---
title: "Algebraic Geometry Problem Sets"
subtitle: "Problem Set 3"
author:
- name: D. Zack Garza
  affiliation: University of Georgia 
  email: dzackgarza@gmail.com 
date: Spring 2022
order: 3
---

# Problem Set 3

## Problem 1

:::{.problem title="Problem 1"}
Let $I$ be an index category, $\mathcal{A}$ an abelian category, and $\mathcal{A}^{I}$ be the category of functors $F: I \rightarrow \mathcal{A}$. Prove that the functor
\[
\cocolim_{i \in I}: \mathcal{A}^{I} \rightarrow \mathcal{A}, \quad F \mapsto \cocolim_{i \in I} F_{i}
\]
is left exact. (By duality, the functor $\colim_{i \in I}$ is right exact.)

What is this functor in the case when $I$ is a poset and $F_{i}$ is a collection of stalks on the space $X=I$ with poset topology?
:::

:::{.solution title="Part 1"}
It suffices to show that $\cocolim_{i\in I}$ is a right adjoint functor, and right adjoints are left exact by general homological algebra.

:::{.claim}
There is an adjunction
\[
\adjunction{\Delta}{\cocolim_{i\in I}}{\cat A}{\cat{A}^{\cat I}}
,\]
where $\Delta$ is the diagonal functor:
\[
\Delta: \cat{A} &\to \cat{A}^{\cat I} \\
X &\mapsto \Delta_X \\
(X \mapsvia{f} Y) &\mapsto (\Delta_X \mapsvia{\eta_f } \Delta_Y)
\]
where 

- The constant functor $\Delta_X: \cat{I} \to \cat{C}$ is defined on objects $i\in \cat I$ as $\Delta_X(i) \da X$ and on morphisms $i\mapsvia{\iota_{ij}} j$ as $\Delta_f(\iota_{ij}) = X \mapsvia{\id_X} X$.
- $\eta_f$ is a natural transformation of functors with components given by $f$:

\begin{tikzcd}
	{\cat{I}} && {\cat{C}} \\
	i && {\Delta_X(i) } && {\Delta_Y(i)} && X && Y \\
	\\
	j && {\Delta_X(j)} && {\Delta_Y(j)} && X && Y
	\arrow[""{name=0, anchor=center, inner sep=0}, "{\iota_{ij}}"', from=2-1, to=4-1]
	\arrow["{\eta_f(i)}", from=2-3, to=2-5]
	\arrow[""{name=1, anchor=center, inner sep=0}, "{\Delta_X(\iota_{ij})}", from=2-3, to=4-3]
	\arrow["{\eta_f(j)}"', from=4-3, to=4-5]
	\arrow[""{name=2, anchor=center, inner sep=0}, "{\Delta_Y(\iota_{ij})}"', from=2-5, to=4-5]
	\arrow[""{name=3, anchor=center, inner sep=0}, "{\id_X}", from=2-7, to=4-7]
	\arrow["{\id_Y}"', from=2-9, to=4-9]
	\arrow["f", from=2-7, to=2-9]
	\arrow["f", from=4-7, to=4-9]
	\arrow[shorten <=13pt, shorten >=13pt, Rightarrow, no head, from=2, to=3]
	\arrow["{\Delta: \cat{I}\to \cat{C}}", shorten <=14pt, shorten >=14pt, Rightarrow, from=0, to=1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

:::

Why this claim is true: this follows immediately from the fact that there is a natural isomorphism
\[
\Hom_{\cat A}(X, \lim F) \iso \Hom_{\cat{A}^{\cat I}}(\Delta_X, F)
,\]
i.e. maps from an object $X$ into the limit of $F$ are equivalent to natural transformations between the constant functor $\Delta_X$ and $F$.
This follows from the fact that a morphism $X\to \lim F$ in $\cat{A}$ is the data of a family of compatible maps $\ts{f_i}_{i\in \cat I}$ over the essential image of $F$:

\begin{tikzcd}
	{\cat{I}} &&& {\cat{A}} \\
	i &&& {F(i)} \\
	&&&&&& {\lim_i F} &&& X \\
	j &&& {F(j)}
	\arrow[""{name=0, anchor=center, inner sep=0}, from=4-1, to=2-1]
	\arrow[""{name=1, anchor=center, inner sep=0}, from=4-4, to=2-4]
	\arrow[from=3-7, to=2-4]
	\arrow[from=3-7, to=4-4]
	\arrow["{f_i}"{description}, curve={height=12pt}, from=3-10, to=2-4]
	\arrow["{f_j}"{description}, curve={height=-18pt}, from=3-10, to=4-4]
	\arrow["{\exists !}", dashed, from=3-10, to=3-7]
	\arrow["F", shorten <=10pt, shorten >=10pt, Rightarrow, from=0, to=1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJpIl0sWzAsMywiaiJdLFswLDAsIlxcY2F0e0l9Il0sWzMsMSwiRihpKSJdLFszLDMsIkYoaikiXSxbMywwLCJcXGNhdHtBfSJdLFs2LDIsIlxcbGltX2kgRiJdLFs5LDIsIlgiXSxbMSwwXSxbNCwzXSxbNiwzXSxbNiw0XSxbNywzLCJmX2kiLDEseyJjdXJ2ZSI6Mn1dLFs3LDQsImZfaiIsMSx7ImN1cnZlIjotM31dLFs3LDYsIlxcZXhpc3RzICEiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbOCw5LCJGIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjEwLCJ0YXJnZXQiOjEwfX1dXQ==)

On the other hand, a natural transformation $\Delta_X \to F$ is precisely the same data:

\begin{tikzcd}
	{\cat{I}} &&& {\cat{A}} \\
	i &&& {F(i)} &&&&&& {\Delta_X(i) = X} \\
	\\
	j &&& {F(j)} &&&&&& {\Delta_X(j) = X}
	\arrow["g", from=4-1, to=2-1]
	\arrow["{F(g)}", from=4-4, to=2-4]
	\arrow["{f_i}"{description}, curve={height=12pt}, from=2-10, to=2-4]
	\arrow["{F(g) = \id_X}", Rightarrow, no head, from=2-10, to=4-10]
	\arrow["{f_j}"{description}, curve={height=-18pt}, from=4-10, to=4-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwxLCJpIl0sWzAsMywiaiJdLFswLDAsIlxcY2F0e0l9Il0sWzMsMSwiRihpKSJdLFszLDMsIkYoaikiXSxbMywwLCJcXGNhdHtBfSJdLFs5LDEsIlxcRGVsdGFfWChpKSA9IFgiXSxbOSwzLCJcXERlbHRhX1goaikgPSBYIl0sWzEsMCwiZyJdLFs0LDMsIkYoZykiXSxbNiwzLCJmX2kiLDEseyJjdXJ2ZSI6Mn1dLFs2LDcsIkYoZykgPSBcXGlkX1giLDAseyJsZXZlbCI6Miwic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs3LDQsImZfaiIsMSx7ImN1cnZlIjotM31dXQ==)


:::

:::{.solution title="Part 2"}
If $\cat{I} = \Open(X)$ where $X$ is given the order topology and $F: \Open{X} \to \cat{A}$ is a functor specified by stalks, $\lim$ sends $F$ to the universal object $\lim F$ living over the essential image of $F$ in $\cat{A}$:

<!--![](figures/2022-04-01_03-23-40.png)-->

\begin{tikzcd}[column sep=tiny]
  & X &&&& {\cat{I} = \Open(X)} &&& {} && {\cat{A}} \\
  & 3 &&&& {X = \ts{1,2,3}} &&&&& {F(X)} &&& \textcolor{rgb,255:red,214;green,92;blue,92}{\lim F} \\
  &&&&& {} \\
  1 && 2 && {\ts{1, 3}} && {\ts{2, 3}} &&& {F_1} && {F_2} \\
  \\
  &&&&& {\ts{3}} &&&&& {F_3} \\
  &&&&& \emptyset &&&&& 0
  \arrow[from=4-1, to=2-2]
  \arrow[from=4-3, to=2-2]
  \arrow[from=6-6, to=4-5]
  \arrow[from=6-6, to=4-7]
  \arrow[from=4-5, to=2-6]
  \arrow[from=4-7, to=2-6]
  \arrow[from=4-10, to=6-11]
  \arrow[from=4-12, to=6-11]
  \arrow[from=7-6, to=6-6]
  \arrow[from=2-11, to=4-10]
  \arrow[from=2-11, to=4-12]
  \arrow["{F\in \cat{A}^{\cat I}}"', Rightarrow, from=1-6, to=1-11]
  \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=2-14, to=7-11]
  \arrow[from=6-11, to=7-11]
  \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=2-14, to=4-12]
  \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=2-14, to=4-10]
  \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=2-14, to=2-11]
  \arrow[squiggly, from=1-2, to=1-6]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

The object corresponding to global sections $F(X) \in \cat{A}$ seems to also satisfies this universal property, so a conjecture would be that this construction recovers $\lim F \cong F(X) \da \globsec{X; F}$.
:::

## Problem 2
:::{.problem title="Problem 2"}
In the category of abelian groups compute $\operatorname{Tor}_{i}^\ZZ \left(\mathbb{Z}_{n}, M\right)$, the left derived functors of $N \mapsto N \otimes_\ZZ M$.
:::

:::{.solution}
\envlist

:::{.claim}
\[
\Tor^{\ZZ}_1(\ZZ/n\ZZ, M) \cong \ker(M \mapsvia{\times n} M)
\cong \ts{m\in M \st nm = 0_M}
,\]
which is the kernel of multiplication by $n$, and $\Tor^{i>1}_{\ZZ}(\ZZ/n\ZZ, M) = 0$.
:::
Why this is true:
in $\rmod$, free implies flat, and $\Tor$ is balanced and can thus be resolved in either variable, so this can be computed by tensoring a free resolution of $\ZZ/n\ZZ$ and using the long exact sequence in $\Tor$:

\begin{tikzcd}[column sep=small]
	&&& 0 & \ZZ && \ZZ && {\ZZ/n\ZZ} & 0 \\
	\\
	{} &&&& {\ZZ \tensor_\ZZ M \cong M} && {\ZZ \tensor_\ZZ M\cong M} && {\ZZ/n\ZZ \tensor_\ZZ M} & 0 \\
	\\
	&&&& \textcolor{rgb,255:red,214;green,92;blue,92}{\Tor_1^{\ZZ}(\ZZ, \ZZ) = 0} && \textcolor{rgb,255:red,214;green,92;blue,92}{\Tor_1^{\ZZ}(\ZZ, \ZZ) = 0} && {\Tor_1^{\ZZ/n\ZZ}(\ZZ, M)} \\
	\\
	&&&& \textcolor{rgb,255:red,214;green,92;blue,92}{\Tor_2^{\ZZ}(\ZZ, \ZZ) = 0} && \textcolor{rgb,255:red,214;green,92;blue,92}{\Tor_2^{\ZZ}(\ZZ, \ZZ) = 0} && {\Tor_2^{\ZZ/n\ZZ}(\ZZ, M)}
	\arrow[from=1-4, to=1-5]
	\arrow[""{name=0, anchor=center, inner sep=0}, "{\times n}", hook, from=1-5, to=1-7]
	\arrow[two heads, from=1-7, to=1-9]
	\arrow[from=1-9, to=1-10]
	\arrow[from=3-9, to=3-10]
	\arrow[two heads, from=3-7, to=3-9]
	\arrow[""{name=1, anchor=center, inner sep=0}, "{(\times n)\tensor \id_M}", from=3-5, to=3-7]
	\arrow[from=5-9, to=3-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=5-7, to=5-9]
	\arrow[from=7-9, to=5-5]
	\arrow[from=7-7, to=7-9]
	\arrow[from=7-5, to=7-7]
	\arrow["{(\wait)\tensor_\ZZ M}", shorten <=9pt, shorten >=13pt, Rightarrow, from=0, to=1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTYsWzMsMCwiMCJdLFs0LDAsIlxcWloiXSxbNiwwLCJcXFpaIl0sWzgsMCwiXFxaWi9uXFxaWiJdLFs5LDAsIjAiXSxbOCw0LCJcXFRvcl8xXntcXFpaL25cXFpafShcXFpaLCBNKSJdLFs0LDIsIlxcWlogXFx0ZW5zb3JfXFxaWiBNIFxcY29uZyBNIl0sWzYsMiwiXFxaWiBcXHRlbnNvcl9cXFpaIE1cXGNvbmcgTSJdLFs4LDIsIlxcWlovblxcWlogXFx0ZW5zb3JfXFxaWiBNIl0sWzksMiwiMCJdLFswLDJdLFs2LDQsIlxcVG9yXzFee1xcWlp9KFxcWlosIFxcWlopID0gMCIsWzAsNjAsNjAsMV1dLFs0LDQsIlxcVG9yXzFee1xcWlp9KFxcWlosIFxcWlopID0gMCIsWzAsNjAsNjAsMV1dLFs0LDYsIlxcVG9yXzJee1xcWlp9KFxcWlosIFxcWlopID0gMCIsWzAsNjAsNjAsMV1dLFs2LDYsIlxcVG9yXzJee1xcWlp9KFxcWlosIFxcWlopID0gMCIsWzAsNjAsNjAsMV1dLFs4LDYsIlxcVG9yXzJee1xcWlovblxcWlp9KFxcWlosIE0pIl0sWzAsMV0sWzEsMiwiXFx0aW1lcyBuIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMyw0XSxbOCw5XSxbNyw4LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbNiw3LCIoXFx0aW1lcyBuKVxcdGVuc29yIFxcaWRfTSJdLFs1LDZdLFsxMiwxMV0sWzExLDVdLFsxNSwxMl0sWzE0LDE1XSxbMTMsMTRdLFsxNywyMiwiKFxcd2FpdClcXHRlbnNvcl9cXFpaIE0iLDAseyJzaG9ydGVuIjp7InNvdXJjZSI6MjAsInRhcmdldCI6MzB9fV1d)

In the resulting long exact sequence, since $\ZZ$ is free, thus flat, thus tor-acyclic, the first two columns vanish in degrees $d\geq 1$.
As a result, in degrees $d\geq 2$, the terms $\Tor_d^\ZZ(\ZZ/n\ZZ, M)$ are surrounded by zeros and thus zero, meaning that only $\Tor_1$ survives.
By exactness, $\Tor_1(\ZZ/n\ZZ, M)$ is isomorphic to the kernel of the next map in the sequence, which is precisely $\ker(M \mapsvia{\times n} M)$ after applying the canonical isomorphism
\[
\ZZ \tensor_\ZZ M &\to M \\
n \tensor m &\mapsto nm
.\]
:::

## Problem 3

:::{.problem title="Problem 3"}
Let $k$ be a field and $R=k[x, y]$. In the category of $R$-modules compute 

- $\operatorname{Ext}_R^{n}(R, m)$
- $\operatorname{Ext}_R^{n}(m, R)$, and 
- $\operatorname{Tor}^R_{n}(m, m)$, 

where $m=(x, y)$ is the maximal ideal at the origin.

:::

:::{.solution title="Problem 3"}
Note that $R$ is a free $R\dash$module, and so $\Ext_R^n(R, M) = 0$ for any $R\dash$module $M$.
This is because $\Ext$ can be computed using a free resolution of either variable.
For $\Ext_R^n(R, m)$, compute this as $\RR\Hom_R(\wait, m)$ evaluated at $R$. 
Take the free resolution
\[
\cdots \to 0 \to R \mapsvia{\id_R} R \to 0
,\]
delete the augmentation and apply the contravariant $\Hom_R(\wait, m)$ to obtain
\[
0 \to \Hom_R(R, m) \cong m \to 0 \to \cdots
,\]
and take homology to obtain 
\[
\Ext_R^0(R, m) \cong m, \qquad \Ext_R^{>0}(R, m) = 0
.\]

Compute $\Ext_R(m, R)$ as $\RR\Hom(m, \wait)$ applied to $R$ proceeds similarly: using the same resolution, applying covariant $\Hom_R(m, \wait)$ yields
\[
0 \to \Hom_R(m, R) \to 0 \to \cdots
,\]
and taking homology yields
\[
\Ext_R^0(m, R) \cong \Hom_R(m, R) \qquad \Ext_R^{>0}(m, R) = 0
.\]



For the $\Tor$ calculation, we can use the Koszul resolution of $m$:
\[
0 \to k[x, y] \mapsvia{\cdot \tv{x, y} } 
k[x, y] \oplus k[x,y] \mapsvia{ \cdot \transp{ \tv{-y, x}} } \gens{x, y} \to 0
,\]
so the differentials are $t\mapsto \tv{tx, ty}$ and $\tv{u, v} \mapsto -uy + vx$ respectively.
More succinctly, this resolution is 
\[
0 \to R \mapsvia{d_1} R\sumpower{2} \mapsvia{d_2}  m \to 0
,\]
so we can delete $m$ and apply $(\wait)\tensor_R m$ to obtain
\[
0 \to R\tensor_R m \mapsvia{d_1 \tensor \id_m}  R\sumpower{2}\tensor_R m \to 0\\
\]
which simplifies to
\[
\complex{C}\da 0 \to m \mapsvia{\tilde d_1 \da  \tv{x, y}}  m \oplus m \to 0 \\
\]
and thus we can compute $\Tor$ as the homology of this complex.
We have
\[
\Tor_0^R(m,m)&=
H^0(\complex C) \\
&= \coker \tilde d_1 \\
&= {m \oplus m \over xm \oplus ym} \\
&\cong {m\over xm} \oplus {m\over ym} \\
&= {\gens{x, y} \over \gens{x^2, y} } \oplus {\gens{x, y} \over \gens{x, y^2}} \\
&= \ts{f(x, y) \da c_1 x \in k[x,y] \st c_1\in k } \oplus \ts{g(x, y) \da c_1 y \in k[x,y] \st c_1\in k } \\
&\cong k \oplus k
\\ \\
\Tor_1^R(m,m)&=
H^1(\complex C) \\
&= \ker \tilde d_1 \\
&= \ts{t\in \gens{x,y} \st \tv{tx, ty} = \tv{0, 0} } \\
&= 0 \\ \\ \\
\Tor{\geq 2}^R(m,m)&=
H^{\geq 2} (\complex C) \\
&= 0
.\]


:::

## Problem 4

:::{.problem title="Problem 4"}
Let $0 \rightarrow F^{\prime} \rightarrow F \rightarrow F^{\prime \prime} \rightarrow 0$ be a short exact triple of sheaves and assume that $F^{\prime}$ is flasque. Prove that the sequence
\[
0 \rightarrow \Gamma\left(F^{\prime}\right) \rightarrow \Gamma(F) \rightarrow \Gamma\left(F^{\prime \prime}\right) \rightarrow 0
\]
of the spaces of global sections is exact.
:::

:::{.solution title="Using cohomology"}
\envlist

:::{.claim}
Flasque sheaves are $F\dash$acyclic for the functor global sections functor $F(\wait) \da \globsec{X; \wait}$.
:::

:::{.proof title="of claim"}
Proved in class.
:::


Applying the functor $\globsec{X; \wait}$ to the given short exact sequence of sheaves produces a long exact sequence of abelian groups in its right-derived functors. 
Using the claim above, we have $\RR^i \globsec{X; \mcf '} = 0$ for $i\geq 1$, and thus we have the following:

\begin{tikzcd}[column sep=small]
	0 && {\mcf'} && \mcf && {\mcf''} && 0 \\
	\\
	0 && {\globsec{X; \mcf'}} && {\globsec{X; \mcf}} && {\globsec{X; \mcf''}} \\
	\\
	&& {\RR^1 \globsec{X; \mcf'} = 0} && {\RR^1 \globsec{X; \mcf}} && {\RR^1 \globsec{X; \mcf''}} \\
	\\
	&& {\RR^2 \globsec{X; \mcf'} = 0} && {\RR^2 \globsec{X; \mcf}} && \cdots
	\arrow[from=1-1, to=1-3]
	\arrow[""{name=0, anchor=center, inner sep=0}, from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=1-7, to=1-9]
	\arrow[from=3-1, to=3-3]
	\arrow[""{name=1, anchor=center, inner sep=0}, from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=3-7, to=5-3]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-7, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow["\sim", from=7-5, to=7-7]
	\arrow["\sim", from=5-5, to=5-7]
	\arrow["{\globsec{X; \wait}}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTUsWzAsMCwiMCJdLFsyLDAsIlxcbWNmJyJdLFs0LDAsIlxcbWNmIl0sWzYsMCwiXFxtY2YnJyJdLFs4LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiXFxnbG9ic2Vje1xcbWNmJ30iXSxbNCwyLCJcXGdsb2JzZWN7XFxtY2Z9Il0sWzYsMiwiXFxnbG9ic2Vje1xcbWNmJyd9Il0sWzIsNCwiXFxSUl4xIFxcZ2xvYnNlY3tcXG1jZid9ID0gMCJdLFs0LDQsIlxcUlJeMSBcXGdsb2JzZWN7XFxtY2Z9Il0sWzYsNCwiXFxSUl4xIFxcZ2xvYnNlY3tcXG1jZicnfSJdLFsyLDYsIlxcUlJeMiBcXGdsb2JzZWN7XFxtY2YnfSA9IDAiXSxbNCw2LCJcXFJSXjIgXFxnbG9ic2Vje1xcbWNmfSJdLFs2LDYsIlxcY2RvdHMiXSxbMCwxXSxbMSwyXSxbMiwzXSxbMyw0XSxbNSw2XSxbNiw3XSxbNyw4XSxbOCw5XSxbOSwxMF0sWzExLDEyXSxbMTIsMTNdLFsxMywxNCwiXFxzaW0iXSxbMTAsMTEsIlxcc2ltIl0sWzE2LDIwLCJcXGdsb2JzZWN7XFx3YWl0fSIsMCx7InNob3J0ZW4iOnsic291cmNlIjoyMCwidGFyZ2V0IjoyMH19XV0=)

In particular, since $\RR^1 \globsec{X; \mcf'} = 0$, the first row forms the desired short exact sequence.
As a corollary, we also obtain $\RR^i \globsec{X; \mcf} \cong \RR^i \globsec{X; \mcf''}$ for all $i\geq 1$.
:::

:::{.solution title="Direct"}
First, we'll modify the notation slightly and give names to the maps involved.
We'll use the following convention for restrictions of sheaf morphisms to opens and stalks:

\begin{tikzcd}
	0 && A && B && C && 0 \\
	& \textcolor{rgb,255:red,92;green,92;blue,214}{0} && \textcolor{rgb,255:red,92;green,92;blue,214}{A(X)} && \textcolor{rgb,255:red,92;green,92;blue,214}{B(X)} && \textcolor{rgb,255:red,92;green,92;blue,214}{C(X)} \\
	{} & \textcolor{rgb,255:red,92;green,92;blue,214}{0} & {} & \textcolor{rgb,255:red,92;green,92;blue,214}{A(U)} & {} & \textcolor{rgb,255:red,92;green,92;blue,214}{B(U)} && \textcolor{rgb,255:red,92;green,92;blue,214}{C(U)} \\
	\\
	0 && {A_x} && {B_x} && {C_x} && 0
	\arrow[from=1-1, to=1-3]
	\arrow["f", from=1-3, to=1-5]
	\arrow["g", from=1-5, to=1-7]
	\arrow[from=1-7, to=1-9]
	\arrow[from=5-1, to=5-3]
	\arrow["{f_x}", from=5-3, to=5-5]
	\arrow["{g_x}", from=5-5, to=5-7]
	\arrow[from=5-7, to=5-9]
	\arrow[curve={height=-12pt}, from=1-3, to=5-3]
	\arrow[curve={height=-12pt}, from=1-5, to=5-5]
	\arrow[curve={height=-12pt}, from=1-7, to=5-7]
	\arrow[from=1-3, to=2-4]
	\arrow[from=1-5, to=2-6]
	\arrow[from=1-7, to=2-8]
	\arrow["F"{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, from=2-4, to=2-6]
	\arrow["G"{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, from=2-6, to=2-8]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=2-2, to=2-4]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-2, to=3-4]
	\arrow["{\ro{F}{U}}"{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, from=3-4, to=3-6]
	\arrow["{\ro{G}{U}}"{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, from=3-6, to=3-8]
	\arrow[from=2-4, to=3-4]
	\arrow[from=2-6, to=3-6]
	\arrow[from=2-8, to=3-8]
	\arrow[from=3-4, to=5-3]
	\arrow[from=3-6, to=5-5]
	\arrow[from=3-8, to=5-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

Given $c\in C(X)$, our goal is to produce a $b\in B(X)$ such that $g(b) = c$, and the strategy will be to use surjectivity at stalks to produce a maximal section of $B$ mapping to $c$, and argue that it must be a section over all of $X$.
This will proceed by showing that if a lift is not maximal, sections over open sets that are missed can be extended using that $A$ is flasque, contradicting maximality.

Write $\ro{c}{x}$ for the image of $c$ in the stalk $C_x$; by surjectivity of $g_x: B_x \surjects C_x$ we can find a germ $b_x$ with $g_x(b_x) = c_x$.
The germ lifts to some set $U\ni x$ and some $b\in B(U)$ with $b\mapsto \ro{c}{U}$ under $\ro{F}{U}: B(U) \to C(U)$.
So define a poset of all such lifts:
\[
P \da \ts{ (U, b \in B(U)) \st \ro{F}{U}(b) = \ro{c}{U} } \\
\qquad \text{ where } (U_1, b_1) \leq (U_2, b_2) \iff U_1 \subseteq U_2 \text{ and} \ro{b_2}{U_1} = b_1
.\]
As noted above, $P$ is nonempty, and every chain $\ts{(U_i, b_i)}_{i\in I}$ has an upper bound given by $(\tilde U, \tilde b)$ where $\tilde U \da \union_{i\in I} U_i$ and $\tilde b$ is the unique glued section of $B$ restricting to all of the $b_i$, which exists by the sheaf property for $B$.
Thus Zorn's lemma applies, and (reusing notation) we can assume $(U, b)$ is maximal with respect to this property.

The claim is that $U$ must be all of $X$.
Toward a contradiction, suppose not -- then pick any $x\in X\sm U$, and again using surjectivity on stalks at $x$, produce an open set $V\ni x$ and a section $b'\in B(V)$ with $\ro{G}{V}(b') = \ro{c}{V}$.
Now on the overlap $W\da U \intersect V$, both $b$ and $b'$ map to $\ro{c}{W}$, and so
\[
\ro{G}{W}(\ro b W - \ro {b'} W) = \ro{c}W \ro c W = 0 \implies b-b'\in \ker \ro{G}{W} = \im \ro{F}{W}
,\]
where we've used exactness in the middle spot in the exact sequence $A(W) \to B(W)\to C(W)$.
So there is some $\alpha \in A(W)$ with $\ro{F}{W}(\alpha) = \ro b W - \ro{b'}{W}$, and since $A$ is flasque this can be extended to a global section $\tilde\alpha\in A(X)$.
Write $\tilde \beta \da F(\tilde \alpha) \in B(X)$ with $\ro{\tilde \beta}{W} = \ro b W- \ro{b'}{W}$ in $B(W)$.
We can now glue $\tilde \beta$ to a section over $U \union V$ which extends the original section $b$: setting $\hat{b} \da \tilde \beta + b'$ yields
\[
\ro{\hat b}{W} = \qty{\ro{b}{W} - \ro{b'}{W}} + b' = \ro{b}{W}
,\]
so this section over $U \union V$ agrees with $b$ on the overlap $W = U \intersect V$, and thus by existence and uniqueness of gluing (using the sheaf property of $B$) $\hat{b} \in B(U \union V)$ is a section extending $b$ over a set that strictly contains $U$.
This contradicts the maximality of the pair $(U, b)$.
:::

## Problem 5

:::{.problem title="Problem 5"}
For a sheaf $F$ on $X$, let
\[
S(F)=\prod_{x \in X}\left(i_{x}\right)_{*} F_{x}, \quad i_{x}: x \rightarrow X
\]
be the sheaf of all, possibly discontinuous section of the étale space of $F$. The canonical flasque resolution of $F$ is
\[
\underline{S}(F) \da 0 \rightarrow F \to S\left(F_{0}\right) \rightarrow S\left(F_{1}\right) \rightarrow S\left(F_{2}\right) \rightarrow \ldots
\]
where $F_{0}=F$ and $F_{i}$ are defined inductively as $F_{i+1}=S\left(F_{i}\right) / F_{i}$. Some books define cohomology groups $\mathbf{H}^{n}(X, F)$ as the cohomology groups of the complex
\[
0 \rightarrow \Gamma\left(S\left(F_{0}\right)\right) \rightarrow \Gamma\left(S\left(F_{1}\right)\right) \rightarrow \Gamma\left(S\left(F_{2}\right)\right) \rightarrow \ldots
\]
Prove that they coincide with the cohomology defined by other means by showing that this gives an exact $\delta$-functor and that $\mathbf{H}^{n}$ are effaceable for $n>0$ through the following steps:

(1) A homomorphism $F \rightarrow G$ induces a canonical homomorphism of resolutions $\underline{S}(F) \rightarrow \underline{S}(G) .$

(2) A short exact triple $0 \rightarrow F^{\prime} \rightarrow F \rightarrow F^{\prime \prime} \rightarrow 0$ induces a short exact triple of complexes $0 \rightarrow \underline{S}\left(F^{\prime}\right) \rightarrow \underline{S}(F) \rightarrow \underline{S}\left(F^{\prime \prime}\right) \rightarrow 0$.

(3) Applying $\Gamma$ to it gives a short exact triple of complexes, i.e. $0 \rightarrow S\left(F_{n}^{\prime}\right) \rightarrow$ $S\left(F_{n}\right) \rightarrow S\left(F_{n}^{\prime \prime}\right) \rightarrow 0$ is exact. (You can assume the previous problem.)

(4) $\left(\mathbf{H}^{n}\right)$ is an exact $\delta$-functor.

(5) For $n>0, \mathbf{H}^{n}(F) \rightarrow \mathbf{H}^{n}(S(F))$ is the zero map.

Conclude by Grothendieck's universality theorem.
:::

:::{.solution title="Part 1"}
This follows readily from the fact that a morphism $f: F\to G$ of sheaves on $X$ induces group morphisms $f_x: F_x \to G_x$ on stalks for every $x\in X$.
Letting $y\in X$ be arbitrary, there is a morphism 
\[
\phi_{y}: \prod_{x\in X} F_x \mapsvia{\pi_y} F_y \mapsvia{f_y} G_y
\]
where $\pi_y$ is the canonical projection out of the product.
By the universal property of the product, the $\phi_y$ assemble to a morphism
\[
S(f): \prod_{x\in X} F_x\to \prod_{y\in X} G_y
.\]

So there is a morphism $S(F_0) \to S(G_0)$ at the first stage of the complex.
This induces a morphism on the quotient sheaves $S(F_0)/F_0 \to S(G_0)/G_0$, and thus by the same argument as above, a morphism on the second stage $S(S(F_0)/F_0) \to S(S(G_0)/G_0)$, i.e. a morphism $S(F_1) \to S(G_1)$.
Continuing inductively yields levelwise morphisms $S(F_i) \to S(G_i)$.
The claim is that these assemble to a chain map

\begin{tikzcd}
	&&&& \textcolor{rgb,255:red,214;green,92;blue,92}{S(F)/F} \\
	0 & F && {S(F_0) = S(F)} && {S(F_1) = S(S(F)/F)} & \cdots \\
	\\
	0 & G && {S(G_0) = S(G)} && {S(G_1) = S(S(G)/G)} & \cdots \\
	&&&& \textcolor{rgb,255:red,214;green,92;blue,92}{S(G)/G}
	\arrow[from=2-2, to=2-4]
	\arrow[from=2-4, to=2-6]
	\arrow[from=4-2, to=4-4]
	\arrow[from=4-4, to=4-6]
	\arrow["{S(f)}", color={rgb,255:red,214;green,92;blue,92}, from=2-4, to=4-4]
	\arrow["{S_1(F)}", from=2-6, to=4-6]
	\arrow[from=2-6, to=2-7]
	\arrow[from=4-6, to=4-7]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, dashed, two heads, from=4-4, to=5-5]
	\arrow["{S(\wait)}"', dashed, hook, from=5-5, to=4-6]
	\arrow["{S(\wait)}", dashed, hook, from=1-5, to=2-6]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, dashed, two heads, from=2-4, to=1-5]
	\arrow[from=2-1, to=2-2]
	\arrow[from=4-1, to=4-2]
	\arrow["f", from=2-2, to=4-2]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, curve={height=-18pt}, dashed, from=1-5, to=5-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTIsWzEsMSwiRiJdLFszLDEsIlMoRl8wKSA9IFMoRikiXSxbNSwxLCJTKEZfMSkgPSBTKFMoRikvRikiXSxbMSwzLCJHIl0sWzMsMywiUyhHXzApID0gUyhHKSJdLFs1LDMsIlMoR18xKSA9IFMoUyhHKS9HKSJdLFs2LDEsIlxcY2RvdHMiXSxbNiwzLCJcXGNkb3RzIl0sWzQsMCwiUyhGKS9GIixbMCw2MCw2MCwxXV0sWzQsNCwiUyhHKS9HIixbMCw2MCw2MCwxXV0sWzAsMSwiMCJdLFswLDMsIjAiXSxbMCwxXSxbMSwyXSxbMyw0XSxbNCw1XSxbMSw0LCJTKGYpIiwwLHsiY29sb3VyIjpbMCw2MCw2MF19LFswLDYwLDYwLDFdXSxbMiw1LCJTXzEoRikiXSxbMiw2XSxbNSw3XSxbNCw5LCIiLDIseyJjb2xvdXIiOlswLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn0sImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFs5LDUsIlMoXFx3YWl0KSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFs4LDIsIlMoXFx3YWl0KSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDgsIiIsMCx7ImNvbG91ciI6WzAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifSwiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEwLDBdLFsxMSwzXSxbMCwzLCJmIl0sWzgsOSwiIiwxLHsiY3VydmUiOi0zLCJjb2xvdXIiOlswLDYwLDYwXSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)

To see this is true, it is enough to show that the first square commutes, i.e. that applying $S(\wait)$ to a morphism of sheaves produces a commuting square. 
This is because every other square has a factorization as indicated, where the square in red naturally commutes since it involves canonically induced maps on quotients/cokernels, and the other half of the square arises by applying the $S$ construction to some morphism of sheaves.

However, this square can be readily seen to commute using the following: first regard the sections of $\mcf$ as continuous sections of its espace étale $\Et_F \mapsvia{\pi} X$ and regarding sections of $S(F)$ as arbitrary (potentially discontinuous) sections of $\pi$.
Then $\mcf \leq S(F)$ is clearly a subsheaf and $F\to S(F)$ is an inclusion of spaces of sections.
:::

:::{.solution title="Part 2" }
By part 1, it is clear there are morphisms $\ul S(F') \to \ul S(F) \to \ul S(F'')$ of complexes of sheaves, yielding a double complex:

\begin{tikzcd}
	&& \vdots && \vdots && \vdots \\
	\\
	&& {S(F'_1)} && {S(F_1)} && {S(F''_1)} \\
	\\
	&& {S(F'_0)} && {S(F_0)} && {S(F''_0)} \\
	\\
	0 && {F'} && F && {F''} && 0 \\
	\\
	&& 0 && 0 && 0
	\arrow[from=9-3, to=7-3]
	\arrow[from=9-5, to=7-5]
	\arrow[from=9-7, to=7-7]
	\arrow[from=7-3, to=5-3]
	\arrow[from=7-5, to=5-5]
	\arrow[from=7-7, to=5-7]
	\arrow[from=5-3, to=3-3]
	\arrow[from=5-5, to=3-5]
	\arrow[from=5-7, to=3-7]
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-5, to=1-5]
	\arrow[from=3-7, to=1-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTcsWzIsNiwiRiciXSxbNCw2LCJGIl0sWzYsNiwiRicnIl0sWzIsOCwiMCJdLFs0LDgsIjAiXSxbNiw4LCIwIl0sWzIsNCwiUyhGJ18wKSJdLFs0LDQsIlMoRl8wKSJdLFs2LDQsIlMoRicnXzApIl0sWzIsMiwiUyhGJ18xKSJdLFs0LDIsIlMoRl8xKSJdLFs2LDIsIlMoRicnXzEpIl0sWzIsMCwiXFx2ZG90cyJdLFs0LDAsIlxcdmRvdHMiXSxbNiwwLCJcXHZkb3RzIl0sWzAsNiwiMCJdLFs4LDYsIjAiXSxbMywwXSxbNCwxXSxbNSwyXSxbMCw2XSxbMSw3XSxbMiw4XSxbNiw5XSxbNywxMF0sWzgsMTFdLFs5LDEyXSxbMTAsMTNdLFsxMSwxNF0sWzE1LDBdLFswLDFdLFsxLDJdLFs2LDddLFs3LDhdLFsyLDE2XSxbOSwxMF0sWzEwLDExXV0=)

It suffices to show injectivity, exactness, and surjectivity respectively along each horizontal row.
Exactness is a local condition, so it suffices to show exactness on stalks.

:::{.claim}
For any open $U$, the following sequence at the first stage of the complex is exact:
\[
0\to S(F')(U) \to S(F)(U) \to S(F'')(U)\to 0
.\]
:::


:::{.proof title="of claim"}
This follows because $S(F')(U) = \prod_{x\in U} F'_x$ and similarly for $F, F''$, and so if $f: F' \to F$ is injective on sheaves, then $f_x: F_x' \to F_x$ is injective on stalks.
:::

Now apply the functor $\colim_{U\ni p}(\wait)$ to this exact sequence and use that taking stalks is exact (despite not generally being a *filtered* colimit) to conclude
\[
0\to S(F')_x \to S(F)_x \to S(F'')_x \to 0
.\]
is exact for all $x\in X$, thus making the following sequence exact:
\[
0\to S(F_0') \to S(F_0) \to S(F_0'')\to 0
\]

Our double complex is now the following:

\begin{tikzcd}
	&& \vdots && \vdots && \vdots \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{?} && {S(F'_1)} && {S(F_1)} && {S(F''_1)} && \textcolor{rgb,255:red,214;green,92;blue,92}{?} \\
	\\
	0 && {S(F'_0)} && {S(F_0)} && {S(F''_0)} && 0 \\
	\\
	0 && {F'} && F && {F''} && 0 \\
	\\
	&& 0 && 0 && 0
	\arrow[from=9-3, to=7-3]
	\arrow[from=9-5, to=7-5]
	\arrow[from=9-7, to=7-7]
	\arrow[from=7-3, to=5-3]
	\arrow[from=7-5, to=5-5]
	\arrow[from=7-7, to=5-7]
	\arrow[from=5-3, to=3-3]
	\arrow[from=5-5, to=3-5]
	\arrow[from=5-7, to=3-7]
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-5, to=1-5]
	\arrow[from=3-7, to=1-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=5-7, to=5-9]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, from=3-1, to=3-3]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, from=3-7, to=3-9]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)


To see that 
\[
0\to S(F_k') \to S(F_k) \to S(F_k'')\to 0
\]
is exact for all $k$, we can truncate this complex:

\begin{tikzcd}
	&& 0 && 0 && 0 \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{?} && {S(F'_0)/F'_0} && {S(F_0)/F_0} && {S(F''_0)/F''_0} && \textcolor{rgb,255:red,214;green,92;blue,92}{?} \\
	\\
	0 && {S(F'_0)} && {S(F_0)} && {S(F''_0)} && 0 \\
	\\
	0 && {F'} && F && {F''} && 0 \\
	\\
	&& 0 && 0 && 0
	\arrow[from=9-3, to=7-3]
	\arrow[from=9-5, to=7-5]
	\arrow[from=9-7, to=7-7]
	\arrow[from=7-3, to=5-3]
	\arrow[from=7-5, to=5-5]
	\arrow[from=7-7, to=5-7]
	\arrow[from=5-3, to=3-3]
	\arrow[from=5-5, to=3-5]
	\arrow[from=5-7, to=3-7]
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-5, to=1-5]
	\arrow[from=3-7, to=1-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=5-7, to=5-9]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-1, to=3-3]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-7, to=3-9]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

The row highlighted in red is exact by the Nine Lemma, regarding each row as a chain complex, and since applying $S(\wait)$ is exact, by applying this to the top row we obtain

\begin{tikzcd}
	&& \vdots && \vdots && \vdots \\
	\\
	0 && {S(F_1')} && {S(F_1)} && {S(F''_1)} && 0 \\
	\\
	0 && {S(F'_0)} && {S(F_0)} && {S(F''_0)} && 0 \\
	\\
	0 && {F'} && F && {F''} && 0 \\
	\\
	&& 0 && 0 && 0
	\arrow[from=9-3, to=7-3]
	\arrow[from=9-5, to=7-5]
	\arrow[from=9-7, to=7-7]
	\arrow[from=7-3, to=5-3]
	\arrow[from=7-5, to=5-5]
	\arrow[from=7-7, to=5-7]
	\arrow[from=5-3, to=3-3]
	\arrow[from=5-5, to=3-5]
	\arrow[from=5-7, to=3-7]
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-5, to=1-5]
	\arrow[from=3-7, to=1-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=5-7, to=5-9]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-7, to=3-9]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMjEsWzIsNiwiRiciXSxbNCw2LCJGIl0sWzYsNiwiRicnIl0sWzIsOCwiMCJdLFs0LDgsIjAiXSxbNiw4LCIwIl0sWzIsNCwiUyhGJ18wKSJdLFs0LDQsIlMoRl8wKSJdLFs2LDQsIlMoRicnXzApIl0sWzIsMiwiUyhGXzEnKSJdLFs0LDIsIlMoRl8xKSJdLFs2LDIsIlMoRicnXzEpIl0sWzIsMCwiXFx2ZG90cyJdLFs0LDAsIlxcdmRvdHMiXSxbNiwwLCJcXHZkb3RzIl0sWzAsNiwiMCJdLFs4LDYsIjAiXSxbMCw0LCIwIl0sWzgsNCwiMCJdLFswLDIsIjAiXSxbOCwyLCIwIl0sWzMsMF0sWzQsMV0sWzUsMl0sWzAsNl0sWzEsN10sWzIsOF0sWzYsOV0sWzcsMTBdLFs4LDExXSxbOSwxMl0sWzEwLDEzXSxbMTEsMTRdLFsxNSwwXSxbMCwxXSxbMSwyXSxbNiw3XSxbNyw4XSxbMiwxNl0sWzksMTBdLFsxMCwxMV0sWzE3LDZdLFs4LDE4XSxbMTksOV0sWzExLDIwXV0=)

The remaining rows are exact by repeating this argument inductively, and regarding the columns as complexes, we obtain the desired exact sequences of complexes by deleting the first row.
:::

:::{.solution title="Part 3"}
Note: there may be a typo in the statement of this problem, so what I will show is that the following sequence of complexes is exact:
\[
0 
\to \globsec{X; \ul S(F')}
\to \globsec{X; \ul S(F)}
\to \globsec{X; \ul S(F'')}
\to 0
.\]

Take the double complex from part (2) and apply the functor $\globsec{X; \wait}$ to obtain the following double complex:

\begin{tikzcd}[column sep=small]
	&& \vdots && \vdots && \vdots \\
	\\
	0 && {\globsec{X; S(F_1')}} && {\globsec{X; S(F_0)}} && {\globsec{X; S(F''_1)}} && \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\
	\\
	0 && {\globsec{X; S(F'_0)}} && {\globsec{X; S(F_0)}} && {\globsec{X; S(F''_0) }} && \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\
	\\
	0 && {\globsec{X; F'}} && {\globsec{X; F}} && {\globsec{X; F''}} && {\RR^1\globsec{X; F'}} \\
	\\
	&& 0 && 0 && 0
	\arrow[from=9-3, to=7-3]
	\arrow[from=9-5, to=7-5]
	\arrow[from=9-7, to=7-7]
	\arrow[from=7-3, to=5-3]
	\arrow[from=7-5, to=5-5]
	\arrow[from=7-7, to=5-7]
	\arrow[from=5-3, to=3-3]
	\arrow[from=5-5, to=3-5]
	\arrow[from=5-7, to=3-7]
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-5, to=1-5]
	\arrow[from=3-7, to=1-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, from=5-7, to=5-9]
	\arrow[from=3-1, to=3-3]
	\arrow[draw={rgb,255:red,214;green,92;blue,92}, from=3-7, to=3-9]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

Here the bottom row continues in the long exact sequence for the right-derived functors of $\globsec{X; \wait}$, i.e. sheaf cohomology.
Since the desired sequence of complexes involved truncating this double complex by deleting the first row, consider everything from row two upward.
That these levelwise maps assemble to a map of complexes is just a consequence of functoriality of $\globsec{X;\wait}$, and left exactness preserves the zeros in the left-most column, so it suffices to show that the right-most column (highlighted in red) is zero as claimed.

However, this follows from the previous problem if the sheaves $S(F_n')$ are all flasque.
This is immediate since they are sheaves of discontinuous sections, and such a section on $U$ can always be extended to a global section by simply assigning any other values on $X\sm U$ -- any choice works, since no compatibility (e.g. continuity) is required.
:::

:::{.solution title="Part 4"}
It is a general theorem in homological algebra that a short exact sequence of chain complexes induces a long exact sequence in cohomology.
In this case, if we take the vertical homology of the above double complex, by the snake lemma there are connecting morphisms:

\begin{tikzcd}[column sep=small]
	&& \vdots && \vdots && \vdots \\
	\\
	0 && {H_2(\globsec{X; \ul{S}(F')})} && {H_2(\globsec{X; \ul{S}(F)})} && {H_2(\globsec{X; \ul{S}(F'')})} && 0\\
	\\
	0 && {H_1(\globsec{X; \ul{S}(F')})} && {H_1(\globsec{X; \ul{S}(F)})} && {H_1(\globsec{X; \ul{S}(F'')})} && 0 \\
	\\
	0 && {H_0(\globsec{X; \ul{S}(F')})} && {H_0(\globsec{X; \ul{S}(F)})} && {H_0(\globsec{X; \ul{S}(F'')})} && 0 \\
	\\
	&& 0 && 0 && 0
	\arrow[from=9-3, to=7-3]
	\arrow[from=9-5, to=7-5]
	\arrow[from=9-7, to=7-7]
	\arrow[from=7-3, to=5-3]
	\arrow[from=7-5, to=5-5]
	\arrow[from=7-7, to=5-7]
	\arrow[from=5-3, to=3-3]
	\arrow[from=5-5, to=3-5]
	\arrow[from=5-7, to=3-7]
	\arrow[from=3-3, to=1-3]
	\arrow[from=3-5, to=1-5]
	\arrow[from=3-7, to=1-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=5-7, to=5-9]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-7, to=3-9]
	\arrow["{\exists \delta_0}"'{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, dashed, from=7-7, to=5-3]
	\arrow["{\exists \delta_1}"'{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, dashed, from=5-7, to=3-3]
	\arrow["{\exists \delta_2}"'{pos=0.3}, color={rgb,255:red,92;green,92;blue,214}, dashed, from=3-7, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

:::

:::{.solution title="Part 5"}
This holds because flasque sheaves are $F\dash$acyclic for $F(\wait) = \globsec{X; \wait}$, so we can conclude that $\mathbf{H}^n(S(F)) = 0$ for $n > 0$ since the sheaves $S(F)$ are always flasque for any sheaf $F$.

> Note: I realized at the last minute that this argument may not actually work, since this $\mathbf{H}^n$ a priori has nothing to do with $\RR \Gamma(X;\wait)$ computed via injective resolutions.

:::