# Monday, February 08

## Transporting Injectives

:::{.remark}
Last time: we had a lemma that for any $M\in \modr$ and $A\in \Ab$ there is an isomorphism
\[
\Hom_{\Ab}( F(M), A) \cong \Hom_{\modr}(M, \Hom_\Ab(R, A) )
,\]
where $F: \modr \to \Ab$ is the forgetful functor.

:::

:::{.definition title="Adjoints"}
A pair of functors $L: \mathcal{A} \to \mathcal{B}$ and $R: \mathcal{B} \to \mathcal{A}$ are **adjoint** is there are natural bijections
\[
\tau_{AB}: \Hom_B(L(A), B) \mapsvia{\sim} \Hom_A(A, R(B) ) && \forall A\in A, B\in B
,\]
where *natural* means that for all \( A \mapsvia{f} A' \) and \( B \mapsvia{g} B' \) there is a diagram

\begin{tikzcd}
	{\Hom_B(LA', B)} && {\Hom_B(LA, B)} && {\Hom_B(LA, B')} \\
	\\
	{\Hom_A(A', RB)} && {\Hom_A(A, RB)} && {\Hom_A(A, RB')}
	\arrow["\tau", from=1-3, to=3-3]
	\arrow["\tau", from=1-1, to=3-1]
	\arrow["{(Lf)^*}"{description}, from=1-1, to=1-3]
	\arrow["{f^*}"{description}, from=3-1, to=3-3]
	\arrow["{g_*}"{description}, from=1-3, to=1-5]
	\arrow["{(Rg)_*}"{description}, from=3-3, to=3-5]
	\arrow["\tau"{description}, from=1-5, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJcXEhvbV9CKExBJywgQikiXSxbMCwyLCJcXEhvbV9BKEEnLCBSQikiXSxbMiwwLCJcXEhvbV9CKExBLCBCKSJdLFsyLDIsIlxcSG9tX0EoQSwgUkIpIl0sWzQsMCwiXFxIb21fQihMQSwgQicpIl0sWzQsMiwiXFxIb21fQShBLCBSQicpIl0sWzIsMywiXFx0YXUiXSxbMCwxLCJcXHRhdSJdLFswLDIsIihMZileKiIsMV0sWzEsMywiZl4qIiwxXSxbMiw0LCJnXyoiLDFdLFszLDUsIihSZylfKiIsMV0sWzQsNSwiXFx0YXUiLDFdXQ==)

In this case we say $L$ is **left adjoint** to $R$ and $R$ is **right adjoint** to $L$ and write $\adjunction{L}{R}{\cat A}{\cat B}$.
:::

:::{.remark}
The lemma thus says that $\Hom_{\Ab}(R, \wait): \Ab \to \modr$ (using that $R\in \rmod$ is a left \(R\dash\)module) is right adjoint to the forgetful functor $\modr\to\ Ab$.
:::

:::{.remark}
Recall that $F$ is **additive** if $\Hom_{\mathcal{B}}(B', B) \to \Hom_{\mathcal{A}}(FB', FB)$ is a morphism of abelian groups for all $B, B' \in \mathcal{B}$.
:::

:::{.proposition title="Right adjoints to exact functors preserve injectives, left adjoints preserve projectives"}
If $R: \mathcal{B} \to \mathcal{A}$ is an additive functor and right adjoint to an exact functor $L: \mathcal{A} \to \mathcal{B}$, then $I\in \mathcal{B}$ injective implies $R(I)\in \mathcal{A}$ is injective.
Dually, if \( \mathcal{L}:A\to B  \) is additive and left adjoint to an exact functor $R: \mathcal{B} \to \mathcal{A}$, then $P\in \mathcal{A}$ projective implies $L(P) \in \mathcal{B}$ is projective.
:::

:::{.corollary title="?"}
If $I\in \Ab$ is injective, then $\Hom_{\Ab}(R, I) \in \modr$ is injective.
:::

:::{.proof title="?"}
This follows from the previous lemma: $\Hom_{\Ab}(R, \wait)$ is right adjoint to the forgetful functor $\rmod\to\Ab$ which is certainly exact.
This follows from the fact that kernels and images don't change, since these are given in terms of set maps and equalities of sets.
:::

:::{.exercise title="2.3.5, 2.3.2"}
Show that $\modr$ has enough injectives, using that $\Ab$ has enough injectives.
:::

:::{.proof title="of proposition"}
It suffices to show that the contravariant functor $\Hom_{\mathcal{A}}(\wait, RI)$ is exact.
We know it's left exact, so we'll show surjectivity.
Suppose we have a SES $0 \to A \mapsvia{f} A'$ which is exact in \( \mathcal{A} \).
Then $0 \to LA \mapsvia{Lf} LA'$ is exact, and we can apply hom to obtain the exact sequence
\[
\Hom_{\mathcal{B} }(LA', I) \mapsvia{(LF)^*} \Hom_{\mathcal{B}}(LA, I) \to 0  
.\]
Applying $\tau$ yields


\begin{tikzcd}
	{\Hom_{\mathcal{B}}(LA', I)} && {\Hom_{\mathcal{B}}(LA, I)} && 0 \\
	\\
	{\Hom_{\mathcal{A}}(A', RI)} && {\Hom_{\mathcal{A}}(A, RI)} && 0
	\arrow["{(Lf)^*}", from=1-1, to=1-3]
	\arrow["{f^*}", from=3-1, to=3-3]
	\arrow["{\tau \sim}"{description}, from=1-1, to=3-1]
	\arrow["{\tau \sim}"{description}, from=1-3, to=3-3]
	\arrow[from=1-3, to=1-5]
	\arrow[dashed, from=3-3, to=3-5]
	\arrow[dashed, from=1-5, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJcXEhvbV97XFxtYXRoY2Fse0J9fShMQScsIEkpIl0sWzIsMCwiXFxIb21fe1xcbWF0aGNhbHtCfX0oTEEsIEkpIl0sWzQsMCwiMCJdLFswLDIsIlxcSG9tX3tcXG1hdGhjYWx7QX19KEEnLCBSSSkiXSxbMiwyLCJcXEhvbV97XFxtYXRoY2Fse0F9fShBLCBSSSkiXSxbNCwyLCIwIl0sWzAsMSwiKExmKV4qIl0sWzMsNCwiZl4qIl0sWzAsMywiXFx0YXUgXFxzaW0iLDFdLFsxLDQsIlxcdGF1IFxcc2ltIiwxXSxbMSwyXSxbNCw1LCIiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMiw1LCIiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=)

- The top sequence is exact since $I$ is injective in \( \mathcal{B}  \) 
- Therefore the bottom map is onto (diagram chase)
:::

## 2.4: Left Derived Functors

:::{.remark}
Goal: define left derived functors of a right exact functor $F$, with applications the bifunctor $\wait \tensor_R \wait$, which is right exact and covariant in each variable.
We're ultimately interested in Hom functors and Ext, but this is slightly more technical since it's covariant in one slot and contravariant in the other, so focusing on this functor makes the theory slightly easier to develop.
This can be fixed by switching \( \mathcal{C}  \) with \( \mathcal{C}\op  \) once in a while.
Everything for left derived functors will have a dual for right derived functors.
:::

:::{.remark}
Let \( \mathcal{A}, \mathcal{B}   \) be abelian categories where \( \mathcal{A}  \) has enough projectives and \( F: \mathcal{A} \to \mathcal{B}   \) is a right exact functor (which implicitly assumes $F$ is additive).
We want to define $L_i F: \mathcal{A} \to \mathcal{B}$ for $i\geq 0$. 
:::

:::{.definition title="Left Derived Functors"}
For \( A \in \mathcal{A}  \), fix once and for all a projective resolution $P \mapsvia{\eps} A$, where $P_{<0} = 0$.
Then define $FP = (\cdots \to F(P_1) \mapsvia{Fd_1 } F(P_0) \to 0$, noting that $A$ no longer appears in this complex.
We can write $H_0(FP) = FP_0 / (Fd_1)(FP_1)$, and define
\[
(L_i F)(A) \da H_i(F P)
.\]

:::

:::{.remark}
Note that $P_2 \mapsvia{d_2} P_1 \mapsvia{d_1} P_0 \mapsvia{\eps} A\to 0$ is exact, and since $F$ is right exact, it can be shown that the following is a SES: $FP_1 \mapsvia{Fd_1} FP_0 \mapsvia{F \eps} FA \to 0$.
We can use this to compute the original homology, despite it not having any homology itself!
From this, we can extra $L_0(A) \da FP_0 / (Fd_1)(FP_1) = FP_0 / \ker F(\eps)$ using exactness at $FP_0$, and by the 1st isomorphism theorem this is isomorphic to the image $FA$ using surjectivity.
So $L_0 F \cong F$.
:::

:::{.theorem title="Left-derived functors are additive"}
$L_i F: \mathcal{A} \to \mathcal{B}$ are additive functors. 
:::

:::{.proof title="?"}
First, $\one_P: P\to P$ lifts $\one_A: A\to A$ since it yields a commuting ladder, and $F(\one) = \one$, so $(L_i f)(\one) = \one$.
Then in the following diagram, the outer rectangle commutes since the inner squares do:


\begin{tikzcd}
	P && A \\
	\\
	{P'} && {A'} \\
	\\
	{P''} && {A''}
	\arrow["f", from=1-3, to=3-3]
	\arrow["g", from=3-3, to=5-3]
	\arrow["{\tilde f}", from=1-1, to=3-1]
	\arrow["{\tilde g}", from=3-1, to=5-1]
	\arrow[from=5-1, to=5-3]
	\arrow[from=3-1, to=3-3]
	\arrow[from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwwLCJBIl0sWzIsMiwiQSciXSxbMiw0LCJBJyciXSxbMCwwLCJQIl0sWzAsMiwiUCciXSxbMCw0LCJQJyciXSxbMCwxLCJmIl0sWzEsMiwiZyJdLFszLDQsIlxcdGlsZGUgZiJdLFs0LDUsIlxcdGlsZGUgZyJdLFs1LDJdLFs0LDFdLFszLDBdXQ==)

So $\tilde g \circ \tilde f$ lifts $g \circ f$ and therefore $g_* f_* = (gf)_*$.
Thus $L_i F$ is a functor.
That they are additive comes from checking the following diagram:


\begin{tikzcd}
	P && A \\
	\\
	Q && B
	\arrow["{\tilde f}", shift left=1, from=1-1, to=3-1]
	\arrow["{\tilde g}"', shift right=2, from=1-1, to=3-1]
	\arrow["f", shift left=2, from=1-3, to=3-3]
	\arrow["g"', shift right=1, from=1-3, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow[from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJQIl0sWzIsMCwiQSJdLFswLDIsIlEiXSxbMiwyLCJCIl0sWzAsMiwiXFx0aWxkZSBmIiwwLHsib2Zmc2V0IjotMX1dLFswLDIsIlxcdGlsZGUgZyIsMix7Im9mZnNldCI6Mn1dLFsxLDMsImYiLDAseyJvZmZzZXQiOi0yfV0sWzEsMywiZyIsMix7Im9mZnNldCI6MX1dLFswLDFdLFsyLDNdXQ==)

Then $\tilde f + \tilde g$ lifts $f+g$, and $H_i$ is an additive functor: $(F \tilde f + F \tilde g)_* = (F\tilde f)_* + (F\tilde g)_*$.
Thus $L_i F$ is additive.
:::