# Is a certain map a quasi-isomorphism?

$$\DeclareMathOperator\Hom{Hom}$$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.

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

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.