# Friday, March 04

:::{.remark}
Idea: regard $A$ as a chain complex supported in degree zero and $A\underset{\coveredby}{\eta} \cocomplex{I}$ an injective resolution, then the induced map $\eta^*: H^*(A)\to H^*(\cocomplex{I})$ is an isomorphism, so $A$ and $\cocomplex{I}$ are quasi-isomorphic.
:::

:::{.exercise title="?"}
Show that if $A\coveredby \cocomplex{I}, \cocomplex{J}$, then there exists a chain homotopy $f: I \homotopic J$.
:::

:::{.remark}
Hints:

\begin{tikzcd}
	A \\
	{X^0} && {I^0} & {X^1} & {I^1} & {X^2} & \bullet \\
	{Y_0} && {J^0} & {Y^1} & {J^1} & {Y^2} & \bullet \\
	B
	\arrow[Rightarrow, no head, from=1-1, to=2-1]
	\arrow[from=2-1, to=3-1]
	\arrow[Rightarrow, no head, from=3-1, to=4-1]
	\arrow[hook, from=2-1, to=2-3]
	\arrow[hook, from=3-1, to=3-3]
	\arrow[two heads, from=3-3, to=3-4]
	\arrow[two heads, from=2-3, to=2-4]
	\arrow[hook, from=2-4, to=2-5]
	\arrow[two heads, from=2-5, to=2-6]
	\arrow[hook, from=2-6, to=2-7]
	\arrow[hook, from=3-4, to=3-5]
	\arrow[two heads, from=3-5, to=3-6]
	\arrow[hook, from=3-6, to=3-7]
	\arrow[from=2-1, to=3-3]
	\arrow["{\exists \text{ since } J^0\text{ is injective}}", dashed, from=2-3, to=3-3]
\end{tikzcd}

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

Given 
\[
f^{n-1} - g^{n-1} = h^n d^{n-1} - d^{n-2} h^{n-1}
,\]
construct $h^{n+1} d_I^n$ such that
\[
(f^n - g^n) d_I^n = ( h^{n+1} d_I^n + d_J^{n-1} h^n ) d_I^n
\]
and extend arbitrarily to $h^{n+1}: I^{n+1} \to J^n$.
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:::{.exercise title="?"}
Prove the Horseshoe lemma.

:::