Cech cohomology commuting with colimits? (Non noetherian confusion.) Suppose that $X$ is a quasicompact, separated $A$-scheme, and $I$ is some directed poset. Suppose that $F_i$ is a system of sheaves on $X$ over $I$.
I am having difficulties proving the claim that $H^i(X, \operatorname{colim} F_j) = \operatorname{colim} H^i(X, F_j)$. (An exercise in Ravi chapter 18.)
Does one need $X$ to be locally Noetherian?
Mainly, the confusion is that because $X$ if is not locally Noetherian, there are conceivably affine opens in some chosen cover were the $\operatorname{colim} (F_i(U)) \not = (\operatorname{colim} F_i)(U)$.
If I could commute these, I know how to finish this, because "filtered colimits commute with cohomology." (The Cech complex for the $\operatorname{colim} F_i$ is the colimit of the Cech complexes of $F_i$, so the cohomology for the Cech complex for $colim F_i$ is the colimit of the cohomology for the Cech complexes of $F_i$.)
I don't think I especially care about non Noetherian things right now, but I feel that I should know how to fix this. 
Edit:
I think this is probably actually very simple, and I am just getting characteristically confused: For the open affine $U$ sets in the covering, I think probably $colim (F_i(U)) = (colim F_i)(U)$. This should be because any cover of $U$ can be refined to a finite cover, and for some abstract nonsense reason filtered colimits commute with finite products ( https://mathoverflow.net/questions/57099/why-do-filtered-colimits-commute-with-finite-limits ) 
It's not completely clear to me how to write that out.
Is it  the case if $X$ is quasi-compact, then $\Gamma(X, \operatorname{colim} F_i) = \operatorname{colim} \Gamma(X,F_i)$? 
**
Edit:
So here is the confusing thing:
Suppose that $U_a$ is some cover of $U$ (quasi compact), and $U_i$ is some finite subcover. Let $F$ be a presheaf, and suppose the exactness of (1) $0 \to F(U) \to \Pi F(U_i) \to \Pi F(U_{i,i'})$. (In the previous set up, this exactness would come from the colimit commuting with finite products.)
Then does it follow that 
(2) $0 \to F(U) \to \Pi F(U_a) \to \Pi F(U_{a, a'})$
is exact? (Which would tell us that the colimit presheaf is already a sheaf.)
The map $F(U) \to \Pi F(U_a)$ is clearly still injective (because $\Pi F(U_i)$ injects in $\Pi F(U_a)$). 
What seems unclear is this: if $f_a$ is in the kernel at the second slot, from the data of the $f_i$ (and the $U_{ij}$) we get an $f \in F(U)$ with the property that $f|_{U_i} = f_i$. However, this $f$ only has $f|_{U_a \cap U_i} = f_a|_{U_a \cap U_i}$, for the cover $U_{ia}$ of $U_a$. But $F$ is not known to be separated, so it is not clear that then $f|_{U_a} = f_a$.
If $U_a$ is quasi-compact, then we can use that $I$ is filtered to get that the canonical map $\operatorname{colim} (F_i(U_a) ) \to ( \operatorname{colim}  F_i) (U_a)$ is injective, i.e. the colimit presheaf is already "separated" over $U_a$. From this $f|_{U_a \cap U_i} = f_a|_{U_a \cap U_i}$ implies that $f|_{U_a} = f_a$, hence (2) is exact.
So here is where we need the space to be Noetherian in order to conclude.
I'm confused.
(Final edit): 
If it is known that all finite covers of $U$ give exact sequences (which would be true in the filtered colimit situation), then one can apply the method of adding $U_a$ back to the finite cover to obtain that $f|_{U_a} = f_a$. The second bolded question should be true (provided the limits are filtered.)
So I think Noetherianity is not necessary, because all of the covers used in computing Cech cohomology are quasi compact, so their sections are the colimits of the sections. I will try to write this out.
 A: Proposition: If $X$ is quasi-compact and separated, and if $F_i$ is system of sheaves, indexed by a direct poset, then $H^j(X, \operatorname{colim} F_i) = \operatorname{colim} H^j(X, F_i)$. (No hypothesis on Noetherianity of $X$.)
This will follow from the following claims:


*

*For $U \subset X$ a quasi-compact open, $\Gamma(X, \operatorname{colim}(F_i)) = \operatorname{colim} \Gamma(X, F_i)$.

*Filtered colimits commute with finite products of abelian groups. More precisely, the natural map (obtained from the colimit of the projection maps $\Pi A_i^j \to A_i$) $\operatorname{colim_i} ( \Pi A^j_i) \to \Pi_J (\operatorname{colim} A_i)^j$ is an isomorphism.

*"Filtered colimits are exact."


One applies these claims to a finite cover of $X$ by affines to obtain the proposition. From the hypothesis about separtedness, each intersection remains affine, hence quasi-compact. So it follows from (2) the the Cech complex for the colimit is the colimit of the Cech complexes, and (3) then implies that the homology is the colimit of the homologies of the individual complexes.
Proofs:
1:
Suppose that $U$ is quasicompact. It suffices to show that the presheaf $F(U) = \operatorname{colim} F_i(U)$ is already a sheaf. Let $U_{a}$, $a \in A$ be some open cover of $U$. 
We claim that $0 \to F(U) \to \Pi F(U_a) \to \Pi F(U_{ab})$ is already exact.
Pass to a finite subcover, $U_j$, $j \in J$. Then using facts $(2)$ and $(3)$, from the exactness of $0 \to F_i(U) \to \Pi F_i(U_j) \to \Pi F_i(U_{j j'})$ we obtain that [1] $0 \to F(U) \to \Pi F(U_j) \to \Pi F(U_{j j'})$ is exact.
Hence $0 \to F(U) \to \Pi F(U_a)$ is exact. Suppose we are given some $f_a \in \Pi F(U_a)$ which is in the kernel. Since the $f_i$ in this product are in the kernel, we obtain from [1] a unique $f \in F(U)$ that restricts to each $f_i$. Moreover, for any particular $a \in A$, $J \cup \{a\}$ is still finite, and the uniqueness of $f$ implies that $f$ also restricts to $f_a$.
2:
Let $\psi: \operatorname{colim_i} ( \Pi A^j_i) \to \Pi_J (\operatorname{colim} A_i)^j$ be the natural map. Of course it is a homomorphism.
Since all involved operations (product and colimit) commute with forgetful passage to the category SET (both the objects and the universal elements continue to represent the "same" functor, and since the diagram involving $\psi$ and the projections will clearly commute, we just have to show that $Forget(\psi)$ is a bijection. This will follow by showing that in SET, filtered colimits commute with finite products.
The linked mathoverflow page in my original question gives the following clever computation:
$(\operatorname{colim_I} A_i) \times (\operatorname{colim_I} A_j) \cong_{nat} (\operatorname{colim_I} A_i \times (\operatorname{colim_I} A_i) \cong_{nat} (\operatorname{colim_{I \times I}} A_i \times A_i) \cong_{nat} (\operatorname{colim_I} A_i \times A_i) $
(We use that $I$ is filtered in the last isomorphism, which is induced by the diagonal inclusion. Being filtered implies that the diagonal is a confinal subset in the directed system $I \times I$.)
I guess that could have been done directly in $Ab$.
Maybe an alternative proof: $colim_I$ is an additive functor from the Abelian category $Ab^I$ (induces homomorphisms of hom sets). Hence it preserves finite products, by abelian category nonsense. Possibly products in $Ab^I$ are not what I want them to be.
3: One repeatedly applies the principle that some element being zero in filtered colimit means that it was eventually zero at some specific term.
(Maybe another nonsense argument could be given by using that filtered limits commute with all limits in SET. Does it follow via the forgetful functor that the same is true in $Ab$?)
