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In this exercise, we have to prove that there is an isomorphism $$\text{Hom}(\text{Tot}^{\oplus}(P\otimes Q),I)\cong \text{Hom}(P,\text{Tot}^{\prod}(\text{Hom}(Q,I))$$ of double complexes.

But if I choose $I$ to be the cochain complex with only $I_0$ (all the other abelian groups of other degree equals to zero) I get $$\text{Hom}(\text{Tot}^{\oplus}(P\otimes Q),I)_{p,q}=\prod_n\text{Hom}(P_{p-n},\text{Hom}(Q_n,I_q))$$ which is $0$ if $q\neq 0$ and $$\text{Hom}(P,\text{Tot}^{\prod}(\text{Hom}(Q,I))_{p,q}=\text{Hom}(P_p,\text{Hom}(Q_{q},I_0))=\text{Hom}(P_p\otimes Q_q,I_0)$$ and they clearly differs.

Where am I wrong ?

brunoh
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1 Answers1

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Actually, the only way I can make this exercise works is by symmetrizing further the two expressions by looking at the $\text{Tot}^{\prod}$ of each terms of the equality.

So we get $$\text{Tot}^{\prod}(\text{Hom}(\text{Tot}^{\oplus}(P\otimes Q),I))\cong \text{Tot}^{\prod}(\text{Hom}(P,\text{Tot}^{\prod}(\text{Hom}(Q,I)))$$

which should be added in the Errata for the H-Book of Prof. Weibel to save us a lot of time !

brunoh
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  • Maybe I am confusing things again, but I think this is right and can - by definition of the Hom-complex as the product-totalization of the Hom-bicomplex - be restated as an isomorphism between Hom-complexes. – Hanno Jan 03 '16 at 21:10
  • @Hanno I think you are right, but in this exercise Prof. Weibel specifically ask to show an isomorphism between the double complex Homs and not the simple complex product-totalization of the Homs bicomplex. Hence my confusion – brunoh Jan 03 '16 at 21:19
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    Yes. My comment was meant to confirm your observations and to only add the note that the isomorphism you state simplifies in terms of hom-complexes. – Hanno Jan 03 '16 at 21:21
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    And in Rotman's "An Introduction to Homological Algebra", Exercise 10.9 (after Section 10.4) is indeed the "$\mathrm{Tot}^{\prod}$ version" instead of Weibel's "double complex version". – Hetong Xu Apr 17 '22 at 14:10
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    ... And there Rotman used this property accompanied by the following Exercise 10.10 there to give a non-spectral-sequence proof of the Kunneth formula in Theorem 10.81. – Hetong Xu Apr 17 '22 at 14:28