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I clarify some definitions. If $j:U\to X$ is the inclusion of an open set, and $\mathscr{F}\in\mathfrak{Ab}(X)$, then we define $$\mathscr{F}_U := j_!(\mathscr{F}\vert_U) = j_! j^{-1}\mathscr{F}$$ and if $i:Z\to X$ is the inclusion of $Z = X\setminus U$, then we define $$\mathscr{F}_Z:= i_*(\mathscr{F}\vert_Z) = i_* i^{-1}\mathscr{F}$$ which of course gives us natural maps $\mathscr{F}_U\to\mathscr{F}$ and $\mathscr{F}\to\mathscr{F}_Z$, which form a short exact sequence $$0\to\mathscr{F}_U\to\mathscr{F}\to\mathscr{F}_Z\to 0$$ (by Hartshorne Ex II.2.19). Now, note that by a similar process, we may define a map $$\mathscr{F}_U\to(\mathscr{F}_U)_{\bar{U}}$$ by defining an inclusion map $\imath:\bar{U}\to X$ and using the natural map $$\mathscr{F}_U\to\imath_*\imath^{-1}\mathscr{F}_U$$ which I wish to show is the identity.

I think that it may be enough if I can show that the restriction map $$\varinjlim_{W\supseteq V}\mathscr{F}_U(W)\to\varinjlim_{W\supseteq V\cap U}\mathscr{F}_U(W)$$ is an isomorphism for all opens $V\subseteq\bar{U}$. Now, using the property that $\mathscr{F}_U\vert_x=0$ for all $x\notin U$, I have succeeded in proving injectivity, but I'm struggling with surjectivity.

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    Check that the map is an isomorphism on stalks: pulling back to UU, we see it's an iso for stalks in $U$, for $p \not \in U$, the stalk of both sheaves is zero. – 2017-01-26
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    I'm not following something here. As defined, $\mathscr{F}_U$ is a sheaf on $X$ but the notation $\mathscr{F}_U\vert_{\bar{U}}$ would typically mean the restriction of $\mathscr{F}_U$ to $\bar U$ (which is also how the notation seems to be used in the last line). So these aren't sheaves on the same space. – 2017-01-27
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    I replaced it with better notation. – 2017-01-27

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