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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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Yes, of course it's not true in general — I asked "when" is it true. I feel it should be true when both complexes consist of either all projective or all injective modules, and are bounded. In the unbounded case it seems false. –  Adeel May 17 '10 at 10:00
    
@adeel: Isn't $\mathbb{Z}$ a projective module? And aren't the above complexes (in Ekedahl's answer) bounded? –  Qfwfq May 17 '10 at 10:41
    
Sorry, I mean complexes $0 \to I_0 \to I_1 \to \cdots$ with $I_i$ injective or $\cdots \to P_1 \to P_0 \to 0$ with $P_i$ projective. Because then you can construct homotopy maps inductively. –  Adeel May 17 '10 at 11:06
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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. –  Torsten Ekedahl May 17 '10 at 15:50

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