# Wednesday, January 27

See phone pic for missed first 10m.

## 1.4: Chain Homotopies

:::{.definition title="Split Exact"}
A complex is called **split** if there are maps $s_n: C_n \to C_{n+1}$ such that $d =dsd$.
In this case, the maps $s_n$ are referred to as the **splitting maps**, and if $C$ is additionally acyclic, we say $C$ is **split exact**. 
:::

:::{.remark}
Note that when $C$ is split exact, we have

\begin{tikzcd}
	&& {C_n} && {C_{n-1}} \\
	\\
	{C_{n+1}} && {C_n}
	\arrow["d", from=3-1, to=3-3]
	\arrow["d", from=1-3, to=1-5]
	\arrow["{s_n}"{description}, from=1-3, to=3-1]
	\arrow["{s_{n-1}}"{description}, from=1-5, to=3-3]
	\arrow["\id"{description}, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJDX24iXSxbNCwwLCJDX3tuLTF9Il0sWzIsMiwiQ19uIl0sWzAsMiwiQ197bisxfSJdLFszLDIsImQiXSxbMCwxLCJkIl0sWzAsMywic19uIiwxXSxbMSwyLCJzX3tuLTF9IiwxXSxbMCwyLCJcXGlkIiwxXV0=)


:::

:::{.example title="Not all complexes split"}
Take 
\[
C = \qty{ 0 \to \ZZ/2\ZZ \mapsvia{d} \ZZ/4\ZZ \to \ZZ/2\ZZ \to 0 }
.\]
Then $\im d = \ts{0, 2} = \ker d$, but this does not split since $\ZZ/2\ZZ^2 \not\cong \ZZ/4\ZZ$: one has an element of order 4 in the underlying additive group.
Equivalently, there is no complement to the image.
What might be familiar from algebra is $ds = \id$, but the more general notion is $dsd = d$.
:::

:::{.example title="?"}
The following complex is not split exact for the same reason:
\[
\cdots \mapsvia{\cdot 2} \ZZ/4\ZZ \mapsvia{\cdot 2} \ZZ/4\ZZ \to \cdots
.\]
:::

:::{.question}
Given $f,g: C\to D$, when do we get equality $f_* = g_*: H_*(C) \to H_*(D)$?
:::

:::{.definition title="Homotopy Terminology for Chains"}
A chain map $f:C\to D$ is **nullhomotopic** if and only if there exist maps $s_n: C_n\to D_{n+1}$ such that $f = ds + sd$:

\begin{tikzcd}
	&& {C_n} && {C_{n-1}} \\
	\\
	{D_{n+1}} && {D_n}
	\arrow["d", from=3-1, to=3-3]
	\arrow["d", from=1-3, to=1-5]
	\arrow["{s_n}"{description}, from=1-3, to=3-1]
	\arrow["{s_{n-1}}"{description}, from=1-5, to=3-3]
	\arrow["\id"{description}, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMiwwLCJDX24iXSxbNCwwLCJDX3tuLTF9Il0sWzIsMiwiRF9uIl0sWzAsMiwiRF97bisxfSJdLFszLDIsImQiXSxbMCwxLCJkIl0sWzAsMywic19uIiwxXSxbMSwyLCJzX3tuLTF9IiwxXSxbMCwyLCJcXGlkIiwxXV0=)

The map $s$ is called a **chain contraction**.
Two maps are **chain homotopic** (or initially: $f$ is chain homotopic to $g$, since we don't yet know if this relation is symmetric) if and only if $f-g$ is nullhomotopic, i.e. $f-g = ds + sd$.
The map $s$ is called a **chain homotopy** from $f$ to $g$.
A map $f$ is a **chain homotopy equivalence** if both $fg$ and $gf$ are chain homotopic to the identities on $C$ and $D$ respectively.
:::

:::{.lemma title="?"}
If map $f:C\to D$ is nullhomotopic then $f_*: H_*(C) \to H_*(D)$ is the zero map.
Thus if $f,g$ are chain homotopic, then they induce equal maps.
:::

:::{.proof title="?"}
An element in the quotient $H_n(C)$ is represented by an $n\dash$cycle $x\in Z_n(C)$.
By a previous exercise, $f(x)$ is a well-defined element of $H_n(D)$, and using that $d(x) = 0$ we have
\[
f(x) = (ds + sd)(x) = d(s(x))
,\]
and so $f[x] = [f(x)] = [0]$.

\begin{tikzcd}
	&& x && {d(x) = 0} \\
	&& {C_n} && {C_{n-1}} \\
	\\
	{D_{n+1}} && {D_n} \\
	&& {d(s(x))}
	\arrow["d", from=4-1, to=4-3]
	\arrow["d", from=2-3, to=2-5]
	\arrow["{s_n}"{description}, from=2-3, to=4-1]
	\arrow["{s_{n-1}}"{description}, from=2-5, to=4-3]
	\arrow["\id"{description}, from=2-3, to=4-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMiwxLCJDX24iXSxbNCwxLCJDX3tuLTF9Il0sWzIsMywiRF9uIl0sWzAsMywiRF97bisxfSJdLFsyLDAsIngiXSxbNCwwLCJkKHgpID0gMCJdLFsyLDQsImQocyh4KSkiXSxbMywyLCJkIl0sWzAsMSwiZCJdLFswLDMsInNfbiIsMV0sWzEsMiwic197bi0xfSIsMV0sWzAsMiwiXFxpZCIsMV1d)

Now applying the first part to $f-g$ to get the second part.
:::

> See Weibel for topological motivations.

## 1.5 Mapping Cones

:::{.remark}
Note that we'll skip *mapping cylinders*, since they don't come up until the section on triangulated categories.
The goal is to see how any two maps between homologies can be fit into a LES.
This helps reduce questions about *quasi-isomorphisms* to questions about split exact complexes.
:::

:::{.definition title="Mapping Cones"}
Suppose we have a chain map $f:B\to C$, then there is a chain complex $\cone(f)$, the **mapping cone of $f$**, defined by
\[
\cone(f)_n = B_{n-1} \oplus C_n
.\]

The maps are given by the following:

\begin{tikzcd}
	{B_{n-1}} && {B_{n-2}} \\
	\oplus && \oplus \\
	{C_n} && {C_{n-1}}
	\arrow["{-d^B}", from=1-1, to=1-3]
	\arrow["{-f}"', from=1-1, to=3-3]
	\arrow["{d^C}", from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJCX3tuLTF9Il0sWzAsMSwiXFxvcGx1cyJdLFswLDIsIkNfbiJdLFsyLDAsIkJfe24tMn0iXSxbMiwyLCJDX3tuLTF9Il0sWzIsMSwiXFxvcGx1cyJdLFswLDMsIi1kXkIiXSxbMCw0LCItZiIsMl0sWzIsNCwiZF5DIl1d)

We can write this down: $d(b, c) = (-d(b), -f(b) + d(c))$, or as a matrix
\[
\begin{bmatrix}
-d^B &  0
\\
-f & d^C
\end{bmatrix}
.\]
:::

:::{.exercise title="?"}
Check that the differential on $\cone(f)$ squares to zero.
:::

:::{.exercise title="Weibel 1.5.1"}
When $f = \id:C\to C$, we write $\cone(C)$ instead of $\cone(\id)$.
Show that $\cone(C)$ is split exact, with splitting map $s(b, c) = (-c, 0)$ for $b\in C_{n-1}, c\in C_n$.
:::

:::{.proposition title="LES in homology of a single chain map using the cone"}
Suppose $f:B\to C$ is a chain map, then the induced maps $f_*: H(B) \to H(C)$ fit into a LES.
There is a SES of chain complexes:

\begin{tikzcd}
	0 && C && {\cone(f)} && {B[-1]} && 0 \\
	&& c && {(0, c)} \\
	&&&& {(b, c)} && {-b}
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=1-7, to=1-9]
	\arrow[from=2-3, to=2-5]
	\arrow[from=3-5, to=3-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMCwwLCIwIl0sWzIsMCwiQyJdLFs0LDAsIlxcY29uZShmKSJdLFs2LDAsIkJbLTFdIl0sWzgsMCwiMCJdLFsyLDEsImMiXSxbNCwxLCIoMCwgYykiXSxbNCwyLCIoYiwgYykiXSxbNiwyLCItYiJdLFswLDFdLFsxLDJdLFsyLDNdLFszLDRdLFs1LDZdLFs3LDhdXQ==)

:::{.exercise title="?"}
Check that these are chain maps, i.e. they commute with the respective differentials $d$.
:::

The corresponding LES is given by the following:

\begin{tikzcd}
	&& \cdots && {H_{n+1}\cone(f)} && {H_{n+1}(B[-1]) = H_n(B)} \\
	\\
	{} & {} & {H_n(C)} && {H_n \cone(f)} && {H_{n}(B[-1]) = H_{n-1}(B)} \\
	\\
	&& \cdots
	\arrow[from=1-3, to=1-5]
	\arrow["{\delta_*}", from=1-5, to=1-7]
	\arrow["\bd", from=1-7, to=3-3, in=180, out=360]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=3-7, to=5-3, in=180, out=360]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbNCwwLCJIX3tuKzF9XFxjb25lKGYpIl0sWzAsMl0sWzYsMCwiSF97bisxfShCWy0xXSkgPSBIX24oQikiXSxbMSwyXSxbMiwyLCJIX24oQykiXSxbNCwyLCJIX24gXFxjb25lKGYpIl0sWzYsMiwiSF97bn0oQlstMV0pID0gSF97bi0xfShCKSJdLFsyLDAsIlxcY2RvdHMiXSxbMiw0LCJcXGNkb3RzIl0sWzcsMF0sWzAsMiwiXFxkZWx0YV8qIl0sWzIsNCwiXFxiZCJdLFs0LDVdLFs1LDZdLFs2LDhdXQ==)

\todo[inline]{Overflowing :(}


:::{.lemma title="?"}
The map $\bd = f_*$
:::

:::{.proof title="?"}
Letting $b\in B_n$ is an $n\dash$cycle.

1. Lift $b$ to anything via $\delta$, say $(-b, 0)$.
2. Apply the differential $d$ to get $(db, fb) = (0, fb)$ since $b$ was a cycle.
3. Pull back to $C_n$ by the map $C\to \cone(f)$ to get $fb$.
4. Then the connecting morphism is given by $\bd[b] = [fb]$.
  But by definition of $f_*$, we have $[fb] = f_* [b]$.
:::

:::