Evaluation map of sheaves $f^{*}(f_{*}\mathcal{F})\rightarrow\mathcal{F}$.

Question

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?

Answer 1

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.