# Monday, March 21

:::{.remark}
Recall free $\implies$ projective and $\rmod$ has enough projectives and enough injectives.
:::

:::{.exercise title="?"}
Show $I$ is injective iff 

\begin{tikzcd}
	0 && J && R \\
	\\
	&&&& I
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow["\exists", dashed, from=1-5, to=3-5]
	\arrow[from=1-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiSiJdLFs0LDAsIlIiXSxbNCwyLCJJIl0sWzAsMV0sWzEsMl0sWzIsMywiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSwzXV0=)


Hint:

<!-- Xournal file: /home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Spring/SheafCohomology/sections/figures/2022-03-21_10-33.xoj -->

![](figures/2022-03-21_10-35-47.png)

Extend to $A' + Ra$ using $1\mapsto a \mapsto i\in I$ under $R\to Ra\to I$.
Take a poset of all $B \subseteq A$ with $g:B\to I$ extending $A'\to I$ and apply Zorn's lemma.
:::

:::{.exercise title="?"}
Show that for $R$ a PID, $M\in \rmod$ is injective iff divisible.
:::

:::{.exercise title="?"}
Show that $\zmod$ has enough innjectives.

> Hint: write $A = \bigoplus \ZZ/K \embeds \bigoplus \QQ/K$.

:::

:::{.remark}
On adjoint functors: 
\[
\adjunction F G {\cat A} {\cat B} \implies \cat{B}(FX, Y) \iso \cat{A}(X, GY)
.\]
Here $F$ is a left adjoint hence right exact, and $G$ is a right adjoint and is left exact.
:::

:::{.exercise title="?"}
Show that if $F$ is left exact then $G$ preserves in injectives, and if $F$ is right exact then $G$ preserves projectives.

Hint:

\begin{tikzcd}
	&& 0 && {A'} && A \\
	\\
	0 && {FA'} && FA && GI \\
	\\
	&&&& I
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow["\exists", dashed, from=3-5, to=5-5]
	\arrow[from=1-5, to=3-3]
	\arrow[from=1-7, to=3-5]
	\arrow[from=3-7, to=5-5]
	\arrow["{\therefore \exists}", dashed, from=1-7, to=3-7]
	\arrow[from=3-3, to=5-5]
	\arrow[from=1-5, to=3-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwyLCIwIl0sWzIsMiwiRkEnIl0sWzQsMiwiRkEiXSxbNCw0LCJJIl0sWzIsMCwiMCJdLFs0LDAsIkEnIl0sWzYsMCwiQSJdLFs2LDIsIkdJIl0sWzQsNV0sWzUsNl0sWzAsMV0sWzEsMl0sWzIsMywiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNSwxXSxbNiwyXSxbNywzXSxbNiw3LCJcXHRoZXJlZm9yZSBcXGV4aXN0cyIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDNdLFs1LDddXQ==)

:::

:::{.remark}
For $f\in \CRing(S\to R)$,
there is an adjunction
\[
\adjunction {M_R\mapsto M_S} { \mods{S}(R, \wait) } {\mods{R}}{\mods{S}}
\]
where $\mods{S}(R, N) \in \mods{R}$ via the action $(rf)(x) \da f(rx)$, sometimes called the *induced $R\dash$module*.
Note that $\rmod(R, N) \iso N$ by $1_R \mapsto n$, and there is an iso
\[
\mods{S}(M_S, N) &\mapstofrom \mods{R}(M_R, \mods{S}(R, N)) \\
\qty{ m\mapsto \psi(m)(1) } &\mapsfrom \psi \\
\phi &\mapsto \qty{m \mapsto \psi(m)(i) \da \psi(im) \da \phi(im) }
.\]
:::

:::{.remark}
Proving $\rmod$ has enough injectives if $\mods{S}$ has enough injectives: use $M_R \cong \rmod(R, M)\embeds \mods{S}(R, M_S) \embeds \mods{S}(R, I)$ where $M_S \embeds I$ embeds into some injective.
Take $R$ arbitrary and $S=\ZZ$ to conclude any $\rmod$ has enough injectives.
:::

:::{.exercise title="?"}
This is a theoretical tool and not particularly practical.
Consider $S\to R \da \QQ\to \CC$ and $M = \CC$.
Then $\mods{\QQ}(\CC, \CC_\QQ) = G\CC_\QQ$.
:::

:::{.remark}
Any $M\in \rmod$ admits a minimal injective hull $M\embeds I$.
:::

:::{.theorem title="?"}
$\Sh(X \to \Ab\Grp)$ and $\mods{\OO_X}$ have enough injectives.
:::

:::{.proof title="?"}
Take
\[
\mcf \embeds \prod_{x\in X} (\iota_x)_* \mcf_x \embeds \prod_{x\in X} I_x
.\]
The claim is that the last term is an injective sheaf.
Using that products of injective are injective, it STS $I_x$ is injective.
For $\iota_x: \ts{x} \embeds X$, use that modules on a point are $\zmod$ and obtain an adjunction
\[
\adjunction {(\iota_x)_*} {(\iota_x)\inv } {\zmod}{\Sh(X\to \Ab\Grp)}
.\]
Finally use that $\zmod$ has enough injectives.
:::