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$\DeclareMathOperator\Hom{Hom}$Assume $F$ and $M$ are respectively right and left modules over a ring $R$ and let $I^\bullet$ be a left-bounded exact complex of $R$-$R$-bimodules. We know there is a natural map of complexes $\varphi: F\otimes_R \Hom(I^\bullet, M)\longrightarrow \Hom_R(\Hom_R(F, I^\bullet), M)$ which is defined in degree $i$ as $(\varphi_i(x\otimes g))(h)=g(h(x))$ for every $x\in F, g\in \Hom_R(I^i, M)$, and $h\in \Hom_R(F, I^i)$. In certain cases, I know the exactness of the complex $\Hom_R(\Hom_R(F, I^\bullet), M)$. I want to know if this implies the exactness of $F\otimes_R \Hom(I^\bullet, M)$. I can also add the hypothesis that $F$ is $R$-flat and $I^\bullet$ is a complex of injective $R$-$R$-bimodules, except for the first nonzero entry ( because that would make $I^\bullet$ split).

I tried to show that this map $\varphi$ is a quasi-isomorphism under the extra assumptions mentioned; however I'm suspicious about this. Might mapping cone arguments be of any help? I appreciate any comment.

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  • $\begingroup$ Do you really mean "injective", or did you mean projective ? $\endgroup$ Nov 22, 2021 at 19:13
  • $\begingroup$ Yes, I'm working with injectives. $\endgroup$
    – H. Ali
    Nov 22, 2021 at 19:15

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This isn’t true when $R=\mathbb{Z}$, $F=\mathbb{Q}$, $M=\mathbb{Z}$ and $I^\bullet$ is the complex $$\cdots\to0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to0\to\cdots$$ with $\mathbb{Z}$ in degree zero.

Then $\varphi$ is a map from $\mathbb{Q}$, as a complex concentrated in degree zero, to the zero complex.

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