Let $f: X \to Y$ be a morphism of ringed spaces, $\mathcal{G}$ a $\mathcal{O}_Y$-module. I have a question about an argument used to construct a adjunct morphism $f^{\#}: \mathcal{G} \to f_*(f^*(\mathcal{G}))$ from $id:f^*(\mathcal{G} \to f^*(\mathcal{G}$ using the adjunction formula $$Hom_{\mathcal{O}_X}(f^*(\mathcal{G},f^*(\mathcal{G})= Hom_{\mathcal{O}_Y}(\mathcal{G}, f_*(f^*(\mathcal{G}))$$
The argument is used in Bosch's "Commutative Algebra and Algebraic Geometry" (see page 270); here the excerpt:
Obviously $f^{\#}$ is induced locally via $$\mathcal{G}(V) \to \mathcal{G}(V) \otimes_{\mathcal{O}(V)} \mathcal{O}_X(f^{-1}(V))$$. My question is why this maps are concretely given by the map $a \mapsto a \otimes 1$?
The author refers to the consructions of $f^{-1}(\mathcal{G}$ and $f^*(\mathcal{G})$ explicitely but I don't see how that imply that this map is given by $a \mapsto a \otimes 1$.
Can anybody explain concretely why this map must be of this shape?
The construction of $f^{-1}(\mathcal{G})$ was explicitely used before to construct the adjunction $Hom_X(f^{-1}(\mathcal{G}),\mathcal{F})= Hom_Y(\mathcal{G}, f_*(\mathcal{G}, f_*(\mathcal{F}))$ as explained here:
Morally the reason is that $1\in\mathcal{O}_X(f^{-1}V)$ is the only "distinguished" nonzero element we can get our hands on so it will appear when we change the ring. More concretely, if you recall the construction of $f^*\mathcal{G}$, the way you get the $\mathcal{O}_X$-module $f^*\mathcal{G}$ from the $f^{-1}\mathcal{O}_Y$-module $f^{-1}\mathcal{G}$ is you tensor over $f^{-1}\mathcal{O}_Y$, i.e., do the $(-)\otimes 1$ on sections and let the $\mathcal{O}_X$ operates on the second factor.