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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?

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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.

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