TITLE: Non-split chain complex which is chain-homotopy equivalent to its homology sequence
QUESTION [8 upvotes]: This is exercise 1.4.4 from Weibel. Consider the homology $H_*(C)$ of chain complex $C$ as a chain complex with zero differentials. It is easy to show that if C is split, then there is a chain homotopy equivalence between $C$ and $H_*(C)$. But how to give an example in which the converse is false?
I think in the counterexample we should have that $H_n(C)$ is a direct summand of $C_n$, because of zero differentials in $H_*(C)$. I thought of something like
$$
0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\oplus\mathbb{Z}/2\rightarrow\mathbb{Z}/2\rightarrow 0,
$$
where the first non-trivial arrow maps $1$ to $(2,0)$, and the second one $(1,0)$ to $1$ and $(0,1)$ to $0$. However, it does not work.

REPLY [6 votes]: This problem in Weibel actually appears to be a mistake. If $C$ and $H_*(C)$ are homotopy equivalent, $C$ is split. I was stuck on the same thing for a while, but resolved it by asking this question. See the question itself for a proof, and the accepted answer for a link to Weibel's website where the mistake is in a list of errata.