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I think I need some help on a problem about sheaf theory.

Suppose that $f: X \rightarrow Y$ is a continuous map of topological spaces, $\mathscr{F}$ is a sheaf on $X$. Prove that the morphism of sheaves $\rho: f^{-1}f_* \mathscr{F} \rightarrow \mathscr{F}$ can be obtained from the morphism of presheaves $Pf^{-1}(f_*\mathscr{F}) \rightarrow \mathscr{F}$.

Here, $f_*$ is the direct image functor, and $f^{-1}$ is the inverse image functor. $Pf^{-1}$ and $f^{-1}$ are connected by sheafification:

Suppose that $f: X \rightarrow Y$ is a continuous map of topological spaces, $\mathscr{G}$ is a sheaf on $Y$. Then for any open subset $U \subset X$, define $$(Pf^{-1} \mathscr{G}) (U) = \varinjlim_{V \supseteq f(U), V \in \mathfrak{O}(Y)} \mathscr{G}(V),$$ then $\{ (Pf^{-1} \mathscr{G})(U) \}$, together with the restriction map, becomes a presheaf. The sheaf associated to $Pf^{-1}\mathscr{G}$ is called the inverse image functor, denoted $f^{-1}\mathscr{G}$.

There are quite a lot of concepts and definitions in sheaf theory. I am afraid I am about to mess them up in my head. As to this problem, although I can find the definition of all things on the book, I don't know what to do.

Would you please give me some help? Thanks in advance!

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The main tool here is the universal property of sheafification, to wit: if $\mathcal E$ is a presheaf on $X$ and $\mathcal F$ a sheaf on $X$, there is a functorial (in $\mathcal E,\mathcal F$) bijection $$ Hom_{Presh_X}(\mathcal E,presh(\mathcal F)) \simeq Hom_{Sh_X}(sh(\mathcal E),\mathcal F)\quad (*)$$

[If you know the terminology, you are in presence of a couple of adjoint functors: the important sheafification functor and the trivial presheafification or forgetful functor]

In order to apply this to your situation, set $\mathcal E=Pf^{-1}(f_*\mathcal F)$ . So the displayed isomorphism $(*)$ tells you that you have to exhibit a morphism $Pf^{-1}(f_*\mathcal F) \to \mathcal F$ .
If you go back to your definition with inductive limits of $Pf^{-1}(f_*\mathcal F)$, you'll have to define ( in the notation of your second greyed box) compatible maps $f_*\mathcal F(V)= \mathcal F(f^{-1}(V)) \to \mathcal F(U)$. And nothing could be easier: take restriction, since $V\supseteq f(U)$ implies $ f^{-1}(V)\supseteq U$ !

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  • $\begingroup$ Could you please give a reference for $(*)$? $\endgroup$ – windsheaf Nov 21 '17 at 7:26
  • $\begingroup$ @windsheaf: With pleasure: Tag 0083 from the Stacks Project, Lemma 6.18.2 $\endgroup$ – Georges Elencwajg Nov 27 '17 at 20:14

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