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Let $X$ be a topological space with a presheaf $\mathcal{F}$ and its sheafification $\varphi \colon \mathcal{F} \to \mathcal{G}$. The sheaf $\mathcal{G}$ is uniquely determined by the property that $\varphi_x \colon \mathcal{F}_x \to \mathcal{G}_x$ is bijective for all $x \in X$. Informally, $\mathcal{G}$ can be understood as the result of identifying all sections of $\mathcal{F}$ which agree locally, and of gluing all compatible sections of $\mathcal{F}$ together.

Although the above seems clear to me, I am not able to prove the following, which seems reasonable to me:

For any section $g \in \mathcal{G}(U)$ there is an open cover $U = \bigcup_i U_i$ and sections $f_i \in \mathcal{F}(U_i)$ such that ${f_i}_{|U_i \cap U_j} = {f_j}_{|U_i \cap U_j}$ for all $i,j$, and $\varphi_{U_i}(f_i) = g_{|U_i}$.

Is this statement even true?

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  • $\begingroup$ You always get such $f_i$, but I think you can only make sure that the stalks of $f_i$ and $f_j$ agree on each point in the intersection. If $\mathcal F$ violates the first sheaf condition (there are hardly natural examples of such), you do not get that the sections agree on the intersection. $\endgroup$ – MooS Dec 5 '16 at 8:34

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