2
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $\mathcal Hom_R(I^\bullet, E^\bullet)\neq 0$.

$\endgroup$
4
  • $\begingroup$ 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 $n$). $\endgroup$ 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. $\endgroup$ Nov 20, 2021 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.