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My question is if homology commutes with taking tensor product. I believe in general it is not true but when the tensor product is with a projective module it is. I would like to take a look at a proof but I havent found one yet.

I was trying to proof the naturality of Kunneth exact sequence when I thought the second part of my question. If indeed homology commutes up to isomorphism, with taking the tensor product with a projective module, is this isomorphism natural?

This is: If $C'$ is a chain complex and $C_i$ is the chain complex which is $C_i$ in dimension, $i$ and zero in every other dimension and $C_i$ is a projective module, is this isomorphism natural?

$\;H_n ( C_i\bigotimes C')\cong C_i\otimes H_{n-i}( C').$

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It's not true for an arbitrary module; the map $H_n(C) \otimes A \to H_n(C\otimes A)$ is injective but in general has a nontrivial cokernel. If $A$ is projective, then $\operatorname{Tor}(*, A) = 0$ (more or less by definition, as it has an obvious projective resolution), and the universal coefficient theorem does give an isomorphism.

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  • $\begingroup$ okay, so the universal coefficient theorem is needed to prove kunneth's theorem? $\endgroup$ – allizdog Oct 25 '17 at 18:55
  • $\begingroup$ They're certainly related, but I don't remember whether one is specifically used to prove the other. One possible method of proving Kunneth's theorem is to set up an appropriate spectral sequence, and the result follows from applying a bunch of homological algebra to it; I just don't remember whether the universal coefficient theorem is part of that general algebraic background. $\endgroup$ – anomaly Oct 25 '17 at 20:04
  • $\begingroup$ Does the same happen if A is an arbitrary chain complex and $C$ is the one that is projetive? $\endgroup$ – allizdog Oct 25 '17 at 20:37
  • $\begingroup$ @allizdog: Just getting back to this. I don't know offhand, but I suspect the result still holds by the symmetry of $\operatorname{Tor}$. $\endgroup$ – anomaly Nov 9 '17 at 17:13

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