# Tuesday, February 08

:::{.remark}
Defining derived functors: for $F$ an additive functor and $M\in \rmod$, take a projective resolution and apply $F$:
\[
\cdots \to C_2 \mapsvia{d_2} C_1 \mapsvia{d_1} C_0 \mapsvia{\eps = d_0} M \to 0 \leadsto F(C_2) \mapsvia{Fd_2} F(C_1) \mapsvia{Fd_1} \cdots
,\]
so $\complex{C} \covers F$.

Define the left-derived functor 
\[
\LL F M \da H_n F\complex{C}
.\]
:::

:::{.remark}
Any $\mu \in \rmod(M, M')$ induces a chain map $\hat \alpha \in \Ch\rmod(H_* F\complex C, H_* F\complex{C}' )$, where $\alpha$ is any lift of $\mu$ to their resolutions.

\begin{tikzcd}
	{\complex{C}} && M \\
	\\
	{\complex{C}'} && {M'}
	\arrow["\eps", Rightarrow, from=1-1, to=1-3]
	\arrow["\mu", from=1-3, to=3-3]
	\arrow["{\eps'}"', Rightarrow, from=3-1, to=3-3]
	\arrow["\alpha"', from=1-1, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJNIl0sWzIsMiwiTSciXSxbMCwyLCJcXGNvbXBsZXh7Q30nIl0sWzAsMCwiXFxjb21wbGV4e0N9Il0sWzMsMCwiXFxlcHMiLDAseyJsZXZlbCI6Mn1dLFswLDEsIlxcbXUiXSxbMiwxLCJcXGVwcyciLDIseyJsZXZlbCI6Mn1dLFszLDIsIlxcYWxwaGEiLDJdXQ==)

:::

:::{.exercise title="?"}
Show that any two lifts $\alpha, \alpha'$ induce the same map on homology.
:::

:::{.remark}
Similarly, $\LL F(M)$ does not depend on the choice of resolution:

\begin{tikzcd}
	{\complex{C}} && M &&&& {F\complex{C}} && {F(M)} \\
	\\
	{\complex{C}'} && M & \leadsto &&& {F\complex{C}'} && {F(M)} \\
	\\
	{\complex{C}} && M &&&& {F\complex{C}} && {F(M)}
	\arrow["{\id_M}", from=1-3, to=3-3]
	\arrow["{\id_M}", from=3-3, to=5-3]
	\arrow["\alpha", from=1-1, to=3-1]
	\arrow["\beta", from=3-1, to=5-1]
	\arrow["\eps", from=5-1, to=5-3]
	\arrow["\eps", from=3-1, to=3-3]
	\arrow["\eps", from=1-1, to=1-3]
	\arrow[from=1-9, to=3-9]
	\arrow[from=3-9, to=5-9]
	\arrow[from=5-7, to=5-9]
	\arrow[from=3-7, to=3-9]
	\arrow[from=1-7, to=1-9]
	\arrow["{F(\alpha)}", from=1-7, to=3-7]
	\arrow["{F(\beta)}"', from=3-7, to=5-7]
	\arrow["{\id_{\complex{C}}}"', curve={height=30pt}, from=1-1, to=5-1]
	\arrow["{\therefore \id_{F \complex{C}}}"', curve={height=30pt}, dashed, from=1-7, to=5-7]
\end{tikzcd}

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

:::

:::{.definition title="Projective resolution of a SES"}
For $0\to M' \to M\to M'' \to 0$ in $\cat{C}$, a **projective resolution** is a collection of chain maps forming projective resolutions of each of the constituent modules:

\begin{tikzcd}
	0 && {\complex{C}'} && {\complex{C}} && {\complex{C}''} && 0 \\
	\\
	0 && {M'} && M && {M''} && 0
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=3-7, to=3-9]
	\arrow[from=1-7, to=1-9]
	\arrow[from=1-5, to=1-7]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-1, to=1-3]
	\arrow[Rightarrow, from=1-3, to=3-3]
	\arrow[Rightarrow, from=1-5, to=3-5]
	\arrow[Rightarrow, from=1-7, to=3-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMiwiMCJdLFsyLDIsIk0nIl0sWzQsMiwiTSJdLFs2LDIsIk0nJyJdLFs4LDIsIjAiXSxbMCwwLCIwIl0sWzIsMCwiXFxjb21wbGV4e0N9JyJdLFs0LDAsIlxcY29tcGxleHtDfSJdLFs2LDAsIlxcY29tcGxleHtDfScnIl0sWzgsMCwiMCJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs4LDldLFs3LDhdLFs2LDddLFs1LDZdLFs2LDEsIiIsMSx7ImxldmVsIjoyfV0sWzcsMiwiIiwxLHsibGV2ZWwiOjJ9XSxbOCwzLCIiLDEseyJsZXZlbCI6Mn1dXQ==)


:::

:::{.exercise title="?"}
Show that such resolutions exist.
This involves constructing $\eps: C_0 \to M$:

\begin{tikzcd}
	0 && {C_0'} && {C \cong C_0' \oplus C_0''} && {C_0''} && 0 \\
	\\
	0 && {M'} && M && {M''} && 0
	\arrow[from=3-1, to=3-3]
	\arrow["\gamma", hook, from=3-3, to=3-5]
	\arrow["\sigma", two heads, from=3-5, to=3-7]
	\arrow[from=3-7, to=3-9]
	\arrow[from=1-7, to=1-9]
	\arrow["{p_0}", two heads, from=1-5, to=1-7]
	\arrow["{\iota_0}", hook, from=1-3, to=1-5]
	\arrow[from=1-1, to=1-3]
	\arrow["\eps"', from=1-3, to=3-3]
	\arrow["{\therefore \exists \eps}"', dashed, from=1-5, to=3-5]
	\arrow["{\eps''}"', two heads, from=1-7, to=3-7]
	\arrow["{\exists \eps^*}"', dashed, from=1-7, to=3-5]
\end{tikzcd}

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

The claim is that $\eps(x, x'') \da \gamma \eps'(x') + \eps^*(x'')$ works.
To prove surjectivity, use the following:
:::

:::{.proposition title="Short Five Lemma"}
Given a commutative diagram of the following form

\begin{tikzcd}
	0 && A && B && C && 0 \\
	\\
	0 && {A'} && {B'} && {C'} && 0
	\arrow[from=3-1, to=3-3]
	\arrow["s", from=3-3, to=3-5]
	\arrow["t", from=3-5, to=3-7]
	\arrow[from=3-7, to=3-9]
	\arrow[from=1-1, to=1-3]
	\arrow["p", from=1-3, to=1-5]
	\arrow["q", from=1-5, to=1-7]
	\arrow[from=1-7, to=1-9]
	\arrow["h"', from=1-7, to=3-7]
	\arrow["g"', from=1-5, to=3-5]
	\arrow["f"', from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTAsWzAsMCwiMCJdLFsyLDAsIkEiXSxbNCwwLCJCIl0sWzYsMCwiQyJdLFs4LDAsIjAiXSxbMCwyLCIwIl0sWzIsMiwiQSciXSxbNCwyLCJCJyJdLFs2LDIsIkMnIl0sWzgsMiwiMCJdLFs1LDZdLFs2LDcsInMiXSxbNyw4LCJ0Il0sWzgsOV0sWzAsMV0sWzEsMiwicCJdLFsyLDMsInEiXSxbMyw0XSxbMyw4LCJoIiwyXSxbMiw3LCJnIiwyXSxbMSw2LCJmIiwyXV0=)

If $g,h$ are mono (resp. epi, resp. iso) then $f$ is mono (resp. epi, resp. iso).
:::

:::{.proof title="of surjectivity, alternative by diagram chase"}
\envlist

- Let $x\in M$
- Set $y=\sigma(x)$
- Find $z\in C_0$ such that $\eps'' p_0 (z) = y$.
- Consider $\eps(z) - x$ and apply $\sigma$:
\[
\sigma(\eps(z) - x) 
&= \sigma \eps(x) - \sigma(x) \\
&= \eps'' p_0(x) - \sigma(x) \\
&= y-y \\
&= 0
.\]

- So $\eps(z) - x\in \ker \sigma = \im \gamma$
- Pull back to $w\in C_0'$ such that $\gamma \eps'(w) = \eps(z) - x$
- Check $\eps i_0 (w) = \gamma \eps'(w) = \eps(z) - x$, so $\eps(i_0(w) - z) = -x$.
:::

:::{.proof title="of existence"}
The setup:

\begin{tikzcd}
	&& 0 && 0 && 0 \\
	\\
	0 && {\ker \eps'} && {\ker \eps} && {\ker \eps''} && 0 \\
	\\
	0 && {C_0'} && {C_0} && {C_0''} && 0 \\
	\\
	0 && {M'} && M && {M''} && 0
	\arrow[from=7-1, to=7-3]
	\arrow["\gamma", hook, from=7-3, to=7-5]
	\arrow["\sigma", two heads, from=7-5, to=7-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=5-7, to=5-9]
	\arrow["{p_0}", two heads, from=5-5, to=5-7]
	\arrow["{\iota_0}", hook, from=5-3, to=5-5]
	\arrow[from=5-1, to=5-3]
	\arrow["{\eps'}"', from=5-3, to=7-3]
	\arrow["{\therefore \exists \eps}"', dashed, from=5-5, to=7-5]
	\arrow["{\eps''}"', two heads, from=5-7, to=7-7]
	\arrow["{\exists \eps^*}"', dashed, from=5-7, to=7-5]
	\arrow[from=3-1, to=3-3]
	\arrow["f", hook, from=3-3, to=3-5]
	\arrow["g", from=3-5, to=3-7]
	\arrow[from=3-7, to=3-9]
	\arrow[hook, from=3-7, to=5-7]
	\arrow[hook, from=3-5, to=5-5]
	\arrow[hook, from=3-3, to=5-3]
	\arrow[from=1-3, to=3-3]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-7, to=3-7]
\end{tikzcd}

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

This is exact and commutative by a diagram chase:

- $f = i \circ \downarrow_{\ker \eps'}$ shows $g(\ker \eps) \subseteq \ker \eps''$
- $g = p \circ \downarrow_{\ker \eps}$ shows $f(\ker \eps') \subseteq \ker \eps$.


To show exactness along the top line:

- $f$ is injective, since it's the restriction of an injective map.
- $g$ is surjective: 
  - Let $x\in \ker \eps''$, so $\eps''(x) = 0$.
  - $\exists y\in C_0$ with $p_0(y) = x$ by surjectivity of $p_0$.
  - Check $\eps''(p_0(y)) = \eps(x) = 0$ in $M''$, so $\sigma\eps(y) = 0$
  - Thus $\eps(y)\in \ker \sigma = \im \gamma$
  - By surjectivity there exists $w \in C_0'$ such that $\gamma( \eps'(w)) = \eps(y)$.
  - Use commutativity to verify 
\[
\eps(i_0(w) - y) 
&= \eps(i_0(w)) - \eps(y) \\
&= \gamma\eps'(w) - \eps(y) \\
&= \eps(y) - \eps(y) \\
&= 0
.\]
  - Then
\[
g(i_0(w) - y) 
&= p_0(i_0 (w)) - g(y) \\
&= -g(y) \\
&= -p_0(y) \\
&= -x
.\]

- Exactness at the middle, i.e. $\im f = \ker g$:
  - $\im f \subseteq \ker g$ by exactness of the second row, so it STS $\ker g \subseteq \im f$.
  - Let $y\in \ker g$, then by commutativity $y\in \ker p_0 = \im i_0$. Note that $y\in \ker \eps$ by definition.
  - Write $y = i_0(x)$ for some $x\in C_0'$
  - Note $\gamma \eps' (x) = \eps i_0(x) = \eps(y) = 0$ since $y\in \ker \eps$.
  - Since $\gamma'$ is mono, $\eps'(x) = 0$, so $y = i_0(x) = f(x)$.

:::

:::{.proposition title="?"}
For $F: \rmod\to\zmod$ additive and a SES
\[
\xi: 0\to M' \mapsvia{f} M \mapsvia{g} M'' \to 0
,\]

note that there are morphisms
\[
\LL F M'' \to \LL F M\to \LL FM'
.\]


There is a connecting morphism 
\[
\Delta: \LL F M'' \to \Sigma^{-1} \LL F M'
,\]
which in components looks like

\begin{tikzcd}
	0 && {\LL_0 F(M'')} && {\LL_0 F(M)} && {\LL_0 F(M')} \\
	\\
	&& {\LL_1 F(M'')} && {\LL_1 F(M)} && {\LL_1 F(M')} \\
	\\
	&& {\LL_2 F(M'')} && \cdots
	\arrow[from=1-3, to=1-1]
	\arrow[from=1-5, to=1-3]
	\arrow[from=1-7, to=1-5]
	\arrow[from=3-3, to=1-7]
	\arrow[from=3-5, to=3-3]
	\arrow[from=3-7, to=3-5]
	\arrow[from=5-3, to=3-7]
	\arrow[from=5-5, to=5-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMCwwLCIwIl0sWzIsMCwiXFxMTF8wIEYoTScnKSJdLFs0LDAsIlxcTExfMCBGKE0pIl0sWzYsMCwiXFxMTF8wIEYoTScpIl0sWzIsMiwiXFxMTF8xIEYoTScnKSJdLFs0LDIsIlxcTExfMSBGKE0pIl0sWzYsMiwiXFxMTF8xIEYoTScpIl0sWzIsNCwiXFxMTF8yIEYoTScnKSJdLFs0LDQsIlxcY2RvdHMiXSxbMSwwXSxbMiwxXSxbMywyXSxbNCwzXSxbNSw0XSxbNiw1XSxbNyw2XSxbOCw3XV0=)

:::