Lemma 31.23.8. Let $f : X \to Y$ be a morphism of locally ringed spaces. Assume that pullbacks of meromorphic functions are defined for $f$ (see Definition 31.23.4).

1. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ Y$-modules. There is a canonical pullback map $f^* : \Gamma (Y, \mathcal{K}_ Y(\mathcal{F})) \to \Gamma (X, \mathcal{K}_ X(f^*\mathcal{F}))$ for meromorphic sections of $\mathcal{F}$.

2. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ Y$-module. A regular meromorphic section $s$ of $\mathcal{L}$ pulls back to a regular meromorphic section $f^*s$ of $f^*\mathcal{L}$.

Proof. Omitted. $\square$

Comment #8691 by on

Typo in (2): it should be “let $\mathcal{L}$ be an invertible $\mathcal{O}_Y$-module.”

Part (1) can be stated more generally as:

Let $f:X\to Y$ be a morphism of locally ringed spaces. Assume pullbacks of meromorphic functions are defined for $f$. Let $\mathcal{G}\in Mod(\mathcal{O}_X)$, $\mathcal{F}\in Mod(\mathcal{O}_Y)$ and suppose we have a morphism $\mathcal{F}\to f_*\mathcal{G}$ of $\mathcal{O}_Y$-modules. Then there is a canonical morphism $\mathcal{K}_Y(\mathcal{F})\to f_*\mathcal{K}_X(\mathcal{G})$ of $\mathcal{K}_Y$-modules. (In particular, setting $\mathcal{G}=f^*\mathcal{F}$, the map $\mathcal{F}\to f_*f^*\mathcal{F}$ to be the unit of $f_*\dashv f^*$ and taking global sections gives the pullback map pullback map $f^* : \Gamma(Y, \mathcal{K}_Y(\mathcal{F})) \to \Gamma(X, \mathcal{K}_X(f^*\mathcal{F}))$.)

The construction is the composite of canonical maps:

I regarded this more general form of part (1) useful for instance if one is working with the sheaf of meromorphic differentials on a $k$-variety $X$, i.e., $\mathcal{K}_X(\Omega_{X/k})$. Pullbacks of meromorphic functions are defined for the normalization $\nu:X'\to X$. Hence one obtains a morphism $\mathcal{K}_X(\Omega_{X/k})\to\nu_*\mathcal{K}_{X'}(\Omega_{X'/k})$. When $X$ is a curve, this last morphism is used by Serre in Algebraic Groups and Class Fields to define the “sheaf of regular meromorphic differentials.”

Comment #8692 by on

Also, in my reformulation of part (1), instead of "canonical" one can say that:

The map $\mathcal{K}_Y(\mathcal{F})\to f_*\mathcal{K}_X(\mathcal{G})$ is the unique morphism of $\mathcal{K}_Y$-modules that makes the following diagram commute:

The fact that the constructed morphism indeed makes the square commute can be seen leveraging the factorization $\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{K}_Y \to f_*\mathcal{G}\otimes_{f_*\mathcal{O}_X}f_*\mathcal{K}_X \to f_*(\mathcal{G}\otimes_{\mathcal{O}_X}\mathcal{K}_X)$.

Uniqueness follows from the fact that by precomposing the $\mathcal{K}_Y$-linear map $can:\mathcal{K}_Y(\mathcal{F})\to f_*\mathcal{K}_X(\mathcal{G})$ with the unit $\mathcal{F}\to\mathcal{K}_Y(\mathcal{F})$ we are obtaining the adjunct $\mathcal{O}_Y$-linear morphism $\mathcal{F}\to f_*\mathcal{K}_X(\mathcal{G})$ of $can$; thus, it is uniquely determined.

Comment #8703 by on

Regarding #8691: Ignore last paragraph. Serre does not use that map in his book. Also, the map $\mathcal{K}_X(\Omega_{X/k})\to\nu_*\mathcal{K}_{X'}(\Omega_{X'/k})$ from the case I explained is just the identity, as $\mathcal{K}_X(\Omega_{X/k})$ and $\mathcal{K}_{X'}(\Omega_{X'/k})$ are the sheaves constantly $\Omega_{X/k,\eta}\cong\Omega_{X'/k,\eta'}$, where $\eta\in X$, $\eta'\in X'$ is the generic point.

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