Let $Y$ be a complex manifold (possibly non-compact) and $\mathcal{F}_n$, $n=1,2,\dots$ be a directed system of sheaves of $\mathcal{O}_Y$-modules. Suppose that the directed system satisfies the following condition: the direct limit of $\mathcal{F}_n$ in the category of presheaves is a sheaf, i.e., for every open $U\subset Y$ we have \begin{equation} (\varinjlim \mathcal{F}_n)(U) = \varinjlim (\mathcal{F}_n(U)). \end{equation} Is it true that \begin{eqnarray} \label{} H^i(Y,\varinjlim \mathcal{F}_n) = \varinjlim H^i(Y, \mathcal{F}_n), \quad \forall i\geq 0 ? \end{eqnarray} If $Y$ is compact then the answer to my question is yes for any directed system. If we look at the proof (in the stack-project [Tag 01FE]) and try to repeat it for non-compact $Y$, then the argument almost works in a sense that if it was true that the functor of taking global sections commutes with direct limits, then higher cohomologies would also commute with direct limits.

However, in general taking global sections does not commute with direct limits, so unless the directed system has some special property we should not expect the answer to my question to be yes. The condition that I imposed is probably the first thing that one could think of, so probably it is too optimistic. Nevertheless, I was wandering if someone knows a counter example or has a better suggestion.


Edit. As t3suji points out, the OP wants an example where $\mathcal{F}_n(U)$ equals $\varinjlim\left(\mathcal{F}_n(U)\right)$, which my example does not satisfy. I suspect that one can make a new example of a sequence of short exact sequences $0\to \mathcal{F}_n\to \mathcal{G}_n\to \mathcal{Q}_n \to 0$, where the sheaves $\mathcal{F}_n$ do satisfy the OP's constraint, yet the sheaves $\mathcal{Q}_n$ do not. In that case, the long exact sequence of cohomology would imply that the groups $H^1(X,\mathcal{F}_n)$ are not compatible with colimits. I will try to find an example; until then, I am leaving my original example. Actually, having failed to construct an example, I tend to agree with Mikhail Bondarko. There is the issue that when you form Čech complexes, you typically need to take infinite products, and these do not commute with colimits. But at least for sheaves that have a finite acyclic cover, it does seem that the Čech complex proves that all cohomology commutes with colimits, so long as the colimit presheaf is a sheaf. As before, I am leaving my original example, which proves that the colimit presheaf is typically not a sheaf.

Let $X$ be the complex manifold $\mathbb{C}^* = \mathbb{C}\setminus\{0\}$ with its usual structure sheaf. For every nonnegative integer $n$, let $e_n:U_n\to X$ be the open inclusion of the submanifold $\mathbb{C} \setminus [0,n]$, i.e., the complement of the closed interval $[0,n]$ in the nonnegative real axis. Let $e:U\to X$ be the open inclusion of $\mathbb{C}\setminus [0,\infty)$, i.e., the complement of the nonnegative real axis. For every integer $n$, define $\mathcal{F}_n = (e_n)_*\mathcal{O}_{U_n}$. Define $\mathcal{F}=e_*\mathcal{O}_U$. Via the sequence of inclusions $U_1\supset U_2 \supset U_3 \supset \dots\supset U$, there is a sequence of restriction homomorphisms $\phi_{n,n+1}:\mathcal{F}_n \to \mathcal{F}_{n+1}$ compatible with the restrictions $\psi_n:\mathcal{F}_n\to \mathcal{F}$. For each $n$, the quotient of $\mathcal{O}_X$ by the subsheaf $\mathcal{F}_n$ is the sheaf $(i_n)_*\mathcal{O}_X|_{Z_n}$ of germs of holomorphic functions on neighborhoods of $

Since stalks commute with colimits, for every $p\in U$, for every $m\geq 0$, $$(\mathcal{F}_m)_p = \varinjlim(\mathcal{F}_n)_p = \mathcal{O}_{U,p}=\mathcal{F}_p.$$ Also, for every $p\in [0,\infty)$, for every $n>p$, $(\mathcal{F}_n)_p\to \mathcal{F}_p$ is an isomorphism. Thus, checking on stalks, the sequence of homomorphisms $(\psi_n:\mathcal{F}_n\to\mathcal{F})$ is the direct limit of the sequence $(\mathcal{F}_n,\phi_{n,n+1})$. On the other hand, $\log(z)$ is a global section of $\mathcal{F}$ that is not a global section of any $\mathcal{F}_n$.

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    $\begingroup$ Doesn't this basically show that $F$ is not the direct limit of $F_n$ in the category of presheaves (which was required in the question)? $\endgroup$ – t3suji Dec 4 '16 at 14:35
  • $\begingroup$ @t3suji: I missed that hypothesis! You are correct. $\endgroup$ – Jason Starr Dec 4 '16 at 14:54

I have a very simple proof; yet I am affraid something is wrong with it.

So, for each hypercover of $X$ the (co)homologies of the corresponding (Moore?) complex for $F$ is the direct limit of those for $F_n$. Passing to the limit with respect to the poset of all hypercovers over X and applying Verdier’s hypercovering theorem (= Theorem 4.10 of https://ncatlab.org/nlab/show/hypercover#properties) one obtains the positive answer to your question.


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