TITLE: Onto function and Power set cardinality.
QUESTION [2 upvotes]: I need some hints for proving that if $f:A\to B$ is onto $B$, then $P(B)\leq P(A)$. And  $|B|\leq |A|$ under axiom of choice.
Thank you!

REPLY [3 votes]: Define $\hat f\colon P(B)\to P(A)$ by setting $\hat f(X)=\{a\in A\mid f(a)\in X\}$. You can show this is an injection.
If the axiom of choice holds simply choose from $\hat f(\{b\})$ to construct the injection from $B$ to $A$.


TITLE: how to find polynominal equation with roots and canonical form
QUESTION [0 upvotes]: Problem
I had to solve a math problem that looks like this:
Find the polynomial equation of degree $2$ for which the roots are $0$ and $4$ and the canonical form is $a(x - 2)^2 + 3.$
Then i got stuck in this equation and couldn't go any further:

$a(x - 2)^2 + 3 = a(x(x-4))$
$a(x^2 - 4x + 4) + 3 = a(x^2+4x)$

REPLY [1 votes]: Plug the roots in the canonical form:
$$a(0-2)^2+3=a(4-2)^2+3=0$$
and solve for $a$.
Even though there is a single parameter but two conditions, there is a solution.


TITLE: Showing the support of a sheaf may not be closed (Liu 2.5)
QUESTION [9 upvotes]: This is question 2.5 of Qing Liu.
I am new in algebraic geometry and really stuck on it and can't do anything to solve it.
The question:
Let $F$ be a sheaf on $X$. Let $\operatorname{Supp} F=\{x\in X:F_x\neq 0\}$. We want to show that in general, $\operatorname{Supp} F$ is not a closed subset of $X$. Let us fix a sheaf $G$ on $X$ and a closed point $x_0\in X$. Let us define a pre-sheaf $F$ by $F(U)=G(U)$ if $x_0\notin U$ and $F(U)= \{s\in G(U):s_{x_0}=0\}$ otherwise. Show that $F$ is a sheaf and that $\operatorname{Supp} F = \operatorname{Supp}G\setminus \{x_0\}$.
I don't know how to solve this question:
To show a pre-sheaf is a sheaf I need to check the "uniqueness" and "gluing local sections".
For the uniqueness: Let $U$ be an open subset of $X$ , $s\in F(U)$, if $x_0\notin U$ , then since $G$ is a sheaf, I don't see a problem for $F$ to be a sheaf.
If $s\in F(U)$ and $x_0 \in U$ and $\{U_i\}_i$ be an open covering of $U$,then there exists an $i_0$ such that $x_0\in U_{i_0}$. the image of $s$ in the stalk $F_{x_0}$ is $s_{x_0}$. $F(U_{i_0})=\{s\in G(U):s_{x_0}=0\}$ by definition. I don't know what to do now? (so sorry and I know this is an easy question...)

REPLY [5 votes]: Side note: a presheaf $F$ has the same stalks as its sheafification $F^\#$. Thus, $\operatorname{supp}(F) = \operatorname{supp}(F^\#)$ and it suffices to find a presheaf $F$ such that $\operatorname{supp}(F)$ is not closed, if all you want to know is that the support of a sheaf may or may not be closed. This doesn't matter here because the problem tells you specifically to prove $F$ is a sheaf, but it's good to know this trick.

$\DeclareMathOperator{res}{res}$
The first thing we need to do when checking that $F$ is a sheaf is to keep in mind the entire datum of the presheaf $F$; not just the objects $F(U)$ for $U$ an open subset of $X$, but also the restriction morphisms $\res_{U, V} : F(U) \to F(V)$ when $V \subseteq U$. In this case, $\res_{U,V}$ is just the restriction (haha) of the restriction morphism $G(U) \to G(V)$ (which we check is well-defined). In this situation (where $F(U)$ is always a subset of $G(U)$ and the restriction morphisms of $F$ are induced by those of $G$), $F$ is called a subpresheaf of $G$. Nothing crazy going on here, but it's important to fully understand the object you're working with. For convenience, I will write $\sigma|_V$ to mean $\res_{U,V}(\sigma)$, since $U$ can always be determined from context.
Next, as you said, we should check "uniqueness". Remember, the uniqueness axioms says that for all open covers $\{U_i\}_{i \in I}$ of an open set $U \subseteq X$ and all $\sigma \in F(U)$, if $\sigma|_{U_i} = 0$ for all $i \in I$, then $\sigma = 0$. We should try to prove this in the most straightforward possible way:
Let $\{U_i\}_{i \in I}$ be an open cover of an open set $U \subseteq X$. Let $\sigma \in F(U)$  be arbitrary. Suppose that $\sigma|_{U_i} = 0$ for all $i \in I$. Since $F$ is a subpresheaf of $G$, we have in particular that $\sigma \in G(U)$ and $\sigma|_{U_i} = 0 \in G(U_i)$ for all $i \in I$. Since $G$ is assumed to be a sheaf, we must have $\sigma = 0$, as desired.
Indeed, this very simple argument shows that any subpresheaf of a sheaf is separated (a.k.a. satisfies the uniqueness axiom): no need to mention anything about $x_0$!
Now you just need to check gluing, which I'll omit because this proof appears in other answers.


TITLE: How to compute the projection of a vector on a plane
QUESTION [1 upvotes]: Can someone check whether my work is correct or not?

Compute the projection of $(1,1,1)$ onto the plane that passes through the points $(1,0,-1), (3,7,-3), (-2,-1,2)$.

My attempt: Let $u = (1,1,1)$. I first try to find a vector normal to the plane given by the three points. By  computing a cross product, I got $n = (19,0,19) = (1,0,1)$.
The projection of $u$ onto the plane is  $$\frac{u.n}{||n||^2}u.$$ Is that correct?

REPLY [1 votes]: A plane is of the form $a_1 x_1 + a_2 x_2 + a_3 x_3 = b$. Since we are given three points on the plane, we obtain a system of $3$ linear equations in $a_1, a_2, a_3, b$. One solution would be $a_1 = a_3 = 1$, $a_2 = 0$ and $b = 0$. Hence, an equation of the plane that passes through the three given points is
$$x_1 + x_3 = 0$$
though any multiple of this equation would also do. Note that this plane also passes through the origin. The vector that is normal to this plane is
$$\mathrm{n} = \begin{bmatrix}1\\ 0 \\ 1\end{bmatrix}$$
We can decompose the point to be projected, $\mathrm{u}$, as the sum of a vector normal to the plane and a vector parallel to the plane
$$\mathrm{u} = \begin{bmatrix}1\\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}1\\ 0 \\ 1\end{bmatrix} + \begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix} = \mathrm{n} + \begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix}$$
Projecting $\mathrm{u}$ onto the plane we obtain only
$$\begin{bmatrix} 0\\ 1 \\ 0\end{bmatrix}$$
as the projection of a vector normal to the plane onto the same plane is the zero vector.


TITLE: Which Has a Larger Volume a Cylinder or a Truncated Cone?
QUESTION [3 upvotes]: The following link describes a programming problem. However, I am unable to work out the maths for the problem.

We are given $r$ – radius of lower base and $s$ – slant height. The figure can be cylinder or truncated cone. You have to find as largest volume as possible to carry oil respect to given information.
You are given two numbers that less than 100: radius $r$ of lower base and slant height $s$. The slant height is the shortest possible distance between the edges of two bases.

EDIT suggested :
Given $ \sqrt{(R-r)^2 + h^2 }$ and $ r,$  find $ h/r $ ratio.
$~~~~~~~~~~~~~~~~$

REPLY [0 votes]: If the constant slant height and base radius are $s, r$ respectively and the variable angle which the slanting face makes with the vertical is $\theta$ then the height and top radius of the frustum are
$$h=s\cos\theta, R=r+s\sin\theta$$
The volume of the frustum is $$V=\frac13 \pi h (R^2+Rr+r^2)$$
Now $$\frac{dh}{d\theta}=-s\sin\theta, \frac{dR}{d\theta}=s\cos\theta$$
Volume is maximum when
$$\frac{dV}{d\theta}=\frac13 \pi [\frac{dh}{d\theta}(R^2+Rr+r^2)+h(2R\frac{dR}{d\theta}+r\frac{dR}{d\theta})=0$$
$$-s\sin\theta (R^2+Rr+r^2)+s\cos\theta(2Rs\cos\theta+rs\cos\theta)=0$$
$$-s\sin\theta (3r^2+3rs\sin\theta+s^2\sin^2\theta)+s^2(1-\sin^2\theta)(3r+2s\sin\theta)=0$$
$$- (3r^2\sin\theta+3rs\sin^2\theta+s^2\sin^3\theta)+(3rs+2s^2\sin\theta-3rs\sin^2\theta-2s^2\sin^3\theta)=0$$
$$3x+(2-3x^2)y-6xy^2-3y^3=0$$
where $x=r/s$ and $y=\sin\theta$.
In general the cubic equation will have to be solved numerically. In the special case $s=r$ so that $x=1$ we have $3-y-6y^2-3y^3=0$. The only real root is $y=\sin\theta\approx 0.56298, \cos\theta\approx 0.82647$. Then the maximum volume is $V\approx 4.33$ as found in the link. Whereas for the cylinder $V=\pi\approx 3.14$.


TITLE: Showing the boundary of a cylinder has Lebesgue-measure zero
QUESTION [3 upvotes]: Let $\partial C=\{(x,y,z) \in \Bbb{R^3}\enspace\vert  \enspace x^2+y^2 = r, \enspace0\leq z\leq h\}.$ I'm trying to show $\partial C$ has Lebesgue-measure zero in $\Bbb{R^3}$.
Let $f: I \to \Bbb{R^2}$, s.t $f=(r\cos x, r\sin x)$, where $I=[0,2\pi]$. Now $f$ is Riemann-integrable and its graph surface $G_f = B(0,r)\subset\Bbb{R^2}.$
Therefore $m_n(G_f)=0$. Then $\partial C=[0,h] \times G_f$. This cartesian product has Lebesgue measure zero, since $[0,h]$ is bounded in $\Bbb{R^2}$. Therefore $m_n(\partial C)=0$. Is this right?

REPLY [0 votes]: Note that $\partial C = \{(x,y)| x^2+y^2 = r \} \times [0,h]$, so
$m_3 (\partial C) = h \cdot m_2 \{(x,y)| x^2+y^2 = r \}$. Hence it suffices to
show that the measure of the circle $rS^1$ is zero.
Note that for fixed $x$ the indicator $y \mapsto 1_{r S^1} ((x,y))$ is zero except for at most 2 points. Hence $\int 1_{S^1} = \int_{x \in \mathbb{R}} \int_{y \in \mathbb{R}} 1_{r S^1} ((x,y)) dy dx = \int_{x \in \mathbb{R}} 0y dx = 0$