Hom and $\otimes$ functors on chain complexes.

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I can't solve the exercise $2.7.3$ from Weibel's book "An Introduction to homological algebra":

Let $P,Q$ be right and left $R$-module chain complexes, $I$ be a cochain complex of abelian groups. Show that there is a natural isomorphism of double complexes:$$ \operatorname{Hom_{Ab}}(\operatorname{Tot}^{\oplus}(P\otimes Q),I) \cong \operatorname{Hom}_R(P,\operatorname{Tot}^{\prod}\operatorname{Hom_{Ab}}(Q,I)). $$

If I'm correct $$ \operatorname{Hom_{Ab}}(\operatorname{Tot}^{\oplus}(P\otimes Q),I)^{p,q}=\operatorname{Hom_{Ab}}({\oplus}(P_i\otimes Q_{p-i}),I^q)={\prod}\operatorname{Hom_{Ab}}(P_i\otimes Q_{p-i},I^q) ,$$ while $$ \operatorname{Hom_R}(P,\operatorname{Tot}^{\prod}\operatorname{Hom_{Ab}(Q,I)})^{p,q}= \operatorname{Hom_R}(P_p,{\prod}\operatorname{Hom_{Ab}(Q_i,I^{q-i})})=\prod \operatorname{Hom_R}(P_p,\operatorname{Hom_{Ab}(Q_i,I^{q-i})})=\prod \operatorname{Hom_{Ab}}(P_p \otimes Q_i,I^{q-i}).$$ But these expressions are not equal. What is wrong?

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You're right.

I think what he means to ask for is a natural isomorphism of complexes $$\operatorname{Tot}^{\prod}\operatorname{Hom_{Ab}}(\operatorname{Tot}^{\oplus}(P\otimes Q),I) \cong \operatorname{Tot}^{\prod}\operatorname{Hom_R}(P,\operatorname{Tot}^{\prod}\operatorname{Hom_{Ab}(Q,I)}).$$