
## 6.26 Morphisms of ringed spaces and modules

We have now introduced enough notation so that we are able to define the pullback and pushforward of modules along a morphism of ringed spaces.

Definition 6.26.1. Let $(f, f^\sharp ) : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces.

1. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. We define the pushforward of $\mathcal{F}$ as the sheaf of $\mathcal{O}_ Y$-modules which as a sheaf of abelian groups equals $f_*\mathcal{F}$ and with module structure given by the restriction via $f^\sharp : \mathcal{O}_ Y \to f_*\mathcal{O}_ X$ of the module structure given in Lemma 6.24.5.

2. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}_ Y$-modules. We define the pullback $f^*\mathcal{G}$ to be the sheaf of $\mathcal{O}_ X$-modules defined by the formula

$f^*\mathcal{G} = \mathcal{O}_ X \otimes _{f^{-1}\mathcal{O}_ Y} f^{-1}\mathcal{G}$

where the ring map $f^{-1}\mathcal{O}_ Y \to \mathcal{O}_ X$ is the map corresponding to $f^\sharp$, and where the module structure is given by Lemma 6.24.6.

Thus we have defined functors

\begin{eqnarray*} f_* : \textit{Mod}(\mathcal{O}_ X) & \longrightarrow & \textit{Mod}(\mathcal{O}_ Y) \\ f^* : \textit{Mod}(\mathcal{O}_ Y) & \longrightarrow & \textit{Mod}(\mathcal{O}_ X) \end{eqnarray*}

The final result on these functors is that they are indeed adjoint as expected.

Lemma 6.26.2. Let $(f, f^\sharp ) : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}_ Y$-modules. There is a canonical bijection

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(f^*\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ Y}(\mathcal{G}, f_*\mathcal{F}).$

In other words: the functor $f^*$ is the left adjoint to $f_*$.

Proof. This follows from the work we did before:

\begin{eqnarray*} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(f^*\mathcal{G}, \mathcal{F}) & = & \mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O}_ X)}( \mathcal{O}_ X \otimes _{f^{-1}\mathcal{O}_ Y} f^{-1}\mathcal{G}, \mathcal{F}) \\ & = & \mathop{Mor}\nolimits _{\textit{Mod}(f^{-1}\mathcal{O}_ Y)}( f^{-1}\mathcal{G}, \mathcal{F}_{f^{-1}\mathcal{O}_ Y}) \\ & = & \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ Y}(\mathcal{G}, f_*\mathcal{F}). \end{eqnarray*}

Here we use Lemmas 6.20.2 and 6.24.7. $\square$

Lemma 6.26.3. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of ringed spaces. The functors $(g \circ f)_*$ and $g_* \circ f_*$ are equal. There is a canonical isomorphism of functors $(g \circ f)^* \cong f^* \circ g^*$.

Proof. The result on pushforwards is a consequence of Lemma 6.21.2 and our definitions. The result on pullbacks follows from this by the same argument as in the proof of Lemma 6.21.6. $\square$

Given a morphism of ringed spaces $(f, f^\sharp ) : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$, and a sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$, a sheaf of $\mathcal{O}_ Y$-modules $\mathcal{G}$ on $Y$, the notion of an $f$-map $\varphi : \mathcal{G} \to \mathcal{F}$ of sheaves of modules makes sense. We can just define it as an $f$-map $\varphi : \mathcal{G} \to \mathcal{F}$ of abelian sheaves such that for all open $V \subset Y$ the map

$\mathcal{G}(V) \longrightarrow \mathcal{F}(f^{-1}V)$

is an $\mathcal{O}_ Y(V)$-module map. Here we think of $\mathcal{F}(f^{-1}V)$ as an $\mathcal{O}_ Y(V)$-module via the map $f^\sharp _ V : \mathcal{O}_ Y(V) \to \mathcal{O}_ X(f^{-1}V)$. The set of $f$-maps between $\mathcal{G}$ and $\mathcal{F}$ will be in canonical bijection with the sets $\mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O}_ X)}(f^*\mathcal{G}, \mathcal{F})$ and $\mathop{Mor}\nolimits _{\textit{Mod}(\mathcal{O}_ Y)}(\mathcal{G}, f_*\mathcal{F})$. See above.

Composition of $f$-maps is defined in exactly the same manner as in the case of $f$-maps of sheaves of sets. In addition, given an $f$-map $\mathcal{G} \to \mathcal{F}$ as above, and $x \in X$ the induced map on stalks

$\varphi _ x : \mathcal{G}_{f(x)} \longrightarrow \mathcal{F}_ x$

is an $\mathcal{O}_{Y, f(x)}$-module map where the $\mathcal{O}_{Y, f(x)}$-module structure on $\mathcal{F}_ x$ comes from the $\mathcal{O}_{X, x}$-module structure via the map $f^\sharp _ x : \mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$. Here is a related lemma.

Lemma 6.26.4. Let $(f, f^\sharp ) : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}_ Y$-modules. Let $x \in X$. Then

$(f^*\mathcal{G})_ x = \mathcal{G}_{f(x)} \otimes _{\mathcal{O}_{Y, f(x)}} \mathcal{O}_{X, x}$

as $\mathcal{O}_{X, x}$-modules where the tensor product on the right uses $f^\sharp _ x : \mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$.

Proof. This follows from Lemma 6.20.3 and the identification of the stalks of pullback sheaves at $x$ with the corresponding stalks at $f(x)$. See the formulae in Section 6.23 for example. $\square$

Hi! I'm a little confused. In section 6.25 is determined by mapping $f^{\sharp}:\mathcal{O}_Y\rightarrow\mathcal{O}_X.$ Here $f^{\sharp}:\mathcal{O}_Y\rightarrow f_*\mathcal{O}_X.$

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