June 9th
Today I learned a little about group homology and chain complexes, from Napkin . Here is our main definition.
Definition. A chain complex is a sequence of (usually abelian) groups $\{A_k\}_{k\in\NN}$ with maps $\varphi_n:A_{n+1}\to A_n$ for which the chain \[\cdots\stackrel{\varphi_{n+1}}\longrightarrow A_{n+1}\stackrel{\varphi_n}\longrightarrow A_n\stackrel{\varphi_{n-1}}\longrightarrow A_{n-1}\to\cdots\] gives $\varphi_n\circ\varphi_{n+1}$ as the trivial group homomorphism.
The geometric way to view this is to take $\{A_k\}$ equal to $k$-chains of a topological space $X$ with $\varphi_\bullet$ equal to the boundary operator $\del.$ Then we can say things like "an $n$-cycle is an element of the kernel of $\del_A:A_n\to A_{n-1}$'' and "an $n$-boundary is an element of the image of $\del_A:A_{n+1}\to A_n.$'' This perspective leads us to the definition of homology groups.
Definition. Given a chain complex $A_\bullet,$ we define the $n$th homology group $H_n(A)$ as the set of $n$-cycles modulo the $n$-boundaries. That is, \[H_n(A_\bullet):=\op{ker}\left(A_n\to A_{n-1}\right)/\op{im}\left(A_{n+1}\to A_n\right).\]
As a more algebraic way to justify homology groups, observe that we can take a long exact sequence\[\cdots\stackrel{\varphi_{n+1}}\longrightarrow A_{n+1}\stackrel{\varphi_n}\longrightarrow A_n\stackrel{\varphi_{n-1}}\longrightarrow A_{n-1}\to\cdots\]satisfying $\op{im}(\varphi_{n+1})=\op{ker}(\varphi_n)$ and split this up into a bunch of connected short exact sequences of the form\[0\to\op{coker}(\varphi_{n+2})\stackrel{\varphi_{n+1}}\to A_{n+1}\stackrel{\varphi_n}\to\op{im}(\varphi_n)\to0.\]The right half of this sequence is exact because $\varphi_n$ is of course surjective onto $\op{im}(\varphi_n).$ However, the main point is that\[\op{coker}(\varphi_{n+2}):=A_{n+2}/\op{im}(\varphi_{n+2}).\]We are not immediately guaranteed that $\varphi_{n+1}:A_{n+2}\to A_{n+1}$ is well-defined over $A_{n+2}/\op{im}(\varphi_{n+2}),$ but it is exactly because $\op{im}(\varphi_{n+2})=\op{ker}(\varphi_{n+1}).$ More explicitly, a coset $a_{n+2}+\op{im}(\varphi_{n+2})$ can simply be taken to $\varphi_{n+1}(a_{n+2})$ because $\varphi_{n+1}(\op{im}(\varphi_{n+2}))=\{e\}.$
To finish, we note that the left half of the sequence is exact because $\varphi_{n+1}$ is injective on $A_{n+2}/\op{im}(\varphi_{n+2})=A_{n+2}/\op{ker}(\varphi_{n+1})$ because it has trivial kernel after modding out by its old kernel. And the image of $\varphi_{n+1}$ doesn't change—we are still more or less pushing all $A_{n+2}$ through—so $\varphi_{n+1}$ continues to surject onto the kernel of $\varphi_n,$ finishing the required exactness.
All of this is to say that long exact sequences are nice: we can study them in terms of short exact sequences, which are simpler because they are shorter. However, our chain complex as defined above does not necessarily have $\op{im}(\varphi_{n+1})=\op{ker}(\varphi_n)$ but merely $\op{im}(\varphi_{n+1})\subseteq\op{ker}(\varphi_n).$ So we define\[H_n(A_\bullet):=\op{ker}(\varphi_{n-1})/\op{im}(\varphi_n)\]to measure how badly we failed around $A_n.$
While we're here, we note that we can turn the set of chain complexes into a category. Here are our morphisms.
Definition. Given a chain complexes $A_\bullet$ and $B_\bullet,$ we define a morphism $f_\bullet:A_\bullet\to B_\bullet$ to be a set of morphisms $\{f_k\}_{k\in\NN}$ that make the following (infinite) diagram commute. img1.png
While natural, the following convinces us that we really do have the correct definition.
Lemma. Given a morphism of chain complexes $f_\bullet:A_\bullet\to B_\bullet,$ we have an induced $H_n(f_\bullet):H_n(A_\bullet)\to H_n(B_\bullet).$
The point is that everything commutes, so the kernels and images behave. Starting from the top, note that we have\[f_n:A_n\to B_n.\]Now, surely $\op{ker}(\varphi_{n-1})\subseteq\op{ker}(f_{n-1}\circ\varphi_{n-1})$ as well, but $f_{n-1}\circ\varphi_{n-1}=\gamma_{n-1}\circ f_n,$ so we see the downward arrow $f_n$ induces a map from $\op{ker}(\varphi_n)$ to $\op{ker}(\gamma_n)$: $\varphi_{n-1}(a)=e$ implies $f_{n-1}(\varphi_{n-1}(a))=e$ implies $\gamma_{n-1}(f_n(a))=e.$
Similarly,\[f_n:A_n\to B_n\]induces a map $f_n:\op{im}(\varphi_n)\to B_n$ by restriction. However, $f_n\circ\varphi_n=\gamma_n\circ\varphi_{n+1},$ so the image of $A_{n+1}\to\op{im}(\varphi_n)\to B_n$ is a subset of $\op{im}(\gamma_n),$ so in fact we may restrict this to a map $f_n:\op{im}(\varphi_n)\to\op{im}(\gamma_n).$
Putting this all together, we saw that $f_n:A_n\to B_n$ restricts to a map\[f_n:\op{ker}(\varphi_{n-1})\to\op{ker}(\gamma_{n-1}).\]We still have $\op{im}(\gamma_n)\subseteq\op{ker}(\gamma_{n-1}),$ so we may safely mod out the domain here by $\op{im}(\gamma_n),$ giving an induced map\[\op{ker}(\varphi_{n-1})\to H_n(B_\bullet).\]We know that $f_n$ takes elements of $\op{im}(\varphi_n)$ to elements of $\op{im}(\gamma_n),$ so they are all going to the kernel of the above induced map. Thus, we may safely mod out by them in the codomain, giving our desired map $H_n(A_\bullet)\to H_n(B_\bullet).$ This finishes. $\blacksquare$