# Wednesday, February 03

:::{.remark}
All rings have 1 in this course!
:::

## Horseshoe Lemma

:::{.proposition title="Horseshoe Lemma"}
Suppose we have a diagram like the following, where the columns are exact and the rows are projective resolutions:


\begin{tikzcd}
	&&&&&&&& 0 \\
	\\
	\cdots && {P_2'} && {P_1'} && {P_0'} && {A'} && 0 \\
	\\
	&&&&&&&& A \\
	\\
	\cdots && {P_2''} && {P_1''} && {P_0''} && {A''} && 0 \\
	&&&&&&&& {} \\
	&&&&&&&& 0
	\arrow["{\iota_A}", from=3-9, to=5-9]
	\arrow["{\eps'}", from=3-7, to=3-9]
	\arrow["{\eps''}", from=7-7, to=7-9]
	\arrow[from=7-9, to=9-9]
	\arrow["{\pi_A}", from=5-9, to=7-9]
	\arrow[from=3-5, to=3-7]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-1, to=3-3]
	\arrow[from=7-1, to=7-3]
	\arrow[from=7-3, to=7-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=3-9, to=3-11]
	\arrow[from=7-9, to=7-11]
	\arrow[from=1-9, to=3-9]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTYsWzgsMiwiQSciXSxbOCw0LCJBIl0sWzgsNiwiQScnIl0sWzgsOCwiMCJdLFsxMCwyLCIwIl0sWzYsMiwiUF8wJyJdLFs0LDIsIlBfMSciXSxbMiwyLCJQXzInIl0sWzAsMiwiXFxjZG90cyJdLFs2LDYsIlBfMCcnIl0sWzQsNiwiUF8xJyciXSxbMiw2LCJQXzInJyJdLFsxMCw2LCIwIl0sWzAsNiwiXFxjZG90cyJdLFs4LDddLFs4LDAsIjAiXSxbMCwxLCJcXGlvdGFfQSJdLFs1LDAsIlxcZXBzJyJdLFs5LDIsIlxcZXBzJyciXSxbMiwzXSxbMSwyLCJcXHBpX0EiXSxbNiw1XSxbNyw2XSxbOCw3XSxbMTMsMTFdLFsxMSwxMF0sWzEwLDldLFswLDRdLFsyLDEyXSxbMTUsMF1d)

Note that if the vertical sequence were split, one could sum together to two resolutions to get a resolution of the middle.
This still works: there is a projective resolution of $P$ of $A$ given by 
\[
P_n \da P_n' \oplus P_n''
\]
which lifts the vertical column in the above diagram to an exact sequence of complexes
\[
0 \to P' \mapsvia{\iota} P \mapsvia{\pi}  P'' \to 0
,\]
where $\iota_n: P_n' \injects P_n$ is the natural inclusion and $\pi_i: P_n \surjects P_n''$ the natural projection.

:::

### Proof of the Horseshoe Lemma

We can construct this inductively:

\begin{tikzcd}
	&& 0 && 0 \\
	\\
	{\ker(\eps')} && {P_0'} && {A'} && 0 \\
	\\
	{\ker(\eps)} && \textcolor{rgb,255:red,92;green,92;blue,214}{P_0} && \textcolor{rgb,255:red,92;green,92;blue,214}{A} && {\coker(\eps)} \\
	\\
	{\ker(\eps'')} && {P_0''} && {A''} && 0 \\
	\\
	&& 0 && 0
	\arrow["{\varepsilon''}", from=7-3, to=7-5]
	\arrow["{\eta''}"', dashed, from=7-3, to=5-5]
	\arrow["\iota"', from=3-3, to=5-3]
	\arrow["\pi"', from=5-3, to=7-3]
	\arrow["{\iota_A}", from=3-5, to=5-5]
	\arrow["{\pi_A}", from=5-5, to=7-5]
	\arrow["{\varepsilon'}", from=3-3, to=3-5]
	\arrow["{\eta'}", from=3-3, to=5-5]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=7-5, to=7-7]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=7-1, to=7-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=7-3, to=9-3]
	\arrow[from=7-5, to=9-5]
	\arrow["\eta", color={rgb,255:red,92;green,92;blue,214}, from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
\end{tikzcd}

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

- $P_0''$ projective and $\pi_A$ surjective implies $\eps''$ lifts to $\eta'': P_0'' \to A$
- Composing yields $\eta' \da \iota_A \circ \eta': P_0' \to A$
- Get $\eps \da \eta' \oplus \eta'':P_0 \da P_0' \oplus P_0'' \to A$.

Flipping the diagram, we can apply the snake lemma to the two columns:


\begin{tikzcd}
	\textcolor{rgb,255:red,214;green,92;blue,92}{0} && 0 && 0 \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\ker(\eps')} && {P_0'} && {A'} && \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\ker(\eps)} && \textcolor{rgb,255:red,92;green,92;blue,214}{P_0} && \textcolor{rgb,255:red,92;green,92;blue,214}{A} && \textcolor{rgb,255:red,214;green,92;blue,92}{\coker(\eps)} \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\ker(\eps'')} && {P_0''} && {A''} && \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\
	\\
	&& 0 && 0
	\arrow["{\varepsilon''}", from=7-3, to=7-5]
	\arrow["{\eta''}"', dashed, from=7-3, to=5-5]
	\arrow["\iota"', from=3-3, to=5-3]
	\arrow["\pi"', from=5-3, to=7-3]
	\arrow["{\iota_A}", from=3-5, to=5-5]
	\arrow["{\pi_A}", from=5-5, to=7-5]
	\arrow["{\varepsilon'}", from=3-3, to=3-5]
	\arrow["{\eta'}", from=3-3, to=5-5]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=7-5, to=7-7]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=7-1, to=7-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=7-3, to=9-3]
	\arrow[from=7-5, to=9-5]
	\arrow["\eta", color={rgb,255:red,92;green,92;blue,214}, from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=3-1]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-1, to=5-1]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=5-1, to=7-1]
	\arrow["\bd", color={rgb,255:red,214;green,92;blue,92}, squiggly, from=7-1, to=3-7, out=65, in=90]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-7, to=5-7]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=5-7, to=7-7]
\end{tikzcd}

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

We can now conclude that

- $\coker \eps = 0$
- $\bd = 0$ since it lands on the zero moduli

So append a zero onto the far left column:

\begin{tikzcd}
	\textcolor{rgb,255:red,214;green,92;blue,92}{0} && 0 && 0 \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\ker(\eps')} && {P_0'} && {A'} && \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\ker(\eps)} && \textcolor{rgb,255:red,92;green,92;blue,214}{P_0} && \textcolor{rgb,255:red,92;green,92;blue,214}{A} && \textcolor{rgb,255:red,214;green,92;blue,92}{\coker(\eps)} \\
	\\
	\textcolor{rgb,255:red,214;green,92;blue,92}{\ker(\eps'')} && {P_0''} && {A''} && \textcolor{rgb,255:red,214;green,92;blue,92}{0} \\
	\\
	0 && 0 && 0
	\arrow["{\varepsilon''}", from=7-3, to=7-5]
	\arrow["{\eta''}"', dashed, from=7-3, to=5-5]
	\arrow["\iota"', from=3-3, to=5-3]
	\arrow["\pi"', from=5-3, to=7-3]
	\arrow["{\iota_A}", from=3-5, to=5-5]
	\arrow["{\pi_A}", from=5-5, to=7-5]
	\arrow["{\varepsilon'}", from=3-3, to=3-5]
	\arrow["{\eta'}", from=3-3, to=5-5]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=7-5, to=7-7]
	\arrow[from=3-5, to=3-7]
	\arrow[from=5-1, to=5-3]
	\arrow[from=7-1, to=7-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=7-3, to=9-3]
	\arrow[from=7-5, to=9-5]
	\arrow["\eta", color={rgb,255:red,92;green,92;blue,214}, from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=3-1]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-1, to=5-1]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=5-1, to=7-1]
	\arrow[from=7-1, to=9-1]
	\arrow["\bd", color={rgb,255:red,214;green,92;blue,92}, squiggly, from=7-1, to=3-7, out=65, in=90]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=3-7, to=5-7]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=5-7, to=7-7]
\end{tikzcd}

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


Thus the LHS column is a SES, and we have the first step of a resolution.
Proceeding inductively, at the next step we have


\begin{tikzcd}
	&&&& 0 \\
	\\
	\cdots && {P_1'} && {\ker(\eps')} && 0 \\
	\\
	&&&& {\ker(\eps)} \\
	\\
	\cdots && {P_1''} & {} & {\ker(\eps'')} && 0 \\
	\\
	&&&& 0
	\arrow[from=1-5, to=3-5]
	\arrow[from=3-5, to=5-5]
	\arrow[from=5-5, to=7-5]
	\arrow[from=7-5, to=9-5]
	\arrow["{d_1'}", from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=7-5, to=7-7]
	\arrow[from=7-1, to=7-3]
	\arrow[from=3-1, to=3-3]
	\arrow["{d_1''}", from=7-3, to=7-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTIsWzQsMCwiMCJdLFs0LDIsIlxca2VyKFxcZXBzJykiXSxbNCw0LCJcXGtlcihcXGVwcykiXSxbNCw2LCJcXGtlcihcXGVwcycnKSJdLFs0LDgsIjAiXSxbNiw2LCIwIl0sWzYsMiwiMCJdLFsyLDIsIlBfMSciXSxbMiw2LCJQXzEnJyJdLFszLDZdLFswLDYsIlxcY2RvdHMiXSxbMCwyLCJcXGNkb3RzIl0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzcsMSwiZF8xJyJdLFsxLDZdLFszLDVdLFsxMCw4XSxbMTEsN10sWzgsMywiZF8xJyciXV0=)

However, this is precisely the situation that appeared before, so the same procedure works.

:::{.exercise title="?"}
Check that the middle complex is exact!
Follows by construction.
:::

## Injective Resolutions

:::{.definition title="Injective Objects"}
Let \( \mathcal{A}  \) be an abelian category, then \( I\in \mathcal{A}  \) is **injective** if and only if it satisfies the following universal property:
$A$ is projective if and only if for every monic $\alpha :A\to I$, any map $f:A\to B$ lifts to a map $B\to I$:


\begin{tikzcd}
	0 && A && B \\
	\\
	&& I
	\arrow[from=1-1, to=1-3]
	\arrow["\alpha", from=1-3, to=3-3]
	\arrow["{\exists \beta}", dashed, from=1-5, to=3-3]
	\arrow["f", hook, from=1-3, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiQSJdLFs0LDAsIkIiXSxbMiwyLCJJIl0sWzAsMV0sWzEsMywiXFxhbHBoYSJdLFsyLDMsIlxcZXhpc3RzIFxcYmV0YSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFsxLDIsImYiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==)

We say \( \mathcal{A}  \) **has enough injectives** if and only if for all $A$, there exists $A\injects I$ where $I$ is injective.
:::

:::{.slogan}
Maps on subobjects extend.
:::

:::{.proposition title="Products of Injectives are Injective"}
If $\ts{I_ \alpha}$ is a family of injectives and $I \da \prod_{\alpha} I_ \alpha \in A$, then $I$ is again injective.
:::

:::{.proof title="?"}
Use the universal property of direct products.
:::

## Baer's Criterion

:::{.proposition title="Baer's Criterion"}
An object $E \in \rmod$ is injective if and only if for every right ideal $J \normal R$, every map $J\to E$ extends to a map $R\to E$.
Note that $J$ is a right $R\dash$submodule.
:::

:::{.proof title="?"}
$\implies$: 
This is essentially by definition.
Instead of taking arbitrary submodules, we're just taking $R$ itself and *its* submodules:

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

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiSiJdLFs0LDAsIlIiXSxbMiwyLCJFIl0sWzAsMV0sWzEsMl0sWzEsM10sWzIsMywiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)


$\impliedby$:
Suppose we have the following:

\begin{tikzcd}
	0 && A && B \\
	\\
	&& E
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow["\alpha", from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiQSJdLFs0LDAsIkIiXSxbMiwyLCJFIl0sWzAsMV0sWzEsMl0sWzEsMywiXFxhbHBoYSJdXQ==)


Let \( \mathcal{E}\da \ts{ \alpha': A' \to E \st A \leq A' \leq B }  \), i.e. all of the intermediate extensions:

\begin{tikzcd}
	0 && A && \textcolor{rgb,255:red,214;green,92;blue,92}{A'} && B \\
	\\
	&& E
	\arrow[from=1-1, to=1-3]
	\arrow["\alpha", from=1-3, to=3-3]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=1-3, to=1-5]
	\arrow[color={rgb,255:red,214;green,92;blue,92}, from=1-5, to=1-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCIwIl0sWzIsMCwiQSJdLFsyLDIsIkUiXSxbNCwwLCJBJyIsWzAsNjAsNjAsMV1dLFs2LDAsIkIiXSxbMCwxXSxbMSwyLCJcXGFscGhhIl0sWzEsMywiIiwwLHsiY29sb3VyIjpbMCw2MCw2MF19XSxbMyw0LCIiLDAseyJjb2xvdXIiOlswLDYwLDYwXX1dXQ==)

Add a partial order to \( \mathcal{E}  \) where \( \alpha ' \leq \alpha'' \) if and only if \( \alpha'' \) extends \( \alpha' \).
Applying Zorn's lemma (and abusing notation slightly), we can produce a maximal \( \alpha': A' \to E \).
The claim is that $A' = B$.
Supposing not, then $A'$ is a proper submodule, so choose a $b\in B \sm A'$.
Then define the set \( J \da \ts{ r\in R \st br \in A' } \), this is a right ideal of $R$ since $A'$ was a right \(R\dash\)module.
Now applying the assumption of Baer's condition on $E$, we can produce a map $f:R\to E$:C

\begin{tikzcd}
	0 && J && R \\
	\\
	&& {A'} \\
	\\
	&& E
	\arrow[from=1-1, to=1-3]
	\arrow["b\cdot\wait"', from=1-3, to=3-3]
	\arrow["{\alpha'}"', from=3-3, to=5-3]
	\arrow[from=1-3, to=1-5]
	\arrow["{\exists f}", dashed, from=1-5, to=5-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwwLCIwIl0sWzIsMCwiSiJdLFsyLDIsIkEnIl0sWzIsNCwiRSJdLFs0LDAsIlIiXSxbMCwxXSxbMSwyLCJiXFxjZG90XFx3YWl0IiwyXSxbMiwzLCJcXGFscGhhJyIsMl0sWzEsNF0sWzQsMywiXFxleGlzdHMgZiIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

Now let $A'' \da A' + bR \leq B$, and provisionally define 
\[
\alpha'': A'' &\to E \\
a + br & \mapsto \alpha'(a) + f(r)
.\]


:::{.remark}
Is this well-defined?
Consider overlapping terms, it's enough to consider elements of the form $br\in A'$. 
In this case, $r\in J$ by definition, and so \( \alpha'(br) = f(r) \) by commutativity in the previous diagram, which shows that the two maps agree on anything in the intersection.
:::

Note that \( \alpha'' \) now extends \( \alpha' \), but \( A' \subsetneq A'' \) since $b\in A''\sm A'$.
But then $A''$ strictly contains $A'$, contradicting its maximality from Zorn's lemma.
:::

:::{.remark}
Big question: what *are* injective modules really? 
These are pretty nonintuitive objects.
:::