# Lecture 4 (Friday, January 22)

## Long Exact Sequences

:::{.remark}
Some terminology: in an abelian category \( \mathcal{A}  \) an example of an **exact complex**  in \( \Ch(\mathcal{A})  \) is
\[
\cdots \to 0 \to A \mapsvia{f} B \mapsvia{g} C \to 0 \to \cdots
.\]

where *exactness* means $\ker = \im$ at each position, i.e. $\ker f = 0, \im f = \ker g, \im g = C$.
We say $f$ is monic and $g$ epic.

As a special case, if $0\to A\to 0$ is exact then $A$ must be zero, since the image of the incoming map must be 0.
This also happens when every other term is zero.
If $0\to A \mapsvia{f} B \to 0$, then $A \cong B$ since $f$ is both injective and surjective (say for \(R\dash\)modules).

:::

:::{.theorem title="Long Exact Sequences"}
Suppose $0\to A\to B \to C \to 0$ is a SES in \( \Ch(\mathcal{A}) \) (note: this is a sequence of *complexes*), then there are natural maps 
\[
\delta: H_n(C) \to H_{n-1}(A)
\]
called **connecting morphisms** which decrease degree such that the following sequence is exact:

\begin{tikzcd}
	& \cdots & {H_{n+1}(C)} \\
	\\
	{H_n(A)} & {H_n(B)} & {H_n(C)} \\
	\\
	{H_{n-1}(A)} & \cdots
	\arrow["{f_* = H_n(f)}", from=3-1, to=3-2]
	\arrow["{g_* = H_n(g)}", from=3-2, to=3-3]
	\arrow["\delta", from=1-3, to=3-1, in=180, out=360]
	\arrow["\delta", from=3-3, to=5-1, in=180, out=360]
	\arrow[from=5-1, to=5-2]
	\arrow[from=1-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwyLCJIX24oQSkiXSxbMSwyLCJIX24oQikiXSxbMiwyLCJIX24oQykiXSxbMiwwLCJIX3tuKzF9KEMpIl0sWzAsNCwiSF97bi0xfShBKSJdLFsxLDQsIlxcY2RvdHMiXSxbMSwwLCJcXGNkb3RzIl0sWzAsMSwiZl8qID0gSF9uKGYpIl0sWzEsMiwiZ18qID0gSF9uKGcpIl0sWzMsMCwiXFxkZWwiXSxbMiw0LCJcXGRlbCJdLFs0LDVdLFs2LDNdXQ==)

This is referred to as the **long exact sequence in homology**.
Similarly, replacing chain complexes by cochain complexes yields a similar connecting morphism that increases degree.

> Note on notation: some books use $\bd$ for homology and $\delta$ for cohomology.

:::

The proof that this sequence exists is a consequence of the *snake lemma*.

:::{.lemma title="The Snake Lemma"}
The sequence highlighted in red in the following diagram is exact:

\begin{tikzcd}[column sep=tiny]
	0 && {{\color{red}\ker(f)}} && {{\color{red}\ker(\alpha)}} && {{\color{red}\ker(\beta)}} && {{\color{red}\ker(\gamma)}} \\
	\\
	&& 0 && A && B && C && 0 \\
	&&&&&&& {} \\
	&& 0 && {A'} && {B'} && {C'} && 0 \\
	\\
	&&&& {{\color{red}\coker(\alpha)}} && {{\color{red}\coker(\beta)}} && {{\color{red}\coker(\gamma)}} && {{\color{red}\coker(g')}} && 0
	\arrow[from=5-3, to=5-5]
	\arrow["{f'}"', from=5-5, to=5-7]
	\arrow["{g'}"', from=5-7, to=5-9]
	\arrow["f", from=3-5, to=3-7]
	\arrow["g", from=3-7, to=3-9]
	\arrow[from=3-3, to=3-5]
	\arrow["\beta"', from=3-7, to=5-7]
	\arrow["\gamma"', from=3-9, to=5-9]
	\arrow["\alpha"', from=3-5, to=5-5]
	\arrow[from=7-5, to=7-7]
	\arrow[from=7-7, to=7-9]
	\arrow[from=7-9, to=7-11]
	\arrow[from=7-11, to=7-13]
	\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=1-9, to=7-5, in=180, out=360, "\exists \delta", color=red, dotted]
	\arrow[from=5-9, to=5-11]
	\arrow[from=3-9, to=3-11]
	\arrow[from=1-5, to=3-5]
	\arrow[from=1-7, to=3-7]
	\arrow[from=1-9, to=3-9]
	\arrow[from=5-5, to=7-5]
	\arrow[from=5-7, to=7-7]
	\arrow[from=5-9, to=7-9]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMjEsWzQsMiwiQSJdLFs2LDIsIkIiXSxbOCwyLCJDIl0sWzQsNCwiQSciXSxbNiw0LCJCJyJdLFs4LDQsIkMnIl0sWzIsNCwiMCJdLFsyLDIsIjAiXSxbNCwwLCJ7XFxjb2xvcntyZWR9XFxrZXIoXFxhbHBoYSl9Il0sWzIsMCwie1xcY29sb3J7cmVkfVxca2VyKGYpfSJdLFs2LDAsIntcXGNvbG9ye3JlZH1cXGtlcihcXGJldGEpfSJdLFs4LDAsIntcXGNvbG9ye3JlZH1cXGtlcihcXGdhbW1hKX0iXSxbNCw2LCJ7XFxjb2xvcntyZWR9XFxjb2tlcihcXGFscGhhKX0iXSxbNiw2LCJ7XFxjb2xvcntyZWR9XFxjb2tlcihcXGJldGEpfSJdLFs4LDYsIntcXGNvbG9ye3JlZH1cXGNva2VyKFxcZ2FtbWEpfSJdLFsxMCw2LCJ7XFxjb2xvcntyZWR9XFxjb2tlcihnJyl9Il0sWzAsMCwiMCJdLFsxMiw2LCIwIl0sWzcsM10sWzEwLDIsIjAiXSxbMTAsNCwiMCJdLFs2LDNdLFszLDQsImYnIl0sWzQsNSwiZyciXSxbMCwxLCJmIl0sWzEsMiwiZyJdLFs3LDBdLFsxLDQsIlxcYmV0YSIsMV0sWzIsNSwiXFxnYW1tYSIsMV0sWzAsMywiXFxhbHBoYSIsMV0sWzEyLDEzXSxbMTMsMTRdLFsxNCwxNV0sWzE1LDE3XSxbMTYsOV0sWzksOF0sWzgsMTBdLFsxMCwxMV0sWzExLDEyXSxbNSwyMF0sWzIsMTldXQ==)


:::

:::{.proof title="of the Snake Lemma: Existence"}
\envlist

- Start with $c\in \ker(\gamma) \leq C$, so $\gamma(c) = 0 \in C'$
- **Choose** $b\in B$ by surjectivity
  - We'll show it's independent of this choice.
- Then $b'\in B'$ goes to $0\in C'$, so $b' \in \ker (B' \to C')$
- By exactness, $b' \in \ker (B' \to C') = \im(A'\to B')$, and now produce a unique $a'\in A'$ by injectivity
- Take the image $[a']\in \coker \alpha$
- Define $\bd(c) \da [a']$.

:::

:::{.proof title="of the Snake Lemma: Uniqueness"}
\envlist

- We chose $b$, suppose we chose a different $\tilde b$.
- Then $\tilde b - b \mapsto c-c = 0$, so the difference is in $\ker g = \im f$.
- Produce an $\tilde a\in A$ such that $\tilde a\mapsto \tilde b - b$
- Then \( \bar a \da \alpha(\tilde a) \), so apply $f'$.
- Define $\beta(\tilde b) = \tilde b ' \in B$.
- Commutativity of the LHS square forces $\tilde a'\mapsto \tilde b' - b'$.
- Then $\bar a + a' \mapsto \tilde b' -b' + b' = \tilde b'$.
- So $\tilde a' + a'$ is the desired pullback of $\tilde b'$
- Then take $[\tilde a'] \in \coker \alpha$; are $a', \tilde a'$ in the same equivalence class?
- Use that fact that $\tilde a = a' + \bar a$, where $\bar a \in \im \alpha$, so $[\tilde a] = [a' + \bar a] = [a'] \in \coker \alpha \da A'/\im \alpha$.


:::

\todo[inline]{A few changes in the middle, redo!}

:::{.proof title="of the Snake Lemma: Exactness"}
\envlist

- Let's show $g: \ker \beta\to \ker \gamma$.
  
  - Let \( b \in \ker \beta \), then consider \( \gamma(g(\beta)) = g'(\beta(b)) = g'(0) = 0 \) and so \( g(b) \in \ker \gamma \).
  
- Now we'll show $\im(\ro{g}{\ker \beta}) \subseteq \ker \delta$
  - Let \( b \in \ker \beta, c = g(b) \), then how is \( \delta(c) \) defined?
  - Use this $b$, then apply \( \beta \) to get \( b' = \beta(b) = 0 \) since \( b \in \ker \beta \).
  - So the unique thing mapping to it $a'$ is zero, and thus $[a'] = 0 = \delta(c)$.

- \( \ker \delta \subseteq \im( \ro{g}{ \ker \beta} ) \) 

  - Let \( c\in \ker \delta \), then \( \delta(c) = 0 = [a'] \in \coker \alpha \) which implies that \( a' \in \im \alpha \).
  - Write \( a' = \alpha(a) \), then \( \beta(b) = b' = f'(a') = f'( \alpha(a)) \) by going one way around the LHS square, and is equal to \( \beta(f(a)) \) going the other way.
  - So \( \tilde b \da b - f(a) \in \ker \beta \), since \( \beta(b) = \beta(f(a)) \) implies their difference is zero.
  - Then $g(\tilde b) = g(b) - g(f(a)) = g(b) = c$, which puts $c\in g(\ker \beta)$ as desired.


:::

:::{.exercise title="?"}
Show exactness at the remaining places -- the most interesting place is at $\coker \alpha$.
Also check that all of these maps make sense.
:::

:::{.remark}
We assumed that \( \mathcal{A}= \rmod  \) here, so we could chase elements, but this happens to also be true in any abelian category \( \mathcal{A}  \) but by a different proof.
The idea is to embed \( \mathcal{A} \to \rmod  \) for some ring $R$, do the construction there, and pull the results back -- but this doesn't quite work!
\( \mathcal{A}  \) can be too big.
Instead, do this for the smallest subcategory \( \mathcal{A}_0  \) containing all of the modules and maps involved in the snake lemma.
Then \( \mathcal{A}_0 \) is small enough to embed into $\rmod$ by the **Freyd-Mitchell Embedding Theorem**.
:::