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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

enter image description here

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

enter image description here

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.

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  • $\begingroup$ 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) $\endgroup$ – Roland Sep 15 '17 at 21:12
  • $\begingroup$ @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 $\endgroup$ – user312648 Sep 16 '17 at 3:23
  • $\begingroup$ Can you please tell me what is the problem with the construction of $G_{\alpha}$? $\endgroup$ – user312648 Sep 16 '17 at 3:27
  • $\begingroup$ 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. $\endgroup$ – Roland Sep 16 '17 at 9:00
  • $\begingroup$ @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 $1$? $\endgroup$ – user312648 Sep 16 '17 at 13:22

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