3
$\begingroup$

In Remark 2.6.9 of Sheaves on Manifolds, Kashiwara and Schapira state that for $i: Z \hookrightarrow X$ a subspace of $X$, and $F$ a sheaf on $X$, that $R\Gamma(Z;F)$ may be different from the functor $R\Gamma(Z; -)$ applied to $F|_Z := i^{-1}F$.

Here, $F \mapsto R\Gamma(Z; F)$ is the right derived functor of $F \mapsto \Gamma(Z; F) := \Gamma(Z; F|_Z)$.

They go on to state that there is a natural isomorphism $R\Gamma(Z; F) \cong R\Gamma(Z; F|_Z)$ in certain situations, such as

1) $Z$ open,

2) $X$ Hausdorff and $Z$ compact,

3) $Z$ closed in a a paracompact open subset $U$ of $X$.

I am confused because these two functors look like they should be the same. After all $i^{-1}$, being an inverse image, is exact and hence descends to the derived category (as stated a few pages earlier on page 109). So it seems to me that in both cases we are dealing with the functor $$R\Gamma(Z; i^{-1}(-))=R(\Gamma(Z; -) \circ i^{-1}) \cong R\Gamma(Z; -) \circ i^{-1}.$$

Can someone explain what is meant by this Remark, and what is wrong with my reasoning?

$\endgroup$
5
  • $\begingroup$ For your reasoning to be correct, one need that $i^{-1}$ sends injective sheaves to acyclic for the functor $\Gamma$. For example, let $F$ be a sheaf, $I^.$ a resolution of $F$. Then $i^{-1}I^.$ is indeed a resolution of $i^{-1}F$ (since $i^{-1}$ is exact), but might not be acyclic for $\Gamma$ and so applying $\Gamma$ will not give $R\Gamma(Z,i^{-1}F)$. $\endgroup$ – Roland Apr 25 '17 at 22:14
  • 2
    $\begingroup$ Here is a curious related fact in étale cohomology. To define the functor $\Gamma_c$ we put $\Gamma\circ j_!$ where $j:X\rightarrow\overline{X}$ is a compactification. Note that $j_!$ is exact and we might be tempted to put $R\Gamma_c$ to be the right derived functor of $\Gamma_c$. But it does not work... Because $j_!$ does not send injective to acyclics, this functor is ill-behaved. The good definition is $R\Gamma\circ j_!$ (which is still denoted by $R\Gamma_c$). $\endgroup$ – Roland Apr 25 '17 at 22:19
  • $\begingroup$ Interesting; I wrongly thought that exact functors preserve injectives. So in the etale cohomology example, I assume $R \Gamma \circ j_! (F)$ is computed by resolving $j_!F$ by injectives, rather than resolving $F$ by injectives, as we would if we had tried to compute $R\Gamma_c(F)$? $\endgroup$ – ಠ_ಠ Apr 25 '17 at 23:25
  • 1
    $\begingroup$ Yes exactly. In particular, it might happen that for an injective sheaf $I$, $R\Gamma_c(X,I)\neq 0$. For derived functors, this thing is impossible. Note also that in the proof of the equality $H^k(X,i_*F)=H^k(Z,F)$ for a closed immersion $i:Z\rightarrow X$, one needs that $i_*$ preserve injective (which is true because it has an exact left adjoint). $\endgroup$ – Roland Apr 26 '17 at 6:33
  • $\begingroup$ @Roland Thanks! I think I get it now. I found a reference for right adjoints with left exact left adjoints preserve injectives as well. If you would like to post your comment(s) as an answer, I would be happy to upvote and accept it. $\endgroup$ – ಠ_ಠ Apr 26 '17 at 11:48
2
$\begingroup$

Let me expend a bit on what I said in the comments.

The phenomenon in the question is a quite subtle point in homological algebra. Derived functors does not always compose very well, even if one of them is exact. More precisely, let $\mathcal{A}\overset{F}\rightarrow\mathcal{B}\overset{G}\rightarrow\mathcal{C}$ be left exact functors between abelian categories. Then, $R(G\circ F)$ is not always $RG\circ RF$. Even under the assumption that $F$ is exact.

Let us assume that $F$ is exact. Then to compute $R^i(G\circ F)(X)$, we need an injective resolution $X\rightarrow I^\bullet$. Then we apply $G\circ F$ to $I^\bullet$ and finally take the $i$-th cohomology.

The thing is, $F(I^\bullet)$ is indeed a resolution of $F(X)$ (because $F$ is exact), but is often no more composed of injective objects, and might even contains objects that are not $G$-acyclic. This means that this resolution cannot be used to compute $RG(F(X))$.

For example, if $i>0$ and $I$ is an injective object, then $R^i(G\circ F)(I)=0$ (every derived functor vanishes on injectives). But $F(I)$ might not be $G$-acyclic. So $R^iG(F(I))$ might be non zero.


Yes this is a subtle point, because exact functors are not always obvious. Here are some examples.

As in your post, if $i:Z\rightarrow X$ is the inclusion of an arbitrary subspace and $F$ a sheaf on $X$, there are ambiguities in the definition of $H^i(Z,F)$. It can mean the derived functor of $\Gamma\circ i^{-1}$, or $H^i(Z,F_{|Z})$. And there are some cases where these functors differ. While I am writting this, I realize that I kind of disagree with the definition of Kashiwara-Shapira. I may be tempted to say that the good one is the latter. Indeed, with $F=\mathbb{Z}$ the constant sheaf on $X$, I would say that $H^i(Z,\mathbb{Z})$ should be $H^i(Z,F_{|Z})$ while the former $R^i(\Gamma\circ i^{-1})(\mathbb{Z})$ might depends on (a neighborhood of $Z$ in) $X$.

As I said in the comments, in étale cohomology we define $\Gamma_c$ to be the functor $\Gamma\circ j_!$ where $j:X\rightarrow \overline{X}$ is any compactification. The functor $j_!$ is exact, but the two functors $R\Gamma\circ j_!$ and $R(\Gamma\circ j_!)$ are really differents. This is the former that gives the good results and which is denoted $R\Gamma_c$, the latter being ill-behaved. In particular, there are injective sheaves where $H^i_c(X,I)\neq 0$.


Sometimes however it works. Here are two examples.

If $i:Z\rightarrow X$ is the inclusion of a closed subset, the functor $i_*$ is exact. And in this particular case $R(\Gamma\circ i_*)$ is indeed isomorphic to $R\Gamma\circ i_*$. This leads to the well-known and heavily used formula $H^k(X,i_*F)=H^k(Z,F)$. Here it works because $i_*$ preserve injective (because its left adjoint $i^{-1}$ is exact).

Another subtle example : let $X$ be a scheme. There are a priori two definitions for $H^i(X,F)$ for $F$ a quasi-coherent sheaf, depending on the category we are working in : $QCoh(X)$ the category of quasi-coherent sheaves on $X$ or $Sh(X)$. Let $U:QCoh(X)\rightarrow Sh(X)$ be the inclusion functor. Then we want to be sure that $R\Gamma\circ U$ and $R(\Gamma\circ U)$ are the same. In this case, $U$ does not preserve injective (take $X=\operatorname{Spec}\mathbb{F}_p$). However, injective quasi-coherent sheaves are flabby, and sheaf cohomology can be computed using flabby sheaves.

I find this last example really subtle (and thankfully there are nothing to worry). But $U$ is never written in practice, so there would be lots of confusions if it did not send injective to acyclics...

$\endgroup$
1
  • $\begingroup$ Haha I have the stack exchange app on my phone so it sent me a notification. Thanks for answering so many of my questions lately; you've been very helpful! $\endgroup$ – ಠ_ಠ Apr 27 '17 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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