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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.