Let $X$ be a scheme, and consider the distinguished affine open base of topology on $X$. That is, the data of all affine opens $\mathrm{Spec}(A)\subset X$ and inclusions only of the form $\mathrm{Spec}A_f\to\mathrm{Spec}A$ for $f\in A$. It makes sense to talk about presheaves and sheaves on this base, as well as presheaves and sheaves of $\mathcal{O}_X$-modules. Moreover, the obvious functor gives an equivalence of categories between sheaves on $X$ and sheaves on this base. For any presheaf $\mathcal{F}$ of $\mathcal{O}_X$-modules on the distinguished affine open base, the restriction morphism
$$\mathcal{F}(\mathrm{Spec A})\to\mathcal{F}(\mathrm{Spec}A_f)$$
factors as the composite
$$\mathcal{F}(\mathrm{Spec A})\to\mathcal{F}(\mathrm{Spec}A)_f\overset{\alpha}{\to}\mathcal{F}(\mathrm{Spec}A_f)$$
for some morphism $\alpha$ of $A_f$-modules. In Vakil's notes, the following very useful theorem is proven:
Theorem. If $\mathcal{F}$ is a sheaf of $\mathcal{O}_X$-modules on the distinguished affine open base, then $\mathcal{F}$ is quasicoherent if and only if $\alpha$ is always an isomorphism.
My question is whethether the following even more useful version of one direction of the theorem is true:
If $\mathcal{F}$ is a presheaf of $\mathcal{O}_X$-modules on the distinguished affine open base such that $\alpha$ is always an isomorphism, then $\mathcal{F}$ is actually a sheaf (and hence is quasicoherent).
This would mean that any construction that assigns a module to a ring and "commutes with localization" would define a quasicoherent sheaf.
Best Answer
This is true and, and it was pointed out that Vakil notes this himself just after Exercise 13.3.D. More precisely, he almost notes it, as exercise 13.3.D (the proof of the theorem in my question) is only stated for sheaves of $\mathcal{O}_X$-modules. However, the proof of the theorem in fact gives an affirmative answer to my question. Indeed, let $\mathcal{F}$ be a presheaf of $\mathcal{O}_X$-modules on the distinguished affine base such that restriction is given by localization. Then on any affine open $\mathrm{Spec}A\subset X$, we have $\mathcal{F}|_{\mathrm{Spec}A}=\widetilde{\mathcal{F}(\mathrm{Spec}A)}$, and the right-hand side a sheaf on $\mathrm{Spec}(A)$. Therefore $\mathcal{F}$ is a sheaf on the distinguished affine base of $X$.