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Let $\mathscr{F}$ be a pre sheaf on $X$, let $\mathscr{F}^+$ be a sheaf on $X$, and let $\varphi:\mathscr{F}\to\mathscr{F}^+$ be a morphism of presheaves such that for all $x\in X$, $\varphi_x$ is an isomorphism. Does this imply that $\mathscr{F}^+$ is the sheafification of $\mathscr{F}$?

Clearly, if $\varphi:\mathscr{F}\to\mathscr{F}^+$ is already the sheafification, then this property is satisfied, but it seems easier to test for the first property than to try using universal properties.

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Yes, this is true. Let $i:\mathscr{F}\to\mathscr{G}$ be the sheafification of $\mathscr{F}$. By the universal property of the sheafification, there exists a map of sheaves $j:\mathscr{G}\to\mathscr{F}^+$ such that $\varphi=ji$. Since $\varphi$ and $i$ both induce isomorphisms on stalks, so does $j$. Thus $j$ is an isomorphism.

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