Let $A,B$ be chain complexes with the $A_n,B_n$ being objects of an abelian category.
Proof that $H_n(A \oplus B) = H_n(A) \oplus H_n(B)$, where $A \oplus B$ shall denote the fact that the product is also the coproduct.
Here is my proof:
We will only use the fact that (co-)limits commute with other (co-)limits.
First, notice that
$$
B_n(A\oplus B)
= \text{im}(d_n^A \oplus d_n^B)
= \text{ker}(\text{coker}(d_n^A \oplus d_n^B))
= \text{ker}(\text{coker}(d_n^A \amalg d_n^B))
= \text{ker}(\text{coker}(d_n^A) \amalg \text{coker}(d_n^B))
= \text{ker}(\text{coker}(d_n^A) \sqcup \text{coker}(d_n^B))
= \text{ker}((\text{coker}(d_n^A)) \sqcup \text{ker}(\text{coker}(d_n^B)))
= B_n(A) \oplus B_n(B) \; .
$$
This should also hint how we proof that $Z_n(A \oplus B) = Z_n(A) \oplus Z_n(B)$.
Now we can use the same game of switching between product and coproduct to get around the quotient (which is a colimit).
I encountered this exercise in Charles Weibel's "Homological algebra" (exercise 1.2.1). He actually wants us to show this fact for chain complexes of $R-$modules, but that seems to be quite an unnecessary restriction. My problem with this proof however is that it is kind of "brutal" and does not give us much inside into $H_n$.
The essential problem that makes this exercise so tiresome seems to be the fact that $H_n$ is defined as a big matryoshka doll of limits and colimits that can not be simplified (or that at least I have not been able to simplify).
Here is my question: Does anyone know a more direct and conceptual proof of this fact? I've seen a few similar ideas to mine in other posts, but I'm not particularly satisfied with either of them.
My optimal proof would be something like this:
- We identify the core properties of $H_n$.
- We show that the properties found in step 1 uniquely determine $H_n$ (so that we can get rid of the definition of $H_n$ that I did not like to begin with).
- The new definition makes our fact trivial OR the new definition at least hints a more conceptual proof.