# Proof of Sheafification in Bosch

In Bosch's textbook, "Algebraic Geometry and Commutative Algebra," he introduces the sheafification through "a rigorous method derived from Cech cohomology."

In particular, he proceeds as follows: Suppose $$\mathcal{F}$$ be a presheaf and $$U$$ be some open set. At the bottom of page 237, he claims that there exists some open cover $$\mathcal{U}=(U_\lambda)_{\lambda\in\Lambda}$$ so that any $$f\in\mathcal{F}^+(U)$$ (defined to be the direct limit of the zeroth cohomology $$H^0(\mathcal{V},\mathcal{F})$$ over all open covers $$\mathcal{V}$$ of $$U$$) can be represented by some $$f'\in H^0(\mathcal{U},\mathcal{F}).$$

He then seems to claim that $$\mathcal{F}(U_\lambda)\cong H^0(\mathcal{U}|_{U_\lambda}, \mathcal{F}).$$ I understand why there is a morphism $$\mathcal{F}(U_\lambda)\to H^0(\mathcal{U}|_{U_\lambda}, \mathcal{F})$$ (just take $$g\mapsto(g|_{V_s})_{s\in S},$$ where the $$V_s$$ are the elements of $$\mathcal{U}|_{U_\lambda}$$), but why is this an isomorphism?

## 1 Answer

Unless $$\mathscr{F}$$ is a sheaf, this morphism need not be an isomorphism. So I think this might be a typo in the book. However, for the proof where this is used, it is only important, that there is a morphism $$\mathscr{F}(U_{\lambda})\rightarrow H^0(\mathscr{U}|_{U_\lambda},\mathscr{F})$$. This is the one, that you have already found.