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Of course if two morphisms of complexes are homotopic their induced maps coincide, but I'm wondering about the converse: if the induced maps on the cohomologies coincide, when does that imply that the morphisms are homotopic?

I've played around with it a bit and I think it might be true for complexes of projective modules? But I'm not sure... are there any well-known results regarding this?

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The answer is no. Consider the complex over the integers $A$ which is $\mathbb Z$ in degree $0$ and $1$ and the only non-trivial differential being multiplication by $2$ and let $B$ be the same complex shifted once to the left (so that it is $\mathbb Z$ in degrees $-1$ and $0$). We have a map of complexes $A \to B$ which is the identity in degree $0$ and (necessarily) zero in all other degrees. This induces zero in cohomology (for trivial reasons) but is not null homotopic. (This is easily seen by an explicit calculation. Abstractly however it has to do with the fact that it realises the non-zero element of $\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/2,\mathbb Z/2)$.)

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  • $\begingroup$ @adeel: Isn't $\mathbb{Z}$ a projective module? And aren't the above complexes (in Ekedahl's answer) bounded? $\endgroup$
    – Qfwfq
    Commented May 17, 2010 at 10:41
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    $\begingroup$ It is enough that the source complex is projective (or the target complex injective) to construct things inductively. If you like you can replace $B$ in my example by $\mathbb Q/\mathbb Z$ in degrees $0$ and $1$ with a suitable map $A \to B$ and get the same counterexample. $\endgroup$
    – Torsten Ekedahl
    Commented May 17, 2010 at 15:50

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