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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 modules over a ring $R$ all of whose terms, except for the first leftmost nonzero one, are injective modules. Let $M$ be an $R$-module and 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.

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Unfortunately no. This is the reason we need injective resolutions. $\mathcal Hom_R(I^\bullet, E^\bullet)$ is actually quasi-isomorphic to $R\mathcal Hom_R(I^\bullet, M)$, that is, the derived functor of $\mathcal Hom_R$. (I just realized that you probably know this, because the plan you describe in the second paragraph is probably based on this).
Anyway, a simple way to see that they are not the same is to take $I^\bullet$ to be defined as $I^0=0$ and $I^i=E^i$ for $i\neq 0$. For this $I^\bullet$, $\mathcal Hom_R(I^\bullet, M)=0$, but $\mathcal Hom_R(I^\bullet, E^\bullet)\neq 0$.

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  • $\begingroup$ Thank you very much Sandor. So what seems to be true is probably that $\mathbb{R}Hom_R(I^\bullet, M)$ is quasi isomorphic to $\mathbb{R}Hom_R(I^\bullet, E^\bullet)$. Doesn't it? $\endgroup$
    – H. Ali
    Nov 17, 2021 at 15:14
  • $\begingroup$ Yes, that's right and for that you don't even need $E^\bullet$ to be an injective resolution. The point is that $R\mathcal Hom$ is a derived functor and as such it preserves quasi-isomorphism essentially by definition. (It is defined on the derived category where quasi-isomorphism=isomorphism). However, the extra info you get from an injective resolution is that for that you don't need the "$R$". In other words, $\mathcal Hom_R(-,E^\bullet)$ computes $R\mathcal Hom_R(-,M^\bullet)$ for any complex $M^\bullet$ that's quasi-isomorphic to $E^\bullet$ (as long as $E^n$ is injective for all $n$). $\endgroup$
    – Sándor Kovács
    Nov 18, 2021 at 5:02
  • $\begingroup$ Right, thanks. But could you please give a hint on what's wrong about the Acyclicity Theorem I mentioned in the statement of the question? I mean, why one can not apply it? $\endgroup$
    – H. Ali
    Nov 20, 2021 at 19:18
  • $\begingroup$ It's for "T-acyclic" objects. In general, $M$ is not $\mathcal Hom_R(I^\bullet, -)$-acyclic, this is a very strong condition. $\endgroup$
    – Sándor Kovács
    Nov 20, 2021 at 23:21

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