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.
• If I recap the construction of $f^*\mathcal{G}$ and the adjunction then I encounter some problems. Now I'm trying to imitate the construction in the second excerpt about: Let $V \subset Y$ open subset of $Y$ and $U \subset Y$ open subset of $X$ satisfying $U = f^{-1}(V)$ then we get following canonical morphism $$\mathcal{G}(V) \xrightarrow{\text{a}} \varinjlim_{f(U) \subset V'}\mathcal{G}(V')\xrightarrow{\text{can}} f^{-1}(\mathcal{G})(U)$$. The first on eis the map to direct system, the second one identification by definition of $f^{-1}(\mathcal{G})$. – KarlPeter Sep 19 '18 at 2:58
• One don’t need much fantasy to concatenate it with canonical map by universal tensor property: $f^{-1}(\mathcal{G})(U) \xrightarrow{\text{b}} f^{-1}(\mathcal{G})(U) \otimes_{ f^{-1}\mathcal{O}_Y(U) }\mathcal{O}_X(U) = f_*( f^{-1}(\mathcal{G})\otimes_{ f^{-1}\mathcal{O}_Y } \mathcal{O}_X )(V)$. Combining these two we get the following locally defined cocatenation: – KarlPeter Sep 19 '18 at 3:00
• $$\mathcal{G}(V) \xrightarrow{\text{a}} \varinjlim_{f(U) \subset V'}\mathcal{G}(V')= f^{-1}(\mathcal{G})(U) \xrightarrow{\text{b}} f^{-1}(\mathcal{G})(U)\otimes_{ f^{-1}\mathcal{O}_Y(U) }\mathcal{O}_X(U) = f_*( f^{-1}(\mathcal{G}) \otimes_{ f^{-1}\mathcal{O}_Y } \mathcal{O}_X) (V)$$ Now the problem: What is $b \circ a$ on level of module elements: Sure, $b$ maps indeed via $r \xrightarrow{\text{b}} r \otimes 1$ since it's belongs to the maps of the tensor product, but what about $a$? Can I see $a$ as "inclusion" $s \xrightarrow{\text{a}} s$ and why? – KarlPeter Sep 19 '18 at 3:01
• You can see $a$ as a sort of restriction/inclusion, because on stalks $a_P=1_P$ if $P\in f(X)$ and $0$ otherwise. – user10354138 Sep 19 '18 at 3:53