ar[rr]\ar[ld]\ar'[d][dd]      && F \ar[dd] \ar[ld]
\\
B_i \ar[rr] \ar[dd]                  && B_{i+1} \ar[dd]
\\
& K_i\ar'[r][rr]\ar[ld]             && {*} \ar[ld]
\\
C_i\ar[rr]                          && C_{i+1}.
}
$$
In this diagram, the bottom square is a homotopy pushout and
the sides are pullbacks.  This assertion is obvious for all squares except the
left side square
$$
\xymatrix{
K_i\cross F \ar[rr]\ar[d]    && B_i \ar[d] \\
K_i \ar[rr] && C_i . \\ }
$$
To see that this  is a pullback square, let $P$ be the
pullback  of $B_i\maprt{}C_i\maplft{}K_i$.  Then $P$ is
also the pullback of $B_{i+1}\maprt{}C_{i+1}\maplft{}C_i\maplft{}K_i$.
Since the composite $C_{i+1}\maplft{}C_i\maplft{}K_i$ is the constant map, the  latter
pullback is $K_i \cross F$.

Now by the Mather's second cube
theorem \cite{Mather}, the top square is a homotopy pushout.
By  Corollary \ref{cor:pushouts}(1),
$\LL_\A(B_i\maprt{} B_{i+1}) \leq \LL_\A ( K_i\cross F \maprt{} F )$.
By Lemma \ref{lem:halfsmash}(1), $\LL_\A(K_i \cross F\maprt{} F)  \leq \cl_\A(F) + 1$.
This proves (1).

The proof of (2) is similar and uses Lemma 6.1(2) and we omit it.
\end{prf}

\bcor\label{cor:5.4}
Let $\A$ be a collection that is closed under wedges and joins
and let $F\maprt{} E\maprt{}B$ be a fibration.  Then
\begin{enumerate}
\item $\cl_\A(E) + 1 \leq (\cl_{\A} (B) + 1)(\cl_\A(F) +1)$;
\item $\cat_\A(E) + 1 \leq (\cat_{\A} (B) + 1)(\cat_\A(F) +1)$.
\end{enumerate}
\ecor

\begin{prf}
We prove (1).  Applying Theorem \ref{thrm:5.3} to the  pullback square
$$
\xymatrix{
F\ar[rr] \ar[d] && E\ar[d]  \\
{*} \ar[rr]     && B,        \\    }
$$
we obtain $\LL_\A(F\maprt{} E) \leq  \cl_\A(B) (\cl_\A(F) + 1)$.
Now the Composition Axiom shows that
$$
\cl_\A(E) \leq \cl_\A(F) + \LL_\A(F\maprt{}E),
$$
so
$$
\begin{array}{rcl}
\cl_\A(E) + 1                               & \leq &
\cl_\A(F) + \LL_\A(F\maprt{}E) + 1  \\      & \leq &
\cl_\A(B) (\cl_\A(F) + 1) + (\cl_\A(F) + 1) \\ & = &
(\cl_{\A} (B) + 1)(\cl_\A(F) +1).
\end{array}
$$
\end{prf}

\brmk {\rm
In the special case $\A=\{ \mathrm{all\ spaces}\}$
we retrieve Varadarajan's result \cite{V}
$$
\cat(E) + 1 \leq (\cat(B) +1) (\cat(F) +1).
$$
Hardie has obtained a further improvement in \cite{Hardie2}, but that
involves a different notion of the category of a map
from the one we consider here \cite{B-G,Fox}.  }
\ermk

\bigskip

\section{Miscellaneous Results and Problems}

In this section we consider several topics.
We first establish some elementary, but useful, facts about
$\LL_\A$ and $\L_\A$.  We then show
that some known results for the collection
$\A = \{{\mathrm{all\ spaces}}\}$ do not hold for an
arbitrary collection $\A$.
Finally, we conclude the section by stating a number of open
questions and discussing them briefly.

We begin with a few elementary results.

\bprp  Let $f:X\maprt{}Y$ and let $\A$ be any
collection.  Then
\begin{enumerate}
\item $\LL_\A(f) = 0$  if and only if $f$ is a homotopy equivalence;
\item $\L_\A(f) = 0$ if and only if $f$ is a homotopy equivalence.
\end{enumerate}
\eprp

\begin{prf}  We   prove (1) and (2) at the same time.  By the axioms, if $f$ is a homotopy
equivalence, then $\L_\A(f) = \LL_\A(f)  = 0$.  Conversely, define a
function $\l_\A$ by
$$
\l_\A (f) = \Biggl\{
\begin{array}{ll}
0 & {\mathrm {if}}\ f\ {\mathrm{is\ a\ homotopy\ equivalence}}\\
1 & {\mathrm {otherwise}}.\\
\end{array}
$$
It is trivial to check that $\l_\A$ satisfies the $\A$-category
axioms, so
$$
\l_\A(f)   \leq \L_\A(f)\leq \LL_\A(f)
$$
for every map $f$.  Consequently, if $f$ is not a homotopy equivalence,
then $\LL_\A(f) \geq\L_\A(f) \geq \l_\A(f) =1$.
\end{prf}

\bprp\label{prop:wdg}
Let $f:X\maprt{}Y$ and $g:X'\maprt{}Y'$ be maps and let
$\A$ be a collection that is closed under wedges.  Then
\begin{enumerate}
\item[(a)]  $\LL_\A(f\wdg g)  \leq \max(\LL_\A(f), \LL_\A(g))$;
\item[(b)]  $\L_\A(f\wdg g) = \max(\L_\A(f), \L_\A(g))$.
\end{enumerate}
\eprp

\begin{prf}
Since both $X \wdg X'$ and $Y\wdg  Y'$ are homotopy pushouts, the inequality
$\L_\A(f \wdg g) \leq \max(\L_\A(f), \L_\A(g))$ is a consequence of Theorem
\ref{thrm:MPT}.  This same argument shows  $\LL_\A(f \wdg g) \leq \max(\LL_A(f), \LL_A(g))$.
The reverse inequality for $\L_\A(f \wdg g)$ follows since $f$ and $g$ are both retracts
of $f \wdg g$.
\end{prf}

An example due to Dupont \cite{dupont} shows that
equality does not generally hold in (a).  Other examples
can be found in \cite{Stanley}, where spaces $X_n$ with
category $n$ and cone length $n+1$ are constructed.
According to an observation of Ganea \cite{Takens}
(see also \cite{Cornea2}), this implies that
there is a space  $A$ such that $\cl( X_n\wdg \s A ) = \cat(X_n)$.
If we let $\A=\{ {\mathrm{all\ spaces}}\}$,
$f:*\maprt{} X_n$ and $g:*\maprt{}\s A$,
then we have
$$
\LL_\A( f ) = \cl (X_n) > \cat(X_n) = \cl (X_n\wdg \s A) = \LL_\A(f\wdg g).
$$

\bcor  Let $X$ and $Y$ be spaces and $\A$ a collection that is
closed under wedges.  Then
\begin{enumerate}
\item[(a)]  $\LL_\A(X\maprt{*}Y) \leq \max ( \kl_\A(X), \cl_\A (Y))$;
\item[(b)]  $\L_\A(X\maprt{*}Y) = \max ( \kit_\A(X), \cat_\A (Y))$.
\end{enumerate}
\ecor

\begin{prf}
The trivial map $X\maprt{*}Y$ is the wedge of the maps $X \maprt{} *$
and $*\maprt{}Y$.  The result follows from Proposition \ref{prop:wdg}.
\end{prf}

Next we turn to some known results for the collection
$\A = \{{\mathrm{all\ spaces}}\}$.   For this collection we
delete the subscript $\A$ and
write $\LL_\A$ as $\LL$, $\L_\A$ as $\L$, etc.

For any map $f:X\maprt{}Y$, it has been proved that $\LL(f) \leq \cl(Y) + 1$
\cite{Marcum} and $\L(f) \leq \cat(Y) + 1$ \cite{Cornea2}.
We show that this may not be true for an arbitrary collection $\A$.

\bexm {\rm  By Corollary \ref{cor:lowerbound},
$\kl_\A(X) \leq \LL_\A(f) + \kl_\A(Y)$.
Thus if $L_\A (f) \le \cl_\A(Y) + 1$ were true, we would have
$$
\kl_\A( X )\leq  \cl_\A(Y) + \kl_\A(Y) + 1,
$$
for {\it any} $X$ and $Y$.  This cannot hold for any collection
$\A$ such that there are spaces $X$ with arbitrarily large
killing length (e.g., for $\A = \S$ or $\s$).
The analogous observation holds for $\L_\A$.}
\eexm

Another classical result concerns the homotopy pushout square
$$
\xymatrix{
A\ar[rr] \ar[d] && B\ar[d]\\
C\ar[rr] && D .\\ }
$$
It has been  shown that $\cl(D) \leq \cl(B) + \cl(C) + 1$  \cite{Hardie}.
We show that this is not true for $\A = \S$, the collection of wedges of spheres.

\bexm {\rm
Consider the homotopy pushout
$$
\xymatrix{
\cp^{t}\ar[rr] \ar[d] && {*}\ar[d]\\
{*}\ar[rr] && \s\cp^{t}.\\ }
$$
As $t$ increases, the length of the longest nontrivial composition  of
Steenrod squares in $H^*(\s \cp^t; \ZZ_2)$ also becomes arbitrarily large.
It follows from \cite[Prop.\thinspace 7.5]{ASS}
that $\cl_\S(\s \cp^t)$ increases as $t$ increases.
This contradicts the $\S$-analog of Hardie's result.  }
\eexm

We conclude the paper by stating and discussing three open problems.

\medskip

\noindent
{\bf Problem 7.6}\quad
We have seen in \cite[Prop.\thinspace 7.3]{ASS} that for certain collections $\A$,
$\wcat(X) \leq 2^{\kl_\A(X)} - 1$,
where $\wcat(X)$ is the weak category of X (see \cite{Gilbert,James}).
Since $\wcat(X) \leq\cat(X) \leq \cat_\A(X)$ for any collection $\A$,
it is reasonable to ask for which collections $\A$ is
$\cat_\A(X) \leq 2^{\kl_\A(X)} - 1$.
Of course $\A$ must not be $\{{\mathrm{all\ spaces}}\}$,
since   $\kl_\A(X) \leq 1$ for every space $X$ in that case.  We note
that the conjecture has been verified in the case $\A = \S$ and
$X = S^n_1 \cross\cdots \cross S^n_r$ \cite[Prop.\thinspace 6.2]{AMS}.
Other evidence for the conjecture in the case $\A = \Sigma$,
the collection of suspensions, has been given in
\cite{AS}, where a weaker form of  this problem has been posed
\cite[\sec 7,\thinspace No. 5]{AS}.

\medskip

\noindent
{\bf Problem 7.7}\quad
Given $f:X\maprt{}Y$.  Is $\LL_\A(f) \leq \kl_\A(X) + \cl_\A(Y)$,
and is $\L_A(f) \leq  \kit_\A(X) + \cat_\A(Y)$?

We discuss the evidence in the case of $\LL_\A$ (the discussion
is analogous for $\L_\A$).
First of all, if $C$ is the cofiber of $f$ it is true that
$\cl_\A(C) \leq \kl_\A(X) + \cl_\A(Y)$
(Corollary \ref{cor:mappingcones}(3a))
and also $\cl_\A(C) \leq \LL_\A(f)$
(Corollary \ref{cor:mappingcones}(1a)).
Secondly, we have that  $\LL_\A(f) \leq \cl_\A(X) + \cl_\A(Y)$
(Corollary \ref{cor:4.10}(1a)) and $\kl_\A(X)\leq \cl_\A(X)$.
Finally, when $\A=\{ \mathrm{all\ spaces}\}$ then $\kl_\A(X)=1$ for
every $X$, and in this case it is known that $L(f) \leq \cl(Y) + 1$
\cite{Marcum}.

\medskip

\noindent
{\bf Problem 7.8}\quad
It is well known that $\cl(X)\leq \cat(X)+1$ \cite{Takens}.
If $\A$ is a collection different from $\{ \mathrm{all\ spaces}\}$,
is there an upper bound for $\cl_\A(X)$ in terms of
$\cat_\A(X)$? This question was asked by Scheerer-Tanr\'e in \cite{S-T}.
Analogously, is there an upper bound for $\kl_\A$ in terms of $\kit_\A$?

We can show that $\kit_\s (X) \leq 1$ implies
$\kl_\s (X) \leq 3$ as follows.  If $\kit_\s(X) \leq 1$
then there is a mapping cone sequence $A\maprt{} X \maprt{*} Y$
with $A\in\s$.  It follows that $\s A = Y \wdg \s X$,
and so there is a retraction
map $\alpha : \s A \maprt{} Y$.
The cofiber of $\alpha $ is $\s^2 X$, and hence we have a
decomposition
$$
\xymatrix{
A\ar[d] &&  \s A \ar[d] && \s^2 X \ar[d]\\
X\ar[rr] && Y \ar[rr] && \s^2 X \ar[rr] && {*}.\\ }
$$
This proves that $\kl_\s(X) \leq 3$.

config: arxiv


TITLE: Noetherian ring with extra condition is a PID
QUESTION [0 upvotes]: Let $R$ be a noetherian integral domain.
Suppose that every finitely generated $R$-module is a direct sum of cyclic $R$-modules. Show that $R$ is a PID.
Here's my attempt:
Let $I$ be an ideal. Since $R$ is noetherian, $I$ can be considered as a finitely generated $R$-module. By the assumption, $I$ is isomorphic to direct sum of cyclic $R$-modules, say $M_1 \otimes \cdots \otimes M_n$. Since $I$ is torsion-free (as $R$ is a domain), it can be shown that $M_j \cong R$ for all $j$, and so $I \cong R^n$.
Now, if I can show $n = 1$, then it is not hard to deduce that $I$ is a principal ideal. But, this is where I am stuck.
I would appreciate any hint.

REPLY [2 votes]: Let $m_1,m_2$ be generators for $M_1,M_2$. We know $M_1 \cap M_2 = 0$, since they are two of the modules in the direct sum decomposition. But $m_1 m_2 \in M_1 \cap M_2$, and because $R$ is a domain, $m_1 m_2 \neq 0$.


TITLE: If $A$ is a semisimple algebra over $\mathbb{C}$, then the codimension of $[A,A]$ in $A$ is the number of isomorphism classes of simple $A$-modules.
QUESTION [2 upvotes]: Claim: (p. 145 in Groups and Representations by Alperin & Bell)
If $A$ is a semisimple algebra over $\mathbb{C}$, then the codimension of $[A,A]$ in $A$ is the number of isomorphism classes of simple $A$-modules, where $[A,A]:=\{ab-ba:a,b\in A\}$.
Pf: What I have thus far is the following:
Consider the matrices $e_{ij}=\delta_{ij}$.
If $i\neq j$, $e_{ij}=e_{ii}e_{ij}-e_{ij}e_{ii}\in [A,A]$
Also, $e_{ii}-e_{jj}=e_{ij}e_{ji}-e_{ji}e_{ij}\in[A,A]$
So,
$[A,A]= \left\{
\begin{bmatrix}
    x_{11}       & x_{12} & x_{13} & \dots & x_{1n} \\
    x_{21}       & x_{22} & x_{23} & \dots & x_{2n} \\
    \vdots &\vdots&\vdots& &\vdots \\
    x_{n1}       & x_{n2} & x_{n3} & \dots & x_{nn}
\end{bmatrix}: x_{11}+\ldots +x_{nn}=0\right\}$
I do not understand which set $\{e_{ij}\}$ belong to.

REPLY [3 votes]: The technological answer here is the following one:
If $A$ is an algebra, the vector space $A/[A,A]$ is called the $0$th Hochschild homology of $A$.

One of the key properties of Hochschild homology is that it is invariant under Morita equivalence: if $A$ and $B$ are algebras which are Morita equivalent, then the vector spaces $A/[A,A]$ and $B/[B,B]$ are isomorphic. In particular, since for every algebra $A$ the algebras $A$ and $M_n(A)$ are Morita equivalent (this is the most basic example of such an equivalence), we always have $$\frac{A}{[A,A]}\cong\frac{ M_n(A)}{[M_n(A),M_n(A)}.$$
On the other hand, Hochschild homology playes well with direct product: if $A$ and $B$ are algebras, then $$\frac{A\times B}{[A\times B,A\times B]}\cong\frac{A}{[A,A]}\times\frac{B}{[B,B]}$$.

What you want follows from these two facts directly, but in fact it follows form the first one alone. That an algebra is semisimple over an algebraically closed field implies easily that its category of modules is equivalent to that of a product of copies of the field —this is a strong form of Schur's lemma—  so one does not really need Wedderburn's theorem.


TITLE: Construct a function $f\in C^{\infty}_{0}(\mathbb{R}^2)$
QUESTION [2 upvotes]: The question request that :
Let $A=[-1,1]\times[-1,1]$, B be a open ball centered at the origin with radius 2. Construct a function $f\in C^{\infty}_{0}(\mathbb{R}^2)$ such that $f(x)=1$ for $x\in A$, $f(x)=0$ for $x\in B^{c}$, and $0<f<1$ on $B\setminus A$.
I attempted to mimic what I learned in constructing a bump function by considering $f(x)=e^{-x^2}$. But then I ran into a trouble that the required function is not radial symmetry. Can anyone give some hints for solving this question? By the way, is there a general method for constructing a smooth function of compact support for arbitrary spaces?

REPLY [0 votes]: Let $g\in C^\infty(\mathbb R)$ satisfy $g=1$ on $[-1,1]$ and $0<g<1$ everywhere else. Note that $g(x)g(y)=1$ on $ [-1,1]^2$ and $0<g(x)g(y)<1$ everwhere else.
Writing $z=(x,y),$ choose $h\in C^\infty(\mathbb R^2)$ that satisfies $h(z)=1$ for $0\le |z| \le \sqrt 2,$ $h(z)=0$ for $|z| \ge 2,$ and $0<h(z)<1$ in between. Note that $h=1$ on on $ [-1,1]^2.$
The function $f(x,y)= g(x)g(y)h(z)$ then does the job.


TITLE: Generating pythagorean triples in this question
QUESTION [4 upvotes]: Find all integer solutions to $x^{2}+8xy+25y^{2}=225 \tag 1$

$(1)$ can be written as $$(x+4y)^2 + (3y)^2 = 15^2$$
This reduces the problem to a matter of generating pythagorean triples. I'm thinking that Euclid's method would be useful, but it's not immediately obvious to me how to use it.

REPLY [1 votes]: Consider the quadratic equation
$$x^2 + 8xy + 25y^2 - 225 = 0$$
where $x$ is the unknown. Compute
$$\frac{\Delta}{4} = 16y^2 - 25y^2 + 225 = 225 - 9 y^2.$$
Then $x$ is an integer if and only if the discriminant is a perfect square.
Notice that $9 \cdot 5^2 = 225 = 0$, therefore $-5 \le y \le 5$. Now check for which $y$ the discriminant is a perfect square, and then compute the solutions for both $x$ and $y$.


TITLE: Tension in a chain around a cone
QUESTION [2 upvotes]: A chain of some mass, forming a circle, is slipped on a smooth cone.
If we consider an infinitesimally small section of the chain, a component of gravity will try to accelerate it along the surface of the cone. So there must be some tension in the string that prevents the acceleration. What will be the direction of the tension on this section of the chain?
Will it be towards the height of the cone, parallel to the ground?

REPLY [1 votes]: Let the cone angle at the apex be $2\alpha$.
Method 1 (Force considerations)
Consider a small arc of the ring subtended by angle $d\phi$. The tension $T$ acts tangential to both ends of this arc. This results in a radially-inward force of $2T \sin{(d\phi /2)}$.
Now consider the entire ring. The forces acting on it are:

$mg$ vertically downards
normal force $N$ all around the ring acting perpendicular to cone surface
radially inward force due to tension

Resolving along vertical direction yields $N = mg / \sin{\alpha}$. The normal force acting along a small arc of the ring subtended by angle $d\phi$ is  $\frac{d\phi}{2\pi} N$. Now resolving along the horizontal direction yields
$$\frac{d\phi}{2\pi} N \cos {\alpha} = 2T \sin{(d\phi /2)}$$
$$\frac{d\phi}{2\pi} mg \cot {\alpha} = 2T \sin{(d\phi /2)}$$
In the limit $d\phi \to 0$, $\sin{(d\phi /2)} \to d\phi /2$, so
$$ T = \frac{mg}{2\pi} \cot \alpha $$
Method 2 (Work-energy)
At equilibrium, the drop in gravitation potential energy by losing a height of $dh$ equals the work done by the tension $T$ in stretching the ring. (The normal force acts perpendicular to the direction of ring movement if the ring slides, so does no work).
A change of height $dh$ changes the ring radius by $dh \tan \alpha$, and thus changes the ring circumference by $2\pi (dh) \tan \alpha$. Therefore
$$T 2\pi (dh) \tan \alpha = mg(dh) $$
$$ T = \frac{mg}{2\pi} \cot \alpha $$


TITLE: performing integration whit default limit?
QUESTION [0 upvotes]: I have this assignment that I cant solve.

I have to solve this integral but I dont know how?

REPLY [0 votes]: You can pull the limit inside the integral.  Note that
$$\lim_{x \to \infty} \left ( 1+\frac1{2 x} \right )^{x} = e^{1/2} $$


TITLE: Why are orbitals are shown bubble shaped in the books when they exist in only in one particular dimension
QUESTION [0 upvotes]: I am utterly confused here: In our books orbitals $s$, $p$, $d$ are shown to be bubble or sphere like but ain't $s$-orbital exist only in $x$-axis so only a particular underline should be shown to denote it, just like when you mark open intervals in maths and the same goes for $p$-orbital for $x$, $y$ and $z$.
So why draw a bubble like shape encompassing all the dimensions/axes/planes?
Secondly, I can understand $d_{xy}$, $d_{xz}$, and $d_{yz}$ but what does $d_z^{2}$ and $d_z^{2} - y^{2}$ means ($x^{2} - y^{2}$ looks like $d_{xy}$, it just seems to be oriented by $90^{\circ}$)?
FYI, I am a high school student.

REPLY [1 votes]: Your teacher may have misinterpreted your question about the s orbitals.  The s orbital never has a subscript.  It is always spherical.
As you move into the p and d orbitals, the subscripts describe the shape of the orbital.  For example $p_z$ is symmetric around the z axis.  This does not mean the electron is confined to the z axis, merely that the probability of finding it at a given place is higher as you get close to the z axis.  The shapes given for the orbitals are defined by probability.  When they show you a shell of an orbital, it is usually rendered as "the electron has a 90% probability of being within this volume at any moment" or some similar construction.
The exact meaning of the subscripts is a bit more complicated.  They deal with what are called Cubic harmonics.  These cubic harmonics are solutions to the angular momentum operator: meaning they keep spin angular momentum conserved.  The subscripts are actually key parts out of the cubic harmonic equations which uniquely identify that solution to the angular momentum operator.  I find they are related to symmetries in the shape of the orbital, but I don't yet know if that's coincidence or something fundamental about how the equations operate.
Also note that these are really probabilistic solutions for a the wave equation describing the electrons.  Wikipedia has a very nice set of animations showing an analogue of these orbitals in 2 dimensions using a drum to show the resonance of this wave.  It may be a helpful tool for developing a more intuitive grasp on what an orbital means.  It's not a 1:1 match to how electrons work, but its close enough that it may help with the visualization (and it comes with a verbal explanation of how the two connect, if you're curious)


TITLE: Checking a solution of a PDE
QUESTION [2 upvotes]: I have the following PDE:
\begin{equation}
  -yu_x + xu_y = 0 \quad\text{where } u(0, y) = f(y)
\end{equation}
I derived a solution as follows:
\begin{align}
  -yu_x + xu_y =& 0 \\
  \iff& \nabla u(x,y)\dot \langle -y, x\rangle = 0 \\
  \implies& \frac{dy}{dx} = \frac{-x}{y} \\
  \iff& ydy = -xdx \\
  \iff& \int ydy = \int -xdx \\
  \iff& y^2 = -x^2 + c_2 \\
  \iff& y^2 + x^2 = c_2
\end{align}
Since $u(x,y)$ is constant along the ODE $\frac{dy}{dx}$, we have:
\begin{align}
  u(x,y) =& c_1 \\
         =& f(c_2) \quad\text{for some function $f$} \\
         =& f(y^2+x^2)
\end{align}
I want to check that this satisfies the PDE. Specifically, any function $f$ should satisfy the pde. My calculus is a bit rusty, and I am not exactly sure how to do this. Here is my reasoning:
Since $u(x,y) = -yu_x + xu_y = 0$ we have to substitute in for $f$ which yields
\begin{equation}
  u(x, y) = -yf_x2x + xf_y2y = 2xyf_y - 2xyf_x
  \end{equation}
We need the above to equal $0$. The $2xy$ and $-2xy$ give me evidence that it should cancel, and that my calculus is off... What is not clear to me is that, we don't know what $f$ is, hence I can't find out what $f_x$ or $f_y$ are. Though, the problem does look symmetrical, and I could see a potential solution involving polar coordinates, but I'm not quite sure how to solve it in this way either.

How can I verify that the above is indeed equal to $0$ and hence satisfies the PDE?

Thanks for all the help!

REPLY [2 votes]: Since we conjecture $u(x,y)=f(x^2+y^2)$, let us (carefully) apply calculus and verify $-yu_x+xu_y=0$.
Note (by chain rule),
$$ u_x = f'(x^2+y^2)*2x, \qquad u_y = f'(x^2+y^2) * 2y. $$
Thus, $-yu_x+xu_y=-2xyf'+2xyf'=0$.
So, you had it, except a small detail with the chain rule. Another way to have caught the mistake -- $f$ is a function of a single variable, so what does $f_x$ or $f_y$ mean?


TITLE: Triangle numbers problem
QUESTION [4 upvotes]: Find the smallest three-digit triangular number that can be
  represented both as the sum of three different triangular
  numbers, and as the sum of two different triangular numbers.

So the answer is $120$, but I found it by trial and error. What is the clever 'mathematical' way of answering this question?

REPLY [1 votes]: I went through it. With $T(x) = (x^2 + x)/2,$ we start with
$$  T(u) = T(v)+T(w) = T(x) + T(y) + T(z).  $$
Multiply by $8.$
Adding $1$ to all three positions and considering the first equality we get
$$ (2u+1)^2 + 1 = (2v+1)^2 + (2w+1)^2.  $$
This is always solvable, except that you want all three letters positive and $v \neq w.$ The short version is that we must have $ (2u+1)^2 + 1$ divisible by two distinct primes $\equiv 1 \pmod 4.$ For example, if $2u+1 = 13,$ we have $170,$ which can indeed be written as $11^2 + 7^2.$
Second thing to check: instead of $1,$ add $2$ to all positions and equate the first position and the third. We have
$$ (2u+1)^2 + 2 = (2x+1)^2 + (2y+1)^2 + (2z+1)^2.  $$
Once again this is always possible, except you want $x,y,z$ positive and distinct. So we strike out as
$$ 171 = 11^2 + 5^2 + 5^2 = 9^2 + 9^2 + 3^2.  $$
When $2u+1 = 21,$ we do get $442 = 17 \cdot 13 \cdot 2,$ and indeed
$$ 21^2 + 1 = 442 =  9^2 + 19^2.  $$ Then
$$  443 = 15^2 + 13^2 + 7^2.  $$
So that works. And the first occasion is actually
$$  \color{magenta}{55 = 10+45 = 6+21+28}.   $$
I see, they want at least $100.$
As $u$ gets larger, the ability to solve this with the full prohibitions should become the norm rather than the exception.


TITLE: Is isomorphism defined between large categories?
QUESTION [2 upvotes]: By definition, an isomorphism between two objects in a category is a morphism so and so.. We know that $\mathbf {Cat}$ is the categories of small categories so that morphisms between objects are functors. Now that we have no category of large categories (or do we have?!), how are we going to define an isomorphism between two large categories?
Or like many other cases (and mainly because of foundational reasons) we again ignore such entities and prefer to talk about small/locally small categories?
Thanks.

REPLY [7 votes]: Everything depends on foundations you work in.
The first way to think about it is that you can define the notion of isomorphism of categories (or any other types of objects) without mentioning a category containing this isomorphism. For example, in the following way:
Definition. Let $\mathcal{A}$ and $\mathcal{B}$ be categories, $T\colon\mathcal{A}\to\mathcal{B}$ be a functor. Then we say that $T$ is an isomorphism iff there exists such functor $S\colon\mathcal{B}\to\mathcal{A}$, that $S\circ T=I_{\mathcal{A}}$ and $T\circ S=I_{\mathcal{B}}$.
The second way is to take appropriate foundations. There is actually no need to deal with the category of all small categories. You may fix a Grothendieck universe $\mathcal{U}$ and define the category $\mathbf{Cat}_{\mathcal{U}}$ of all $\mathcal{U}$-small categories. Then $\mathbf{Cat}_{\mathcal{U}}$ is not a $\mathcal{U}$-small category, but it is a $\mathcal{U'}$-small category for some universe $\mathcal{U'}$, such that $\mathcal{U}\in\mathcal{U'}$. In such case, for instance, the isomorphism of duality
$$
\text{op}_{\mathcal{U}}\colon\mathbf{Cat}_{\mathcal{U}}\to\mathbf{Cat}_{\mathcal{U}}
$$
such that $\text{op}(\mathcal{A})=\mathcal{A}^{op}$, is an isomorphism in $\mathbf{Cat}_{\mathcal{U'}}$.


text:
\section{Dense core sampling}\label{sec:kron_sampling}

In this section, we focus on the most general situation when $\mathbfcal{G}$ is an unstructured dense tensor. Our objective is to design the sampling sets $\{\mathcal{L}_i\}_{i=1}^R$ by solving the discrete optimization problem~\eqref{eq:sparsesensing}.

We formulate the sparse tensor sampling problem using the frame potential as a performance measure. Following the same rationale as in \cite{frame_potential}, but for multidomain signals, we will argue that the frame potential is a tight surrogate of the MSE. By doing so, we will see that when we impose a Kronecker structure on the sampling scheme, as in \eqref{eq:separable_phi}, the frame potential of $\matx{\Psi}$ can be factorized in terms of the frame potential of the different domain factors. This allows us to propose a low complexity algorithm for sampling tensor data.

Throughout this section, we will use tools from submodular optimization theory. Hence, we will start by introducing the main concepts related to submodularity in the next subsection.
\subsection{Submodular optimization}
Submodularity is a notion based on the law of diminishing returns \cite{submodular_book} that is useful to obtain heuristic algorithms with near-optimality guarantees for cardinality-constrained discrete optimization problems. More precisely, submodularity is formally defined as follows.

\begin{definition}[Submodular function \cite{submodular_book}]\label{def:submodular}
  A set function $f:2^\mathcal{N}\rightarrow\R$ defined over the subsets of $\mathcal{N}$ is submodular if, for every $\mathcal{X}\subseteq\mathcal{N}$ and $x,y\in\mathcal{N}\setminus\mathcal{X}$, we have
  \begin{equation*}
    f(\mathcal{X}\cup\{x\})-f(\mathcal{X})\geq f(\mathcal{X}\cup\{x,y\})-f(\mathcal{X}\cup\{y\}).
  \end{equation*}
  A function $f$ is said to be supermodular if $-f$ is submodular.
\end{definition}

Besides submodularity, many near-optimality theorems in discrete optimization require functions to be also monotone non-decreasing, and normalized.

\begin{definition}[Monotonicity]
  A set function $f:2^\mathcal{N}\rightarrow\R$ is monotone non-decreasing if, for every $\mathcal{X}\subseteq\mathcal{N}$,
  \begin{equation*}
    f(\mathcal{X}\cup\{x\})\geq f(\mathcal{X})\quad \forall x\in\mathcal{N}\setminus\mathcal{X}
  \end{equation*}
\end{definition}

\begin{definition}[Normalization]
  A set function $f:2^\mathcal{N}\rightarrow\R$ is normalized if $f(\varnothing)=0$.
\end{definition}

In submodular optimization, matroids are generally used to impose constraints on an optimization, such as the ones in~\eqref{eq:sparsesensing}. A matroid generalizes the concept of linear independence in algebra to sets. Formally, a matroid is defined as follows.

\begin{definition}[Matroid \cite{discrete_optimization}]\label{def:matroid}
  A finite matroid $\mathcal{M}$ is a pair $(\mathcal{N},\mathcal{I})$, where $\mathcal{N}$ is a finite set and $\mathcal{I}$ is a family of subsets of $\mathcal{N}$ that satisfies: 1) The empty set is independent, i.e., $\varnothing\in\mathcal{I}$; 2) For every $\mathcal{X}\subseteq\mathcal{Y}\subseteq\mathcal{N}$, if $\mathcal{Y}\in\mathcal{I}$, then $\mathcal{X}\in\mathcal{I}$; and 3) For every $\mathcal{X},\mathcal{Y}\subseteq\mathcal{N}$ with $|\mathcal{Y}|>|\mathcal{X}|$ and $\mathcal{X},\mathcal{Y}\in\mathcal{I}$ there exists one $x\in\mathcal{Y}\setminus\mathcal{X}$ such that $\mathcal{X}\cup\{x\}\in\mathcal{I}$.

\end{definition}

In this paper we will deal with the following types of matroids.
\begin{example}[Uniform matroid \cite{discrete_optimization}]
  The subsets of $\mathcal{N}$ with at most $K$ elements form a uniform matroid $\mathcal{M}_\mathrm{u}=(\mathcal{N},\mathcal{I}_\mathrm{u})$ with $\mathcal{I}_\mathrm{u}=\{\mathcal{X}\subseteq\mathcal{N}:|\mathcal{X}|\leq K\}$.
\end{example}

\begin{example}[Partition matroid \cite{discrete_optimization}]\label{ex:partition_matroid}
  If $\{\mathcal{N}_i\}_{i=1}^R$ form a partition of $\mathcal{N}=\bigcup_{i=1}^R\mathcal{N}_i$ then $\mathcal{M}_\mathrm{p}=(\mathcal{N},\mathcal{I}_\mathrm{p})$ with $\mathcal{I}_\mathrm{p}=\{\mathcal{X}\subseteq\mathcal{N}:|\mathcal{X}\cap\mathcal{N}_i|\leq K_i\quad i=1,\dots,R\}$ defines a partition matroid.
\end{example}

\begin{example}[Truncated partition matroid \cite{discrete_optimization}]
  The intersection of a uniform matroid $\mathcal{M}_\mathrm{u}=(\mathcal{N},\mathcal{I}_\mathrm{u})$ and a partition matroid $\mathcal{M}_\mathrm{p}=(\mathcal{N},\mathcal{I}_\mathrm{p})$ defines a truncated partition matroid $\mathcal{M}_\mathrm{t}=(\mathcal{N},\mathcal{I}_\mathrm{p}\cap\mathcal{I}_\mathrm{u})$.
\end{example}

The matroid-constrained submodular optimization problem
\begin{equation}
  \maximize_{\mathcal{X}\subseteq\mathcal{N}} f(\mathcal{X})\quad \text{subject to}\quad \mathcal{X}\in\bigcap_{i=1}^T\mathcal{I}_i\label{eq:submodular_matroid}
\end{equation}
can be solved near optimally using Algorithm~\ref{alg:greedy_matroid}. This result is formally stated in the following theorem.

\begin{algorithm}[t]
\begin{algorithmic}[1]
\Require{$\mathcal{X}=\varnothing$, $K$, $\{\mathcal{I}_i\}_{i=1}^T$}
\For{$k\gets 1$ \textbf{to} $K$}
\State $s^\star=\argmax_{s\notin\mathcal{X}}\{f(\mathcal{X}\cup\{s\}):\mathcal{X}\cup\{s\}\in\bigcap_{i=1}^T\mathcal{I}_i\}$
\State $\mathcal{X}\leftarrow\mathcal{X}\cup\{s^\star\}$
\EndFor
\State\Return $\mathcal{X}$
\end{algorithmic}
\caption{Greedy maximization subject to $T$ matroid constraints}
\label{alg:greedy_matroid}
\end{algorithm}

\begin{theorem}[Matroid-constrained submodular maximization\cite{submodular_2}]\label{thm:submodular_matroid}
  Let $f:2^\mathcal{N}\rightarrow\R$ be a monotone non-decreasing, normalized, submodular set function, and $\{\mathcal{M}_i=(\mathcal{N},\mathcal{I}_i)\}_{i=1}^T$ be a set of matroids defined over $\mathcal{N}$. Furthermore, let $\mathcal{X}^\star$ denote the optimal solution of \eqref{eq:submodular_matroid}, and let $\mathcal{X}_\text{greedy}$ be the solution obtained by Algorithm~\ref{alg:greedy_matroid}. Then
  \begin{equation*}
    f(\mathcal{X}_\text{greedy})\geq \cfrac{1}{T+1}\;f(\mathcal{X}^\star).
  \end{equation*}
\end{theorem}

\subsection{Greedy method}\label{sec:dense_fp}

The frame potential \cite{mse_fp} of the matrix $\matx{\Psi}$ is defined as the trace of the Grammian matrix

$
  \FP{\matx{\Psi}}\coloneqq\tr{\matx{T}^H\matx{T}}
$

with $\matx{T}=\matx{\Psi}^H\matx{\Psi}$. The frame potential can be related to the MSE,

$
  \MSE(\matx{\Psi}(\mathcal{L}))=\tr{\matx{T}^{-1}(\mathcal{L})},
 $

using~\cite{frame_potential}
\begin{equation}
  c_1\cfrac{\FP{\matx{\Psi}(\mathcal{L})}}{\lambda_\mathrm{max}^2\{\matx{T}(\mathcal{L})\}}\leq\MSE(\matx{\Psi}(\mathcal{L}))\leq c_2\cfrac{\FP{\matx{\Psi}(\mathcal{L})}}{\lambda_\mathrm{min}^2\{\matx{T}(\mathcal{L})\}},\label{eq:MSE_bound}
\end{equation}
where $c_1$, and $c_2$ are constants that depend the data model.

From the above bound, it is clear that by minimizing the frame potential of $\matx{\Psi}$ one can minimize the MSE, which is otherwise difficult to minimize as it is neither convex, nor submodular.

The frame potential of $\matx{\Psi}(\mathcal{L})\coloneqq\matx{\Psi}_1(\mathcal{L}_1)\otimes\dots\otimes\matx{\Psi}_R(\mathcal{L}_R)$  can be expressed as the frame potential of its factors
$\matx{\Psi}_i(\mathcal{L}_i)\coloneqq\matx{\Phi}_i(\mathcal{L}_i)\matx{U}_i$. To show this, recall the definition of the frame potential as
\begin{align*}
  \FP{\matx{\Psi}(\mathcal{L})}&=\tr{\matx{T}^H(\mathcal{L})\matx{T}(\mathcal{L})}\\

  &=\tr{\matx{T}_1^H\matx{T}_1\otimes\dots\otimes\matx{T}_R^H\matx{T}_R},\numberthis\label{eq:kron_fp_incomplete}
\end{align*}
where $\matx{T}_i=\matx{\Psi}_i^H\matx{\Psi}_i$.
Now, using the fact that for any two matrices $\matx{A}\in\C^{K_A\times K_A}$ and $\matx{B}\in\C^{K_B\times K_B}$ we have $\tr{\matx{A}\otimes\matx{B}}=\tr{\matx{A}}\tr{\matx{B}}$, we can expand \eqref{eq:kron_fp_incomplete} as
\begin{equation*}
  \FP{\matx{\Psi}(\mathcal{L})}=\prod_{i=1}^R\tr{\matx{T}_i^H\matx{T}_i}=\prod_{i=1}^R \FP{\matx{\Psi}_i(\mathcal{L}_i)}.
\end{equation*}
For brevity, we will write the above expression alternatively as an explicit function of the selection sets $\mathcal{L}_i$:
\begin{align}
  F(\mathcal{L})&\coloneqq\FP{\matx{\Psi}(\mathcal{L})}=\prod_{i=1}^RF_i(\mathcal{L}_i) \coloneqq \prod_{i=1}^R \FP{\matx{\Psi}_i(\mathcal{L}_i)}\label{eq:kron_fp}
\end{align}
Expression \eqref{eq:kron_fp} shows again the advantage of working with a Kronecker-structured sampler: instead of computing every cross-product between the columns of $\matx{\Psi}$ to compute the frame potential, we can arrive to the same value using the frame potential of $\{\matx{\Psi}_i\}_{i=1}^R$.

\subsubsection{Submodularity of $F(\mathcal{L})$}

Function $F(\mathcal{L})$ as defined in \eqref{eq:kron_fp} does not directly meet the conditions [cf. Theorem~\ref{thm:submodular_matroid}] required for near optimality of the greedy heuristic, but it can be modified slightly to satisfy them. In this sense, we define the function $G:2^\mathcal{N}\rightarrow\R$ on the subsets of $\mathcal{N}$ as
\begin{equation}
\label{eq:set_func}
  G(\mathcal{S})\coloneqq F(\mathcal{N})- F(\mathcal{N}\setminus\mathcal{S})

\end{equation}
where recall that $F(\mathcal{N}) = \prod_{i=1}^R F_i(\mathcal{N}_i)$, $F(\mathcal{N}\setminus\mathcal{S}) = \prod_{i=1}^RF_i(\mathcal{N}_i\setminus\mathcal{S}_i)$, and
$
  \mathcal{S}=\bigcup_{i=1}^R\mathcal{S}_i, \quad \mathcal{S}_i\cap\mathcal{S}_j=\varnothing\text{ for }i\neq j$. Therefore, $\{\mathcal{S}_i\}_{i=1}^R$ form a partition of $\mathcal{S}$.

It is clear that if we make the change of variables from $\mathcal{L}$ to $\mathcal{S}$ maximizing $G$ over $\mathcal{S}$ is the same as minimizing the frame potential over $\mathcal{L}$. However, working with the complement set results in a set function that is submodular and monotone non-decreasing, as shown in the next theorem. Consequently, $G$ satisfies the conditions of the near-optimality theorems.
\begin{theorem}\label{thm:submodular_G}
  The set function $G(\mathcal{S})$ defined in \eqref{eq:set_func} is a normalized, monotone non-decreasing, submodular function for all subsets of $\mathcal{N}=\bigcup_{i=1}^R\mathcal{N}_i$.
\end{theorem}
\begin{proof}
See Appendix~\ref{ap:proof_G}.
\end{proof}

With this result we can now claim near-optimality of the greedy algorithm that solves the cardinality constrained maximization of $G(\mathcal{S})$. However, as we said, minimizing the frame potential only makes sense as long as \eqref{eq:MSE_bound} is tight. In particular, whenever $\matx{T}(\mathcal{L})$ is singular we know that the MSE is infinity, and hence \eqref{eq:MSE_bound} is meaningless. For this reason, next to the cardinality constraint in \eqref{eq:sparsesensing} that limits the total number of sensors, we need to ensure that $\matx{\Psi}(\mathcal{L})$ has full column rank, i.e., $L_i\geq K_i$ for $i=1,\dots,R$. In terms of $\mathcal{S}$, this is equivalent to
\begin{equation}
  |\mathcal{S}_i|=|\mathcal{N}_i\setminus\mathcal{L}_i|\leq N_i-K_i\quad i=1,\dots,R,\label{eq:identifiability_constraint}
\end{equation}
where this set of constraints forms a partition matroid $\mathcal{M}_p=(\mathcal{N},\mathcal{I}_p)$ [cf. Example~\ref{ex:partition_matroid} from Definition~\ref{def:matroid}]. Hence, we can introduce the following submodular optimization problem as surrogate for the minimization of the frame potential
   \begin{equation}
   \label{eq:submodular_max_kron_matroid}
    \maximize_{\mathcal{S}\subseteq\mathcal{N}} \, G(\mathcal{S}) \,\,\, \text{s. to} \,\,\,  \mathcal{S} \in \mathcal{I}_\mathrm{u}\cap\mathcal{I}_p
   \end{equation}

with $\mathcal{I}_\mathrm{u}=\{\mathcal{A}\subseteq\mathcal{N}:|\mathcal{A}|\leq N-L\}$ and $\mathcal{I}_p=\{\mathcal{A}\subseteq\mathcal{N}:|\mathcal{A}\cap\mathcal{N}_i|\leq N_i-K_i\;\;i=1,\dots,R\}$.
Theorem~\ref{thm:submodular_matroid} gives, therefore, all the ingredients to assess the near-optimality of Algorithm~\ref{alg:greedy_matroid} applied on \eqref{eq:submodular_max_kron_matroid}, for which the results are particularized as the following corollary.

\begin{corollary}\label{corollary:bound_G}
  The greedy solution $\mathcal{S}_\text{greedy}$ to \eqref{eq:submodular_max_kron_matroid} obtained from Algorithm~\ref{alg:greedy_matroid}) is $1/2$-near-optimal, i.e.,
$G(\mathcal{S}_\text{greedy})\geq \frac{1}{2}G(\mathcal{S}^\star).$
\end{corollary}

\begin{proof}
  Follows from Theorem~\ref{thm:submodular_matroid}, and since \eqref{eq:submodular_max_kron_matroid} has $T=1$ (truncated-partition) matroid constraint.
\end{proof}

Next, we compute an explicit bound with respect to the frame potential of $\matx{\Psi}$, which is the objective function we initially wanted to minimize. This bound is given in the following theorem.

\begin{theorem}\label{thm:approx_fp}
  The greedy solution $\mathcal{L}_\text{greedy}$ to \eqref{eq:submodular_max_kron_matroid} obtained from Algorithm~\ref{alg:greedy_matroid} is near optimal with respect to the frame potential as
$ F(\mathcal{L}_\text{greedy})\leq \gamma F(\mathcal{L}^\star)$
  with $\gamma=\cfrac{1}{2}\left(\cfrac{K}{L_{\text{min}}^2}\prod_{i=1}^R F_i(\mathcal{N}_i)+1\right)$, and $L_\text{min}=\min_{i\in\mathcal{L}}\norm{\vec{u}_i}^2_2$, being $\vec{u}_i$ the $i$th row of $(\matx{U}_1\otimes\dots\otimes\matx{U}_R)$.
\end{theorem}
\begin{proof}
Obtained similar to the bound in \cite{frame_potential}, but specialized for \eqref{eq:kron_fp} and 1/2-near-optimality; see~\cite{thesis} for details.

\end{proof}

As with the $R=1$ case in~\cite{frame_potential}, $\gamma$ is heavily influenced by the frame potential of $(\matx{U}_1\otimes\dots\otimes\matx{U}_R)$. Specifically, approximation gets tighter when $F(\mathcal{N})$ is small or the core tensor dimensionality decreases.

\subsubsection{Computational complexity}

The running time of Algorithm~\ref{alg:greedy_matroid} applied to solve \eqref{eq:submodular_max_kron_matroid} can greatly be reduced by precomputing the inner products between the rows of every $\matx{U}_i$ before starting the iterations. This has a complexity of $\mathcal{O}(N_i^2K_i)$ for each domain. Once these inner products are computed, in each iteration we need to find $R$ times the maximum over $\mathcal{O}(N_i)$ elements. Since we run $N-L$ iterations, the complexity of all iterations is $\mathcal{O}(N^2_\text{max})$, with $N_\text{max}=\max_i N_i$. Therefore, the total computational complexity of the greedy method is $\mathcal{O}(N_\text{max}^2K_\text{max})$ with $K_\text{max}=\max_i K_i$.

\subsubsection{Practical considerations}\label{sec:practical_fp}

Due to the characteristics of the greedy iterations, the algorithm tends to give solutions with a very unbalanced cardinality. In particular, for most situations, the algorithm chooses one of the domains in the first few iterations and empties that set till it hits the identifiability constraint of that domain. Then, it proceeds to another domain and empties it as well, and so on.  This is due to the objective function, which is a product of smaller objectives. Indeed, if we are asked to minimize a product of two elements by subtracting a value from them, it is generally better to subtract from the smallest element. Hence, if this minimization is performed multiple times we will tend to remove always from the same element.

The consequences of this behavior are twofold. On the one hand, this greedy method tends to give a sensor placement that yields a very small number of samples $\tilde{L}$, as we will also see in the simulations. Therefore, when comparing this method to other sensor selection schemes that produce solutions with a larger $\tilde{L}$ it generally ranks worse in MSE for a given $L$. On the other hand, the solution of this scheme tends to be tight on the identifiability constraints for most of the domains, thus hampering the performance on those domains. This implication, however, has a simple solution. By introducing a small slack variable $\alpha_i>0$ to the constraints, we can obtain a sensor selection which is not tight on the constraints. This amounts to solving the problem
\begin{align*}
  &\maximize_{\mathcal{S}\subseteq\mathcal{N}}  G(\mathcal{S})\numberthis\\
  &\text{s. to } \quad |\mathcal{S}|= N-L,
   |\mathcal{S}\cap\mathcal{N}_i|\leq N_i-K_i-\alpha_i,  i=1,\dots,R.
\end{align*}
Tuning $\{\alpha_i\}_{i=1}^R$ allows to regularize the tradeoff between compression and accuracy of the greedy solution.

We conclude this section with a remark on an alternative performance measure.
\begin{remark} As a alternative performance measure, one can think on maximizing the set function $\log \det{\matx{T}(\mathcal{L})}$. Although this set function can be shown to be submodular over all subsets of $\cal{N}$~\cite{thesis}, the related greedy algorithm cannot be  constrained to always result in an identifiable system after subsampling. Thus, its applicability is more limited than the frame potential formulation; see~\cite{thesis} for a more detailed discussion.
\end{remark}


TITLE: Using the law of large numbers to get a dominating sequence
QUESTION [1 upvotes]: Let $X_1, X_2, \dots $ be an iid sequence of random variables such that $EX_1 = \mu$, $E \vert X_1\vert^p < \infty$ for some $1<p<2$ and $E \log X_1 < 0$. The goal is to compute
\begin{equation}
E(S_\infty) = E \left( \sum_{n=1}^\infty \prod_{k=1}^n X_k \right).
\end{equation}
My idea was to consider
\begin{align}
\prod_{k=1}^n X_k = & \exp \left( \sum_{k=1}^n \log X_k \right) \\
= & \exp \left( (\sum_{k=1}^n \log X_k ) / n \right)^n \\
= & \exp \left( E \log X_1 + o (n^{1/p -1} \right)^n \\
= & \exp \left(n E \log X_1 + o (n^{1/p} \right) \\
\leq & \exp (-k \epsilon)
\end{align}
for some $\epsilon >0$. I here used Theorem 2.5.12 from Durrett's book Probability: Theory and Examples. Then I got a dominating sequence that converges so I could use dominated convergence to take the exptecation inside the sum in above equation to get
\begin{equation}
E (S_\infty) = \sum_{n=1}^\infty (EX_1)^n
\end{equation}
Is this correct? The part with the dominating sequence feels weird and the conclusion too... I am fairly sure that this is false but I can't find my mistake.
Here a screenshot of the Theorem

REPLY [1 votes]: If the random variables are non-negative (which is suggested by the fact you are considering logarithms), then you can always put the expectation sign under the sum thanks to Tonelli's theorem, in particular, you don't have ot assume that the logarithm has negative expectation. In the case $\mathrm E[X_1]\ge 1$, the series is not integrable (and in the case where the the logarithm has positive expectation it even diverges almost surely), but then the both sides of the equality are infinite.


TITLE: How does temperature constant in isothermal process?
QUESTION [0 upvotes]: 1 mole of an ideal gas (at 25°C and 1 atm pressure) is placed in a cylinder with a piston. Ambient temperature is 25°C and atmospheric pressure is 1 atm. Now if heat is applied to the system (ideal gas) in an isothermal process, its temperature will not rise but pressure and volume will change. How is this possible? Why does it keep the temperature constant instead of pressure? Is this isothermal process actually possible?

REPLY [1 votes]: The gas pushes the piston outwards, doing work against the piston. In an isothermal process for an ideal gas the work is done by the gas at the same rate that heat flows into the gas, so, according to the first law of thermodynamics:
Gain in internal energy = Heat flow into gas – Work done by gas,
there will be no change in the internal energy, $U$, of the gas and (since the gas is ideal and $U=nc_vT$) no change in its temperature.
On a molecular scale, the heat flow into the gas occurs because gas molecules hitting the cylinder wall bounce back with, on average, more kinetic energy than on approach, owing to the enhanced vibration of the  molecules of the slightly higher temperature wall. But gas molecules hitting the piston that is moving outwards bounce back with less kinetic energy than on approach.
"Is this isothermal process actually possible?" For the process to be reversible, the temperature difference between the cylinder walls and the gas must approach zero, and the speed of motion of the piston must approach zero, so reversible isothermal expansion is indeed an ideal, and is not strictly practicable.


TITLE: Resources to investigate rational numbers
QUESTION [2 upvotes]: I have been told that resources like Mathematica's Number Recognition (which I've never tried myself) and the Inverse Symbolic Calculator  (ISC)  can be used to find possible closed forms for real numbers (with many known decimal digits).
But I would like to know if there are similar resources (preferably free ones) that work for rational numbers or integers (even large integers). In particular, I'm searching for some program which can:

find "complex" ways to express a number (for example, by using particular constants or functions);
find expressions of a number in terms of sums of squares, squares roots, cube, cube roots, sum of prime numbers, and other things.

Note: I know that Wolfram Alpha can do such things, but the limited computation time doesn't allow to have many results.

REPLY [1 votes]: For example, what is the output given 133245?
$ 168^2+179^2+188^2+194^2=133245 $
$ 6^3+13^3+18^3+50^3=133245 $
Those are just two examples of what can be done but I used Mathematica instead.