A bijection $\operatorname{Hom}(f^{-1}G,F)= \operatorname{Hom}(G,f_{*}F)$ preserves the induced map on stalks?

Question

Let $f:X \to Y$ be a continuous map,let $F$ be a sheaf on $X$ and $G$ be a sheaf on $Y$. There is a bijection $\operatorname{Hom}(f^{-1}G,F)= \operatorname{Hom}(G,f_{*}F)$. Let $\phi \in \operatorname{Hom}(G,f_{*}F)$,and corresponding element of $\operatorname{Hom}(f^{-1}G,F)$ be $\phi^{\sharp}$. In this situation,for each $x\in X$ $\phi$ and $\phi^{\sharp}$ induce the map $\phi_{x} , \phi^{\sharp}_{x}:G_{f(x)} \to F_{x}$. Are $\phi_{x}$ and $\phi^{\sharp}_{x}$ the same map?

Answer 1

Yes it is true. But first recall a few things about stalks and how are the maps defined.

If $x\in X$ is any point and $F$ any sheaf on $X$, then we will write again $x$ for the inclusion map $\{x\}\rightarrow X$. With this convention $x^{-1}F$ is a sheaf on $\{x\}$ and can be identified with the set of its global section, and this is just $F_x$. In the sequel, we will write $x^{-1}F$ instead of $F_x$ making it clearer how things compose.

For example $G_{f(x)}$ becomes $f(x)^{-1}G=x^{-1}f^{-1}G$.

Finally, recall that the map $\phi_x:G_{f(x)}\rightarrow F_x$ is defined as the following : $$\phi_x:G_{f(x)}\overset{\phi}\longrightarrow (f_*F)_{f(x)}=(f^{-1}f_*F)_x\overset\epsilon\longrightarrow F_x$$ where $\epsilon$ is the counit of the adjunction $(f^{-1},f_*)$.

Now let us start with $\phi:G\rightarrow f_*F$. The construction of $\phi^\#$ is the following : $$\phi^\#:f^{-1}G\overset{f^{-1}\phi}\longrightarrow f^{-1}f_*F\overset{\epsilon}\longrightarrow F$$ where $\epsilon$ is the counit of adjunction. And by definition, the map $\phi^\#_x$ is the image of $\phi^\#$ under the functor $x^{-1}$. Hence $\phi^\#_x$ is the composite : $$\phi^\#_x:x^{-1}f^{-1}G\overset{x^{-1}f^{-1}\phi}\longrightarrow x^{-1}f^{-1}f_*F\overset{x^{-1}\epsilon}\longrightarrow x^{-1}F $$

This is exactly the same formula as the definition of $\phi_x$. Hence these maps are equal.