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.
This is almost true.
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.