# Direct limit of Sheaf cohomology groups on Noetherian space

This is a proposition in Hartshorne's Algebriac geometry.

Let $X$ be a noetherian topological space, and let $\{\mathcal{F}_\alpha\}$ be a direct system of abelian sheaves. Then there are natural isomorphisms for each $i\geq 0$ $$\varinjlim H^i(X,\mathcal{F}_\alpha)\rightarrow H^i(X,\varinjlim \mathcal{F}_\alpha).$$

Idea $1$ : To get map $\varinjlim H^i(X,\mathcal{F}_\alpha)\rightarrow H^i(X,\varinjlim \mathcal{F}_\alpha)$ we use the idea of universal property related to $\varinjlim$. Fix $i\geq 0$, suppose there exists maps $H^i(X,\mathcal{F}_\beta)\rightarrow H^i(X,\varinjlim \mathcal{F}_\alpha)$ satisfying following commutative diagram

then, by universal property, there exists a unique map $\varinjlim H^i(X,\mathcal{F}_\alpha)\rightarrow H^i(X,\varinjlim \mathcal{F}_\alpha)$. It remains to get a map $H^i(X,\mathcal{F}_\beta)\rightarrow H^i(X,\varinjlim \mathcal{F}_\alpha)$, forget about commutative diagram. We know that morphism of sheaves induces morphism on sheaf cohomology. We have natural maps $\mathcal{F}_\beta\rightarrow \varinjlim \mathcal{F}_\alpha$ for each $\beta$. Thus, we have induced morphisms on sheaf cohomology. We do have following commutative diagram

which induces commutative diagram of sheaf cohomology that we have asked for. So, there is a unique morphism $\varinjlim H^i(X,\mathcal{F}_\alpha)\rightarrow H^i(X,\varinjlim \mathcal{F}_\alpha)$ satifying some conditions and It remains to prove that this is an isomorphism. I do not see an obvious next step towards proving that this is an isomorphism.

Idea $2$ : In case of $i=0$, it means $\varinjlim \Gamma(X,\mathcal{F}_\alpha)\rightarrow \Gamma(X,\varinjlim \mathcal{F}_\alpha)$ is an isomorphsim. This is easy. I guess induction would help to see isomorphism for $i\geq 1$. Now, do we have a relation between $H^0(X,\varinjlim \mathcal{F}_\alpha)$ and $H^1(X,\varinjlim \mathcal{F}_\alpha)$?? Exact sequence induces long exact sequences of sheaf cohomology. So, we are expected to have following sequence

$$H^0(X,\varinjlim \mathcal{F}_\alpha)\rightarrow H^0(X,\text{Some thing})\rightarrow H^0(X,\text{another thing})\rightarrow H^1(X,\varinjlim \mathcal{F}_\alpha).$$ This should come from $$0\rightarrow \varinjlim \mathcal{F}_\alpha\rightarrow \text{Some thing}\rightarrow \text{another thing}\rightarrow 0.$$ As direct limit is an exact functor, we expect to get above map from $$0\rightarrow \mathcal{F}_\alpha\rightarrow \text{Some thing}^*\rightarrow \text{another thing}^*\rightarrow 0.$$ Do we have such exact sequence given an abelian sheaf? Yes. As sheaves of abelian groups has enough injectives, there exists an injective sheaf $\mathcal{G}$ and an injective morphism of sheaves $\mathcal{F}\rightarrow \mathcal{G}$. Considering quotient, we have exact sequence $$0\rightarrow \mathcal{F}_\alpha\rightarrow \mathcal{G}\alpha\rightarrow \mathcal{H}_\alpha\rightarrow 0.$$ As every injective sheaf is a flasque sheaf, $\mathcal{G}_\alpha$ is flasque. Corresponding long exact sequence gives $$0\rightarrow H^0(X,\mathcal{F}_\alpha)\rightarrow H^0(X, \mathcal{G}_\alpha)\rightarrow H^0(X,\mathcal{H}_\alpha)\rightarrow H^1(X,\mathcal{F}_\alpha)\rightarrow 0.$$

I do not really see how do we use that $\mathcal{G}_\alpha$ is flasque to show that $H^1(X,\varinjlim\mathcal{F}_\alpha)=\varinjlim H^1(X,\mathcal{F}_\alpha)$. But, assuming this, it follows easily. As $\mathcal{G}_\alpha$ is flasque, we have $0\rightarrow H^1(X,\mathcal{H}_\alpha)\rightarrow H^2(X,\mathcal{F}_\alpha)\rightarrow 0$ i.e., $H^1(X,\mathcal{H}_\alpha)\cong H^2(X,\mathcal{F}_\alpha)$. Considering direct limits, we have $$\varinjlim H^1(X,\mathcal{H}_\alpha)\cong \varinjlim H^2(X,\mathcal{F}_\alpha).$$

We have $\varinjlim H^1(X,\mathcal{H}_\alpha)\cong H^1(X,\varinjlim \mathcal{H}_\alpha)$. So, $H^1(X,\varinjlim \mathcal{H}_\alpha)\cong \varinjlim H^2(X,\mathcal{F}_\alpha)$.

As direct limit is an exact functor, we have $$0\rightarrow \varinjlim \mathcal{F}_\alpha\rightarrow \varinjlim \mathcal{G}_\alpha \rightarrow \varinjlim \mathcal{H}_\alpha\rightarrow 0.$$

As direct limit of flasque sheaves is flasque, $\varinjlim \mathcal{G}_\alpha$ is flasque. So, corresponding long exact sequence of sheaf cohomology groups is $$0\rightarrow H^1(X,\varinjlim \mathcal{H}_\alpha)\rightarrow H^2(X,\varinjlim \mathcal{F}_\alpha)\rightarrow 0$$ i.e., $H^1(X,\varinjlim \mathcal{H}_\alpha)\cong H^2(X,\varinjlim \mathcal{F}_\alpha)$. So, we have $$H^2(X,\varinjlim \mathcal{F}_\alpha)\cong H^1(X,\varinjlim \mathcal{H}_\alpha)\cong \varinjlim H^2(X,\mathcal{F}_\alpha).$$ Thus, $H^2(X,\varinjlim \mathcal{F}_\alpha)\cong \varinjlim H^2(X,\mathcal{F}_\alpha).$ Similar result holds for higher sheaf cohomology groups as well.

So, It remains to prove this in case of $H^1$. Any suggestion regarding this and any comment on what I have done is welcome.

• What is wrong with the proof in Hartshorne ? There is a problem with your construction of $\mathcal{G}_\alpha$. You need the fact that the category of directed system has enough injectives (which is true), but you can also follows Hartshorne who construct an directed system using a functorial resolution (by flasque sheaves, not injective ones) – Roland Sep 15 '17 at 21:12
• @Roland: There is nothing wrong with Hartshorne's proof. I was just trying to think of my own proof... :D To be frank, I do not really understand the proof. I thought I will just try something, If I fail, I had to come back to his proof anyways... I will loose nothing If I try, except possibly some misconception :D :D – user312648 Sep 16 '17 at 3:23
• Can you please tell me what is the problem with the construction of $G_{\alpha}$? – user312648 Sep 16 '17 at 3:27
• Sure it does not hurt to try on your own. The problem with your construction of $G_\alpha$ is (unless I missed something) that you don't get an direct system : what are the maps $G_\alpha\rightarrow G_\beta$ and (more importantly !) how can you be sure that they fits into a commutative diagram like the one in a previous question of yours. – Roland Sep 16 '17 at 9:00
• @Roland : Yes, for get about commutativity, there is not even a natural choice for $G_\alpha\rightarrow G_\beta$ given maps $F_\alpha\rightarrow F_\beta, F_\alpha\rightarrow G_\alpha, F_\beta\rightarrow G_\beta$.. I have to go back and read that proof in Hartshorne. Do you have any thing to say on Idea $1$? – user312648 Sep 16 '17 at 13:22