I have a question about a step in the proof of Thm. 2.7 in Hartshorne (p. 210):
Why and how the sheaf $\mathcal{F}_U := i_!(\mathcal{F}|_U)$ regarded as a sheaf on the closere $\bar{U}$ concretely?
Intuitively I would guess by restricting but $i_!(-)$-operation implies that $\mathcal{F}_U$ comes from a sheafification of a sheaf on $U$, so I'm not sure if ones come back to the same sheaf on $U$ by restriction a sheafification...
Let $h: \overline U \hookrightarrow X$ be the closed immersion. They consider $h^{-1}\mathcal F_U$ and notice that this does not change anything, more concretely:
$\mathcal F_U$ is supported on $\overline U$, hence the canonical map $\mathcal F_U \to h_*h^{-1}\mathcal F_U$ is an isomorphism (*), thus $$H^j(X,\mathcal F_U)=H^j(X,h_*h^{-1}\mathcal F_U)=H^j(\overline U,h^{-1}\mathcal F_U).$$
Now they use induction on the number of irreducible components to deduce that the RHS is zero whenever $j > \dim X$. Note that it does not matter how exactly $\mathcal F_U$ was defined. The only important fact is that its support is contained in $\overline U$.
(*) Of course we can check this on the stalks. This easy because you know how to compute the stalks after $h^{-1}$ (for every map) and you also know how to compute the stalks after $h_*$ (not in general, but if $h$ is a closed immersion).
Summarizing, we have shown the following easy-to-memorize statement: You can compute the cohomology of a sheaf on the closure of its support.