Question 43 in section 2.2 of Hatcher’s Algebraic Topology

Question

I'm not sure about a definition in this problem. Here is the question:

$\textbf{43.}$ (a) Show that a chain complex of free abelian groups $C_n$ splits as a direct sum of subcomplexes $0 \to L_{n+1} \to K_n \to 0$ with at most two nonzero terms. [Show the short exact sequence $0 \to \text{Ker } \partial \to C_n \to \text{Im } \partial \to 0$ splits and take $K_n = \text{Ker } \partial$.]

So I was able to do the hint, but I don't know what it means for a chain complex to split as the direct sum of other complexes. I was hoping someone could clarify what this means. My apologies if this definition is in Hatcher, I tried finding it but I couldn't.

Answer 1

There is a bit of a clash of variables in this problem.

Let me restate the problem somewhat differently, peeling apart the notation and indexing things a bit differently.

Given any chain complex $\mathcal C = (C_n)$ of free abelian groups, show for each $m$ there exists subgroups $K_m \subset C_m$ and $L_{m+1} \subset C_{m+1}$, such that $1 \mapsto L_{m+1} \mapsto K_m \mapsto 1$ is a subcomplex that we shall denote $\mathcal D_m$, and such that $\mathcal C$ is the direct sum $\oplus_{m} \mathcal D_m$.

The thing to keep in mind here is that the concept of "direct sum" has a meaning in the category of chain complexes that respects the structure of the chain complex, in particular the terms of the direct sum must themselves be "subobjects", which is to say sub chain complexes. So, in essence, the comment of @Berci is correct, if taken together with the prescription stated in the problem that $1 \mapsto L_{m+1} \mapsto K_m \mapsto 1$ is a subcomplex.