I know that in general, the stalks of the pushforward of a sheaf need not be the same as the original stalk. That is, $(f_{*}\mathcal{F})_{f(p)}=\mathcal{F}_p$ is not true in general. But when $X \subset Y$ and $f$ is the inclusion map, is it true that the above holds for $p \in X$ and that the stalks are zero(or the terminal object) for points not in X?
I am asking this because I was able to prove this for any subset, but in Hartshorne Chapter 2, exercise 19, he asks you to prove that this holds for closed subsets and that a different construction(denoted $f_{!}\mathcal{F}$) gives these stalks for open subsets.
No it is not true, take $Y$ to be the unit disc in $\mathbb{C}$ and $X\subset Y$ to be the punctured unit disc and $\mathcal{F}$ to be the constant sheaf of fiber say $\mathbb{Z}$ over $X$. The inclusion will be denoted by $i$, and $0$ will be the origin of the disc. Then the fiber of $i_*\mathcal{F}$ at $0$ will be... $\mathbb{Z}$.