Why is the flat cotorsion pair actually a cotorsion pair? I asked this question some while ago on Stack Exchange but didn't get an answer (link), so I am trying it here as well.
Fix a ringed space $(X,\mathcal{O})$ and denote by $\mathcal{F}$ the class of flat modules. A module $C$ is called cotorsion if $Ext^1(F,C)=0$ for all flat modules $F$.
I want to see that a module $F$ is flat if and only if $Ext^1(F,C)$ for all cotorsion modules $C$.
 A: The proof is given on page 3 of Mark Hovey's paper "Cotorsion Pairs and Model Categories" http://homepages.math.uic.edu/~bshipley/hovey.pdf
After writing "It is not at all obvious that this is a cotorsion pair", he gives the proof, using pure short exact sequences. It suffices to show that if $Ext^1(D,E) = 0$ for all cotorsion $E$ then $D$ is flat. This means $Ext^1(D,M^+)=0$ for all $M$, where $M^+ = Hom_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$. To finish the proof, Hovey shows that $Tor^1(D,M)^+ = 0$ for all $M$ (which implies $D$ is flat), using that $\mathbb{Q}/\mathbb{Z}$ is injective as an abelian group (and using derived Hom-Tensor adjointness). For full details, see Hovey's paper linked above. Note: the setting here is the category of $R$-modules for some ring $R$. But, you can generalize the proof to ringed spaces (indeed, this is discussed a bit in the last section of the same paper).
A: The category of sheaves on a ringed space is a Grothendieck category. Let $Q$ denote an injective cogenerator. We write $hom$ for internal and $Hom$ for external hom.
Let $F^+:=hom(F,Q)$. Then for any sheaf $F$ the sheaf $F^+$ is pure injective. This follows just from the adjunction $\otimes \dashv \hom$ and the fact that $Q$ is injective.

Recall that a short exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is called pure if it stays exact after taking the tensor product with any element. We also just say that $A\rightarrow B$ is a pure injection. A sheaf is called pure injective if it is injective for the family of pure injections.
Next we have to see that $F^+$ is cotorsion for any $F$.
Lemma Pure injective modules are cotorsion 
Proof Let $M$ be a pure injective sheaf, $F$ a flat sheaf and $0\rightarrow M\rightarrow E\rightarrow F\rightarrow 0$ be an extension. Apply $Hom(-,M)$ to find a splitting $E\rightarrow M$. Hence the extension is trivial and we see that $Ext^1(F,M)=0$
Now we can finally show the final result
Proposition $(\mathcal{F},\mathcal{F}^\perp)$ is a cotorsion pair 
Proof Let $F\in {}^\perp(\mathcal{F}^\perp)$ and $M$ be any sheaf. It is $Ext^1(F,M^+)=0$ by assumption. But we also have $Ext^1(F,hom(M,Q))=Hom(Tor_1(F,M),Q)=0$. This uses that $Q$ is injective. Since $Q$ is also a cogenerator we can conclude $Tor_1(F,M)=0$, hence $F$ is flat as desired.
