# Wednesday, March 02

:::{.remark}
For $F\in \Ab\Cat(\cat A, \cat B)$ left exact, assuming $\cat A$ has enough injectives, there is a right derived functor $\RR F$ so that a SES $0\to A\to B\to C\to 0$ admits a LES with a connecting morphism $\delta$:
\[
0\to \RR FA\to \RR F B\to \RR F C \mapsvia{\delta} \Sigma^1 \RR F A \to \cdots
.\]
Note that $\delta$ depends on the triple appearing in the SES.
:::

:::{.theorem title="Grothendieck"}
$\RR F$ and $\delta$ are universal among $\delta\dash$functors.
:::

:::{.remark}
Injectives will be acyclic and homology will measure how things are glued.
Analogy: simplicial or cellular homology uses contractible objects (with trivial homology) to measure how spaces are glued from simplices or spheres.
:::

:::{.remark}
Recall the definitions of projective and injective objects, which require existence (but not uniqueness) of certain lifts.
In $\rmod$, free implies projective, so free resolutions usually suffice and one can study generators, relations, syzygies, etc.

We'll show that $\cat{A} \da \Sh(X; \Ab\Grp)$ has enough injectives, but usually won't have enough projectives.
Recall that this means that every $A\in \cat{A}$ admits a monomorphism $A\injects I$ for $I$ an injective object.
If there are enough injectives, every object admits an injective resolution, and any two such resolutions are homotopy equivalent.
:::

:::{.remark}
Recall that
\[
\cocomplex{\RR} F(X) = \complex{H}( F( X \coveredby \cocomplex{I} ))
\]
and $\RR^{i\geq 1} F(I) = 0$ if $I$ is itself injective.
:::

:::{.remark}
Recall the Horseshoe lemma:

\begin{tikzcd}
	0 && 0 && 0 \\
	A && {I^1_A} && {I^2_A} && \cdots \\
	B && \textcolor{rgb,255:red,92;green,92;blue,214}{\exists I^1_B \da I^1_A \oplus I^1_C} && \textcolor{rgb,255:red,92;green,92;blue,214}{\exists I^2_B \da I^2_A \oplus I^2_C} && \textcolor{rgb,255:red,92;green,92;blue,214}{\cdots} \\
	C && {I^1_C} && {I^2_C} && \cdots \\
	0 && 0 && 0
	\arrow[from=4-3, to=4-1]
	\arrow[from=4-5, to=4-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=3-7, to=3-5]
	\arrow[dashed, from=4-7, to=4-5]
	\arrow[dashed, from=2-7, to=2-5]
	\arrow[from=2-5, to=2-3]
	\arrow[from=2-3, to=2-1]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-5, to=3-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=3-1]
	\arrow[from=1-3, to=2-3]
	\arrow[from=2-3, to=3-3]
	\arrow[from=3-3, to=4-3]
	\arrow[from=4-3, to=5-3]
	\arrow[from=1-1, to=2-1]
	\arrow[from=2-1, to=3-1]
	\arrow[from=3-1, to=4-1]
	\arrow[from=4-1, to=5-1]
	\arrow[from=1-5, to=2-5]
	\arrow[from=2-5, to=3-5]
	\arrow[from=3-5, to=4-5]
	\arrow[from=4-5, to=5-5]
\end{tikzcd}

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

Note that the complex in the middle is not the direct sum of the two outer complexes, just the terms -- the differential $d_B$ on $\cocomplex{I}_B$ will be of the form 
\[
d_B = \matt {d_A} * 0 {d_C}
.\]
:::

:::{.exercise title="?"}
Prove this, using that additive functors preserve direct sums.
Conclude using that this construction yields a SES of complexes $0\to F\cocomplex{I}_A \to F\cocomplex{I}_B\to F\cocomplex{I}_C\to 0$.
:::

:::{.exercise title="?"}
Prove that if $I$ is injective then $0\to I\to B\to C\to 0$ splits by explicitly constructing a left and right splitting to show that $B$ satisfies the universal property of the biproduct.
Show also that the same conclusion holds for $0\to A\to B\to P\to 0$ with $P$ projective.
:::