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This question has raised from my current research; the terminology and notation comes from either of C. Weibel's "introduction to homological algebra" or "Methods of Homological algebra" by S. Gelfand and Y. Manin. Let $I^\bullet$ be an exact left-bounded complex of injective modules over a ring $R$ and let $M$ be an $R$-module. Take an injective resolution $$0\rightarrow M\rightarrow E^0\rightarrow E^1\rightarrow\cdots$$ of $M$; this can be described in terms of a quasi isomorphism $M\rightarrow E^\bullet$ where I view $M$ as a complex concentrated in degree zero and $E^\bullet$ is just the deleted resolution $0\rightarrow E^0\rightarrow E^1\rightarrow\cdots$. My question is that does the functor $Hom_R^\bullet(I^\bullet, -)$, as defined in the aforementioned books, preserve this quasi isomorphism? That is to say, I want to know if the complexes $Hom_R^\bullet(I^\bullet, M)$ and $Hom_R^\bullet(I^\bullet, E^\bullet)$ are quasi isomorphic.

My plan is to use Acyclicity Theorem, mentioned on page 54 of B. Iversen's "Cohomology of sheaves". For, I need to show that both $M$ and $E^\bullet$ are $Hom_R^\bullet(I, -)$-acyclic in the sense that for $i\geq 1$, the $i$-th right derived functor ${\mathbb R}^i Hom_R(I^\bullet,-)$ vanishes both over $M$ and $E^\bullet$. It seems to me that this holds because $I^\bullet$ is exact and $E^\bullet$ is left-bounded. Am I true? Any comment is appreciated.

H. Ali
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  • The complex $I^\cdot$ is left bounded exact complex of injective modules implies that $I^\cdot$ is split. So $Hom_R(I^\cdot, X)$ is also split exact for any complex $X$. – Jian Nov 16 '21 at 08:50
  • Thank you Jian. What happens if I^\bullet has not all its terms injective. Actually the complex I have is left bounded exact and all its terms, except for the first leftmost nonzero entry, is injective. – H. Ali Nov 16 '21 at 11:39
  • I don't know about this. Maybe two complexes you mentioned are not quasi-isomorphic? – Jian Nov 16 '21 at 12:44

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