I was revising some old postgraduate notes of mine in homological algebra (written during a postgrad course on the topic, I had taken more than ten ;) years ago) and I came accross the following problem: Weibel's book "An introduction to homological algebra" (which had been among my textbooks by that time), states the following exercise: (ex. $1.4.4$):

Consider the homology $H_*(C)$ of $C$ as a chain complex with zero differentials. Show that if the complex $C$ is split, then there is a chain homotopy equivalence between $C$ and $H_*(C)$. Give an example in which the converse fails.

I've solved the exercise but i've failed to find a counterexample for the converse. In fact, I think the converse also holds, but I haven't yet found the time to clear it up. Can somebody help ?

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    $\begingroup$ Consider any short exact sequence as a complex $C$ concentrated in three consecutive degrees. If this complex is null homotopic, the same is true after applying any additive functor $F$. Yet typically the homology of $F(C)$ is nonzero (and this is the main motivation behind the development of homological algebra). $\endgroup$ – Jason Starr Jan 29 '17 at 22:15
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    $\begingroup$ @JasonStarr how does this address the question? $\endgroup$ – Gabriel C. Drummond-Cole Jan 30 '17 at 2:59
  • $\begingroup$ @GabrielC.Drummond-Cole. I misread the OP's question. Sorry about that. $\endgroup$ – Jason Starr Jan 30 '17 at 8:45

You are correct, the converse also holds. This is a mistake in Weibel, see this answer.

  • $\begingroup$ aahh... ok, i also saw the link to the corrections to the book. Thank you! $\endgroup$ – Konstantinos Kanakoglou Jan 30 '17 at 5:24

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