For sheaves of sets, given a map $f:X\rightarrow Y$, there's the usual $f^{-1}\dashv f_\ast$ adjunction. In the context of sheaves of modules, there's supposedly an adjunction $f^\ast \dashv f_\ast$ where $f^\ast(-)=f^{-1}(-)\otimes \mathcal O_X$. Here, $f^\ast$ is a functor $\mathcal O_Y$-$\mathsf{Mod}\rightarrow \mathcal O_X$-$\mathsf{Mod}$.
I don't really understand how $f^\ast$ is a left adjoint of $f_\ast$. What has changed in this context to make for a different left adjoint to what seems to be the same functor $f_\ast$. Where do morphisms of ringed spaces come into play? I thought the proof should be one line using the $f^{-1}\dashv f_\ast$ and $-\otimes B\dashv \mathsf{hom}(B,-)$ adjunctions.
Here's my attempt: $$\begin{aligned}\mathsf{Hom}\left(f^{-1}(G)\otimes\mathcal O_{X},F\right) & \cong\mathsf{Hom}\left(\mathcal O_{X},\mathsf{hom}(f^{-1}G,F)\right)\\ & \cong\mathsf{Hom}\left(\mathcal O_{X},\mathsf{hom}(G,f_{\ast}F)\right)\\ & \cong\mathsf{Hom}\left(\mathcal O_{X}\otimes G,f_{\ast}F\right)\\ & \cong\mathsf{Hom}\left(G,f_{\ast}F\right) \end{aligned} $$
Here are my doubts:
- Is there really an iso $\mathsf{hom}(f^{-1}G,F)) \cong \mathsf{hom}(G,f_{\ast}F))$ of internal hom sheaves? I mean, the adjunction only yields isos of hom-sets, so why should it lift to an iso of sheaves?
- I remember reading that this adjunction depends on the notion of a morphism of ringed spaces, and yet, that notion does not seem to play any role in the proof...
There is a mistake in your proof. You don't have an isomorphism $\underline{\operatorname{hom}}(f^{-1}F,G)=\underline{\operatorname{hom}}(F,f_*G)$ of internal homs because the first is a sheaf on $X$ while the second is a sheaf on $Y$. Instead, there exist an isomorphism $f_*\underline{\operatorname{hom}}(f^{-1}F,G)=\underline{\operatorname{hom}}(F,f_*G)$.
But for the adjunction you want, you does not need this adjuction. Instead, you need the sheaf version of the restriction/extension of scalar adjunction. That is, if $A\rightarrow B$ is a morphism of rings, then any $B$-module $N$ can be seen as a $A$-module by restriction, and there is an isomorphism $$\operatorname{Hom}_B(B\otimes_B M,N)=\operatorname{Hom}_A(M,N).$$
Now, the functor $f_*:Sh(Y)\rightarrow Sh(X)$ has $f^{-1}$ as a left adjoint : $$\operatorname{Hom}_X(f^{-1} F,G)=\operatorname{Hom}_Y(F,f_* G).$$ Endow, $X$ and $Y$ with a sheaf of rings $\mathcal{O}_X$ and $\mathcal{O}_Y$. And let $f^\#:\mathcal{O}_Y\rightarrow f_*\mathcal{O}_X$ be a morphism of sheaves of rings. If $G$ is a sheaf of $\mathcal{O}_X$ modules on $X$, then $f_*G$ has a structure of $\mathcal{O}_Y$-module via the map $$\mathcal{O}_Y\otimes f_*G\overset{f^\#\otimes id}\longrightarrow f_*\mathcal{O}_X\otimes f_*G\longrightarrow f_*(\mathcal{O}_X\otimes G)\longrightarrow f_*G$$ Hence there is a subgroup of $\operatorname{Hom}_Y(F,f_*G)$ consisting of maps of $\mathcal{O}_Y$-module. Call it $\operatorname{Hom}_{\mathcal{O}_Y}(F,f_*G)$.
Conversely, if $F$ is a $\mathcal{O}_Y$-module on $Y$, then $f^{-1}F$ has a structure of $f^{-1}\mathcal{O}_Y$-module via $$f^{-1}\mathcal{O}_Y\otimes f^{-1}F\simeq f^{-1}(\mathcal{O}_Y\otimes F)\rightarrow f^{-1}F$$ hence there is a subgroup of $\operatorname{Hom}_Y(f^{-1}F,G)$ consisting of maps of $f^{-1}\mathcal{O}_Y$-modules. Call it $\operatorname{Hom}_{f^{-1}\mathcal{O}_Y}(f^{-1}F,G)$.
Now, this is true that this two Hom-set are naturally isomorphic. So that you have an adjunction $f^{-1},f_*$ between sheaves of $f^{-1}\mathcal{O}_Y$-modules on $X$ and $\mathcal{O}_X$-modules on $Y$.
However, sheaves of $f^{-1}\mathcal{O}_Y$ on $X$ are note $\mathcal{O}_X$ modules. For this, you will need the extension of scalar I talk above. Indeed, the adjoint of $f^\#$ is a map of rings $f^{-1}\mathcal{O}_Y\rightarrow \mathcal{O}_X$, hence $\mathcal{O}_X$ is a $f^{-1}\mathcal{O}_Y$ algebra. The extension/restriction adjonction is then $$\operatorname{Hom}_{\mathcal{O}_X}(f^{-1}F\otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X,G)=\operatorname{Hom}_{f^{-1}\mathcal{O}_Y}(f^{-1}F,G)=\operatorname{Hom}_{\mathcal{O}_X}(F,f_*G)$$