# Abstract Nonsense

## Exact Sequences and Homology (Pt. IV)

Point of Post: This is a continuation of this post.

Ok, now on to the general case. What the above really told us was that to prove that we get a well-defined map $H_n(\mathbf{C})\to H_n(\mathbf{D})$ we really need to show that $f_n(\text{im }\partial_{n+1})\subseteq\text{im }d_{n+1}$ and that $f_n(\ker\partial_n)\subseteq\ker d_n$. Of course, there is no containment in general abelian categories, and thus we need to replace these containments by finding canonical maps $\text{im }\partial_{n+1}\to\text{im }d_{n+1}$ and canonical maps $\ker\partial_n\to\ker d_n$.

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We begin by constructing a map $\ker \partial_n\to\ker d_n$. To do this we consider the diagram

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$\begin{array}{ccccc}\ker d_n & \xrightarrow{\ell} & D_n & \xrightarrow{d_n} & D_{n-1}\\ & & \big\uparrow_{f_n} & & \\ & & C_n & & \\ & & \big\uparrow_{v} & & \\ & & \ker\partial_n & & \end{array}$

Where it should be obvious what the maps are. Now, note that $d_n\circ f_n\circ v=f_{n-1}\circ\partial_n\circ v=0$ and so by the definition of the kernel we get a unique arrow $\ker\partial_n\xrightarrow{q_n}\ker d_n$ such that $\ell\circ q_n=f_n\circ v$.

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To construct the arrow $\text{im }\partial_{n+1}\to\text{im }d_{n+1}$ we use the trick that since we are dealing with abelian categories we only have to define a map $\text{coim }\partial_{n+1}\to\text{coim }d_{n+1}$. To do this we consider the diagram

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$\begin{array}{ccccc} \ker \partial_{n+1} & \xrightarrow{\alpha} & C_{n+1} & \xrightarrow{\beta} & \text{coim }\partial_{n+1}\\ & & \big\downarrow_{f_{n+1}} & & \\ & & D_{n+1} & & \\ & & \big\downarrow_\gamma & & \\ & & \text{coim }d_{n+1} & & \end{array}$

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So, if we can show that $\gamma\circ f_{n+1}\circ\alpha=0$. That said, we can put in the following maps

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$\begin{array}{ccccc} \ker \partial_{n+1} & \xrightarrow{\alpha} & C_{n+1} & \xrightarrow{\beta} & \text{coim }\partial_{n+1}\\ \big\downarrow_{q_{n+1}}& & \big\downarrow_{f_{n+1}} & & \\ \ker d_{n+1} & \xrightarrow{j} & D_{n+1} & & \\ & & \big\downarrow_\gamma & & \\ & & \text{coim }d_{n+1} & & \end{array}$

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Note then that by construction we have commutativity of that square and so $\gamma\circ f_{n+1}\circ\alpha=\gamma\circ j\circ q_{n+1}=0$ since $\gamma\circ j=0$ by definition. Thus, we get a well-defined arrow $\text{coim }\partial_{n+1}\xrightarrow{p_n}\text{coim }d_{n+1}$ and thus using the natural isomorphisms $\text{coim }\xrightarrow{\approx}\text{im}$ which comes from the definition of an abelian category we get maps $\text{im }\partial_{n+1}\to\text{im }d_{n+1}$. Now, finally we consider the following diagram

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$\begin{array}{ccccccc}\text{coim }\partial_{n+1} & \xrightarrow{\approx} & \text{im }\partial_{n+1} & \to & \ker \partial_n & \to & H_n(\mathbf{C})\\ \big\downarrow & & \big\downarrow & & \big\downarrow{q_n} & \\ \text{coim }d_{n+1} & \to & \text{im }d_{n+1} & \to & \ker d_n & \\ & & & & \big\downarrow & & \\ & & & & H_n(\mathbf{D}) & & \end{array}$

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This diagram is commutative, whicn enables one to quickly check that the composition

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$\text{im }\partial_{n+1}\to\ker\partial_n\xrightarrow{q_n}\ker d_n\to H_n(\mathbf{D})$

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is zero, which (by definition of cokernels) gives us a map $H_n(\mathbf{C})\to H_n(\mathbf{D})$ which we shall denote $H_n(f)$, $f_{n,\ast}$, or just $f_{\ast}$ when the $n$ is clear.

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One can check, of course, that if we are dealing with $\mathscr{A}=R\text{-}\mathbf{Mod}$ that $H_n(f)(x+\text{im }\partial_{n+1})=f_n(x)+\text{im }d_{n+1}$ as one would hope.

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It’s easy to run through the construction of $H_n(f)$ to find that $H_n(g\circ f)=H_n(g)\circ H_n(f)$ and $H_n(1_\mathbf{C})=1_{H_n(\mathbf{C})}$. Moreover, note that to compute $H_n(\mathbf{C}\oplus\mathbf{D})$ one has to find $\text{coker}(\text{im }(\partial_{n+1}\oplus\ d_{n+1})\to\ker(\partial_n\oplus d_n))$, but it’s easy to check that this is just

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$\text{coker}(\text{im }\partial_{n+1}\to \ker\partial_n)\oplus\text{coker}(\text{im }d_{n+1}\to\ker d_n)$

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Which tells us that $H_n(\mathbf{C}\oplus\mathbf{D})=H_n(\mathbf{C})\oplus H_n(\mathbf{D})$. Finally, noting the obvious fact that $H_n(\mathbf{0})=0$ we may conclude that:

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Theorem: $H_n$ is an additive functor $\mathbf{Ch}(\mathscr{A})\to\mathscr{A}$.

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We call $H_n$ the $n^{\text{th}}$ homology functor.

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Remark: The above was absolutely horrible. Namely, while doing the construction this way is nice in that it is a) actually explicit in the sense that we produce the arrows we are looking for and b) shows a very theoretic, categorical way in which $H_n(f)$ is defined it is just a horrid diagram chase. So, why did I actually do it? Due to the metatheorem concerning abelian categories one can “theoretically” just prove this result for $R\text{-}\mathbf{Mod}$ for all rings $R$, note that this construction only involves “local” ideas, and say that we can therefore pull the result from $R\text{-}\mathbf{Mod}$. While this is an invaluable tool/idea, one which we shall be using often, it is good to see (at least once) that one can (if one is determined enough) actually “construct” the maps that the metatheorem guarantees us. Anyways, besides the somewhat messy nature of the chase it is actually quite simple if one merely keeps the $R\text{-}\mathbf{Mod}$ case in mind and follows their nose.

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References

[1] Weibel, Charles A. An Introduction to Homological Algebra. Cambridge [England: Cambridge UP, 1994. Print.

[2] Schapira, Pierre. “Categories and Homological Algebra.” Web. <http://people.math.jussieu.fr/~schapira/lectnotes/HomAl.pdf&gt;.

[3] Rotman, Joseph. An Introduction to Homological Algebra. Dordrecht: Springer, 2008. Print.