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!


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$ !

  • $\begingroup$ 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 $\endgroup$ Nov 27 '17 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.