# The structure of contractible complexes

Let $$\mathcal{C}$$ be an additive category. We denote the category of cochain complexes in $$\mathcal{C}$$ by $$CH(\mathcal{C})$$. For $$X, Y \in CH(\mathcal{C})$$, define $$X \oplus Y$$ to be the complex where each term is the direct sum of the corrsponding terms of $$X$$ and $$Y$$.

A complex $$X \in CH(\mathcal{C})$$ is contractible if the identity morphism of $$X$$ is null-homotopic or, equivalently, $$X$$ is isomorphic to the zero complex in $$H(\mathcal{C})$$, where $$H(\mathcal{C})$$ is the homotopy category of $$\mathcal{C}$$.

We can see complexes of the form $$\cdots \rightarrow A \overset{1_A}{\rightarrow} A \rightarrow 0 \rightarrow \cdots$$ are contracible. Denote the set of complexes of the above form by $$S$$. My question is that whether each contractible complex is the direct sum of some complexes in $$S$$? Thank you for your help.

It is true if idempotents split in $$\mathcal{C}$$ (i.e., for every idempotent endomorphism $$e$$ of an object $$M$$ of $$\mathcal{C}$$, $$M$$ decomposes as a direct sum $$\text{im}(e)\oplus\ker(e)$$). For if $$X$$ is contractible with contracting homotopy $$h$$, then $$h^{n+1}d^n$$ is an idempotent, and so $$X^n=\ker(h^{n+1}d^n)\oplus\text{im}(h^{n+1}d^n)$$. It is also easy to check that $$\ker(d^n)=\ker(h^{n+1}d^n)$$, so that $$X$$ is the direct sum of complexes $$\dots\to0\to\text{im}(h^{n+1}d^n)\stackrel{d^n}{\to}\ker(d^{n+1})\to0\to\dots,$$ where the nonzero differential is an isomorphism, with inverse induced by $$h^{n+1}$$.
If idempotents don't split, then it's easy to construct counterexamples, but only because of the lack of splitting. For example, if $$\mathcal{C}$$ is the category of even-dimensional vector spaces over a field $$k$$, then $$\dots\to k^2\to k^2\to k^2\to k^2\to\dots,$$ with all differentials given by $$\pmatrix{0&1\\0&0}$$, is a counterexample.