# preserving quasi isomorphisms

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.

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 $$\mathcal Hom_R(I^\bullet, E^\bullet)\neq 0$$.
• 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? Nov 17, 2021 at 15:14
• 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 $n$). Nov 18, 2021 at 5:02
• It's for "T-acyclic" objects. In general, $M$ is not $\mathcal Hom_R(I^\bullet, -)$-acyclic, this is a very strong condition. Nov 20, 2021 at 23:21