# Why is the tensored up inverse image left adjoint to the direct image?

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 $$\operatorname{Hom}_{\mathcal{O}_X}(f^{-1}F\otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X,G)=\operatorname{Hom}_{f^{-1}\mathcal{O}_Y}(f^{-1}F,G)=\operatorname{Hom}_{\mathcal{O}_X}(F,f_*G)$$

• What is the map $f_*\mathcal{O}_X\otimes f_*G\longrightarrow f_*(\mathcal{O}_X\otimes G)$? Also, aren't the maps $f_\ast (\mathcal O_X\otimes G)\rightarrow f_\ast G$ and $f^{-1}(\mathcal O_Y \otimes F)\rightarrow f^{-1}F$ isos? Since the ring is the unit of the monoidal structure of every category of modules... – user153312 Nov 8 '15 at 18:13
• Let $\otimes^p$ be the tensor product of presheaves. You have an map $f_*\mathcal{O}_X\otimes^p f_*G\rightarrow f_*(\mathcal{O}_X\otimes^p G)\rightarrow f_*(\mathcal{O}_X\otimes G)$ where the first map is the obvious one on each open set, and the second come from the sheaffification. You get the desired map by the the universal property of sheaffification, because $f_*\mathcal{O}_X\otimes f_*G$ is the sheaf associated to the presheaf $f_*\mathcal{O}_X\otimes^p f_*G$. – Roland Nov 8 '15 at 18:21
• Altenatively, if you know that $f^{-1}$ commutes with tensor product, this map correspond by adjunction to the map $f^{-1}(f_*\mathcal{O}_X\otimes G)=f^{-1}f_*\mathcal{O}_X\otimes f^{-1}f_*G\rightarrow \mathcal{O}_X\otimes G$ where the last map is given by the counit of the $(f^{-1},f_*)$-adjunction. – Roland Nov 8 '15 at 18:24
• Sorry, It's not obvious to me what the map $f_*\mathcal{O}_X\otimes^p f_*G\rightarrow f_\ast (\mathcal O_X \otimes ^p G)$ is. Also, why aren't the maps $f_\ast (\mathcal O_X\otimes G)\rightarrow f_\ast G, f^{-1}(\mathcal O_Y \otimes F)\rightarrow f^{-1}F$ isos? – user153312 Nov 8 '15 at 18:30
• The map $\mathcal{O}_X\otimes G\rightarrow G$ is not an iso. This is the module structure on $G$. (I did not take the tensor product over $\mathcal{O}_X$). A module structure on a sheaf is given by maps like these, but they are not iso in general. Maybe I should have replaced the tensor product by a usual product, and said bilinear map instead. – Roland Nov 8 '15 at 18:39