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Let $f:X\rightarrow Y$ be a projective morphism of algebraic varieties and $\mathcal{F}$ be a coherent sheaf on $X$. Then some people say that there is a canonical evaluation map $$ f^{*}(f_{*}\mathcal{F})\rightarrow\mathcal{F}, $$ which I don't quite understand the definition. I do understand both pushforward $f_{*}$ and pullback $f^{*}$ of sheaves, but how does one define "evaluation" map above?

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2 Answers 2

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There is an adjunction between $f^*$ and $f_*$ that says that morphisms $f^* f_* \mathscr{F} \to \mathscr{F}$ are in bijective correspondence with morphisms $f_* \mathscr{F} \to f_* \mathscr{F}$. The "evaluation" morphism is the counit of this adjunction and corresponds to $\textrm{id} : f_* \mathscr{F} \to f_* \mathscr{F}$.

Less abstractly, recall that $f^* f_* \mathscr{F}$ is defined to be the sheaf $a \mathscr{G}$ associated with the presheaf $\mathscr{G}$ defined by $$U \mapsto \mathscr{O}_X (U) \otimes_{f^{-1} \mathscr{O}_Y (U)} \varinjlim_{V \supseteq f U} f_* \mathscr{F} (V) = \varinjlim_{V \supseteq f U} \mathscr{O}_X (U) \otimes_{\mathscr{O}_Y (V)} \mathscr{F} (f^{-1} V)$$ There is an evident presheaf morphism $\mathscr{G} \to \mathscr{F}$ induced by the restriction maps: after all, if $V \supseteq f U$, then $f^{-1} V \supseteq f^{-1} f U \supseteq U$, and each $\mathscr{F}(U)$ is a $\mathscr{O}_X(U)$-module, so there is a map $\mathscr{G} (U) \to \mathscr{F} (U)$ induced by the two universal properties. The universal property of associated sheaves then assures us that $\mathscr{G} \to \mathscr{F}$ factors through the universal presheaf-to-sheaf morphism $\mathscr{G} \to a \mathscr{G} = f^* f_* \mathscr{F}$, and the "evaluation" morphism is precisely the sheaf morphism $f^* f_* \mathscr{F} \to \mathscr{F}$ so obtained.

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Dear @Zhen, The presheaf you have described is the presheaf version of $f^{-1}f_*\mathscr{F}$, not $f^*f_*\mathscr{F}$, which is $f^{-1}f_*\mathscr{F}\otimes_{f^{-1}\mathscr{O}_Y}\mathscr{O}_X$. You've produce a morphism $f^{-1}f_*\mathscr{F}\rightarrow\mathscr{F}$ from an $f^{-1}\mathscr{O}_Y$-module to an $\mathscr{O}_X$-module, so the morphism $f^*f_*\mathscr{F}\rightarrow\mathscr{F}$ comes by the universal property of base change along the comorphism $f^{-1}\mathscr{O}_Y\rightarrow\mathscr{O}_X$. –  Keenan Kidwell Sep 16 '12 at 4:33
    
Blah, I keep forgetting about the extra complications of $\mathscr{O}_X$-modules... –  Zhen Lin Sep 16 '12 at 5:09
    
Thank you for the detailed answer, Zhen. Then name "evaluation mop" comes form the adjunction bijection evaluated at $id$. I have seen this bijection before. The actual map is quite complicated and I never really understand the map from the bottom of my heart. –  M. K. Sep 16 '12 at 7:07

Let $U \subset X$ be an open subset. A section $s \in (f^{\ast}f_{\ast}\mathcal{F})(U)$ is a function $s: U \rightarrow \bigsqcup_{p \in U} (f_{\ast}\mathcal{F})_{f(p)} \otimes_{\mathcal{O}_{Y, f(p)}} \mathcal{O}_{X, p}$ with certain properties (see Hartshorne's book, Proposition-Definition II.1.2). We have a natural projection homomorphism $\varphi: (f_{\ast}\mathcal{F})_{f(p)} = \varinjlim_{V \ni f(p)} \mathcal{F}(f^{-1}(V)) \rightarrow \varinjlim_{V' \ni p} \mathcal{F}(V') = \mathcal{F}_{p}$. If $s(p) = \sum_{i} t_{i} \otimes s_{i} \in (f_{\ast}\mathcal{F})_{f(p)} \otimes_{\mathcal{O}_{Y, f(p)}} \mathcal{O}_{X, p}$, then the evaluation morphism of sheaves takes $s$ to $\tilde{s}: U \rightarrow \bigsqcup_{p \in U} \mathcal{F}_{p}$, given by $\tilde{s}(p) = \sum_{i} s_{i}\varphi(t_{i})$.

In my opinion, this is correct, but I would like to read comments. One curious and natural question is: what are the kernel and image of such morphism?

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