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Suppose that $X$ is a topological space, $\mathcal{F}$ is a sheaf (of abelian groups, rings, ideals, modules) on $X$ and $Y \subset X$. Do you know a natural way to get an induced sheaf on $Y$ from $\mathcal{F}$?

If $Y$ is open, the answer is straight forward. I will be happy with an answer assuming the following extra conditions: $Y$ is closed and $X$ is irreducible.

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What do you mean by "induced"? Is the pushforward not considered acceptable? – user27126 Apr 25 '13 at 3:44
Sorry, I don't get what you mean. You have an inclusion $Y \hookrightarrow$ X and a sheaf $\mathcal{F}$ on $X$, you don't have a sheaf on $Y$, what do you "pushforward"? – mr.bigproblem Apr 25 '13 at 3:59
Sorry, I misread your problem. What about pullback? – user27126 Apr 25 '13 at 4:02
Uhm, this probably works. I remember seeing the "particular" construction (by particular, I mean with a direct formula) from some text but I cannot recall exactly what it is right now. Working out the sheafification in the pullback construction, perhaps one would get a direct formula. I'll check this tomorrow and response to your answer. Thanks! – mr.bigproblem Apr 25 '13 at 4:07
Yes, you can get a direct "formula" for the set of sections, but this is useless in all situations I know of. Better think of pullback as left adjoint of pushforward, and the latter is easy to describe. – Martin Brandenburg Apr 25 '13 at 7:46

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