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Let $\mathcal{F}, \mathcal{G}$ be sheaves on topological spaces $X, Y$ respectively and $f: X \to Y$ a continuous map. How to show that $\operatorname{Hom}_X(f^{-1}\mathcal{G}, \mathcal{F}) = \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})$?

By definition, for an open set $U \subseteq Y$, $$ f_*f^{-1}\mathcal{G}(U) = \lim_{\rightarrow}_{V \supseteq f(f^{-1}(U))} \mathcal{G}(V). \qquad (1) $$ Since $U \supseteq f(f^{-1}(U))$, $U$ is one of $V$'s on the right hand side of (1). We take $g$ as the inclusion map. Then we have a presheaf. We make sheafification then we obtain a natrual map of sheaves $$\phi: \mathcal{G} \to f_*f^{-1}\mathcal{G}.$$

By definition, $$f^{-1}f_{*}\mathcal{F}(U)=\lim_{\rightarrow}_{V \supseteq f(U)} \mathcal{F}(f^{-1}(V)).$$

If $V \supseteq f(U)$, then $U \subseteq f^{-1}(V)$. Since $\mathcal{F}$ is a sheaf, we have the restriction map: $\operatorname{res}_{f^{-1}(V), U}: \mathcal{F}(f^{-1}(V)) \to \mathcal{F}(U)$. By using the definition of direct limits, there is a natrual map $\lim_{\rightarrow}_{V \supseteq f(U)} \mathcal{F}(f^{-1}(V))$ to $\mathcal{F}(U)$. Therefore we obtain a presheaf. We make sheafification then we obtain a natrual map of sheaves $$ \psi: f^{-1}f_*\mathcal{F} \to \mathcal{F}. $$

Let $\alpha: \operatorname{Hom}_X(f^{-1}\mathcal{G}, \mathcal{F}) \to \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})$ be the map given by $g \mapsto (f_*g)\circ \psi$ and $\beta: \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F}) \to \operatorname{Hom}_X(f^{-1}\mathcal{G}, \mathcal{F}) $ be the map given by $g \mapsto \phi\circ (f^{-1}g)$. How to verify that $\alpha \circ \beta = 1_{\operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})}$ and $\beta \circ \alpha = 1_{\operatorname{Hom}_X(f^{-1}\mathcal{G}, \mathcal{F})}$? Thank you very much.

LJR
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  • What do you mean by "...we have a presheaf. We make sheafification..."? To define a morphism of sheaves $\mathcal G\to f_\ast f^{-1}\mathcal G$ it suffices to define it on open sets in a way compatible with restriction. I don't think sheafification comes into it. – Andrew Aug 29 '13 at 15:23
  • You have to check that the triangle identities are satisfied. But this is easy, just using the definitions. What have you tried? – Martin Brandenburg Aug 29 '13 at 16:37
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    Equation (1) is not correct because it is not true in general that $ (f^{-1}\mathcal{G})(U)=\lim_{\rightarrow}_{V \supseteq f(U)} \mathcal{G}(V) $. You have to sheafify the presheaf given by the right hand side. – Georges Elencwajg Aug 29 '13 at 17:16
  • Ah, I see, I was thrown off by the equality in (1). You are defining a morphism to the inverse image presheaf, and then you compose to the inverse image sheaf. – Andrew Aug 29 '13 at 18:54
  • @MartinBrandenburg, thank you very much. What are the triangle identities explicitly? – LJR Aug 30 '13 at 02:27
  • @GeorgesElencwajg, thank you very much. How to write $f_*f^{-1}\mathcal{G}(U)$ explicitly? – LJR Aug 30 '13 at 02:33
  • @LJR. By usual description of sheafification . Also, use that morphisms into sheaves are the same from presheaf as from sheafification. – Georges Elencwajg Aug 30 '13 at 06:49
  • @GeorgesElencwajg, thank you very much. – LJR Aug 30 '13 at 11:49

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