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Let $C$, $C'$, $D$, $D'$ be chain complexes, $f$, $f'\colon C\to C'$ and $g$, $g'\colon D \to D'$ two pairs of homotopic chain maps. How to show $f \otimes g$ and $f' \otimes g' \colon C\otimes D\to C'\otimes D'$ are homotopic? I have tried many formulas for such a homotopy, but none of them worked.

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First do the case $g = g' = 1_{D'}$, so show that $f \otimes 1_{D'} \simeq f' \otimes 1_{D'}$ (or even better $(f-f') \otimes 1_{D'} \simeq 0$), then use the formula $f \otimes g = (f \otimes 1_{D'}) \circ (1_{C} \otimes g)$ together with the fact that compositions of homotopic maps are homotopic. From there it should be easy to figure out the formula for the homotopy (if you really need it). – t.b. May 22 '12 at 17:09

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