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I am simply asking for a definition for something everyone uses but nobody defines. Really, this is used in class and in Hartshorne, and I have tried to look for a definition in Hartshorne, Qing Liu, Wikipedia, nothing comes up, so I am wondering whether somebody on this planet knows a definition of this.

Let $X,Y$ be topological spaces and $F, G$ be sheaves of modules over $X,Y$ respectively.

The pull-back of a sheaf is very well-documented and defined everywhere with high precision:

If $f:X\rightarrow Y$ is a continuous map, then

$f^*G=f^{-1}G\otimes_{f^{-1}O_Y} O_X$

So I know what $f^*G$ and what $(f^*G)(U)$ are (with $U \subset X$).

But what is $f^*s$ if $s\in G(Y)$, or more generally $s \in G(U)$ where $U \subset Y$ is some open subset of $Y$?

I know there is already a discussion in this thread and apparently the definition is given in a comment for affine schemes (it is just the image by the induced ring map), but I don't find it particularly enlightening. Could somebody please provide a straightforward definition for the pull-back of a section of a sheaf of modules on a general scheme? Can it be defined in a simple way (with e.g. a formula) without using high-powered, unintelligible stuff? In particular I don't know what adjunction correspondance is...

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There is a pretty geometric answer, if $X, Y$ are schemes and $G = \mathcal{O}_Y$ is the structure sheaf. In that case, a global section $s \in \Gamma(Y, \mathcal{O}_Y)$ is the same as a morphism $Y \to \mathbb{A}^1_\mathbb{Z}$. The pull-back $f^*s$ then corresponds to the composed morphism $$X \xrightarrow{f} Y \to \mathbb{A}^1_\mathbb{Z},$$ and defines in that way a global section $f^*s \in \Gamma(X, \mathcal{O}_X)$.

For abritrary ringed spaces $X,Y$, and sheaf $G$, verify that a global section $s \in \Gamma(Y, G)$ is the same as a $\mathcal{O}_Y$-module homomorphism $s: \mathcal{O}_Y \to G$. Pulling-back that morphism yields $$f^*s: \mathcal{O}_X = f^*\mathcal{O}_Y \to f^*G,$$ which defines the global section $f^*s \in \Gamma(X, f^*G)$. Here it is important to verify, that $f^*$ is a functor $\operatorname{Mod}(Y) \to \operatorname{Mod}(X)$.

Also verify that this yields the first construction in the case $G = \mathcal{O}_Y$.

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  • $\begingroup$ So the pull-back of a section is only defined for global sections on the structure sheaf? Why is $s \in \Gamma(Y,O_Y)$ equivalent to a morphism $Y \rightarrow \operatorname{Spec}A[T]$? And what is $A$ in this case? $\endgroup$ – Evariste Dec 28 '19 at 20:49
  • $\begingroup$ I added the general case. Here $\mathbb{A}^1 = \operatorname{Spec} \mathbb{Z}[T]$. The equivalence $Y \to \mathbb{A}^1$ and $s \in \Gamma(Y, \mathcal{O}_Y)$ is easy, if $Y$ is affine. I guess the non-affine case can be done by gluing the map (or the section) obtained on an affine cover. $\endgroup$ – red_trumpet Dec 28 '19 at 20:53
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    $\begingroup$ In the easy affine case, is the morphism just the one corresponding to $\mathbb{Z}[T] \rightarrow A$ which evaluates the polynomial at $s \in A$ if $A = \Gamma(Y,O_Y)$? It is the only obvious one that comes to mind that could be associated to a global section $\endgroup$ – Evariste Dec 28 '19 at 21:01
  • $\begingroup$ @Evariste Exactly. $\endgroup$ – red_trumpet Dec 28 '19 at 21:02
  • $\begingroup$ Ok, thanks a lot. $\endgroup$ – Evariste Dec 28 '19 at 21:02
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red_trumpet's answer is quite categorical, choosing to talk about morphisms and apply functoriality in place of talking about elements as much as possible. This is a good perspective, but it can also be useful to understand what's going on directly at the level of elements, so I thought I'd add another answer. In fact, there's a fairly satisfying "only thing it could be" definition of $f^*s$.

Suppose we have: $f\colon X\to Y$ a morphism of schemes, $G$ a sheaf of $\mathcal{O}_Y$-modules, $U\subseteq Y$ open, and $s\in G(U)$.

First of all, what should $f^*s$ be? The most natural answer is that we should have $f^*s \in f^*G(f^{-1}U)$. Ok, now let's unpack the definitions. Set $V = f^{-1}(U)$.

First, $f^*G = f^{-1}G\otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X$. This tensor product is the sheafification of the tensor product presheaf $f^{-1}G\otimes^{\text{p}}_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X$ (that p is for "presheaf"). And there is a map of presheaves from any presheaf to its sheafification. So to give an element of $f^*G(V)$, it suffices to give an element of $$(f^{-1}G\otimes_{f^{-1}\mathcal{O}_Y} ^{\text{p}}\mathcal{O}_X)(V) = f^{-1}G(V)\otimes_{f^{-1}\mathcal{O}_Y(V)}\mathcal{O}_X(V).$$ Now given an $R$-module $M$ and an extension of rings $R\to S$, there is a natural map $m\mapsto m\otimes_R 1_S$ from $M$ to the "base changed" $S$-module $M\otimes_R S$ So to give an element of the module above, it suffices to give an element of the module $f^{-1}G(V)$.

The sheaf $f^{-1}G$ is the sheafification of the presheaf $f^{-1}_\text{p}G$ defined by $$f^{-1}_\text{p}G(V) = \text{colim}_{W\supseteq f(V)} G(W).$$ Again, there is a map from any presheaf to its sheafification, so to give an element of $f^{-1}G(V)$, it suffices to give an element of $f^{-1}_\text{p}G(V)$.

We have $U\supseteq f(f^{-1}U) = f(V)$, so $G(U)$ is one of the modules included in the colimit diagram. That is, we have a map $$G(U) \to \text{colim}_{W\supseteq f(V)} G(W) = f^{-1}_\text{p}G(V).$$

The image of $s\in G(U)$ under this long chain of maps $$G(U)\to f^{-1}_\text{p}G(V) \to f^{-1}G(V) \to f^{-1}G(V)\otimes_{f^{-1}\mathcal{O}_Y(V)} \mathcal{O}_X(V)\to f^*G(V)$$ is the desired element $f^*s\in f^*G(V)$.

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  • $\begingroup$ Yep, thats more explicit. One note: In general, a map $M \to M \otimes_R N$ involves choices, e.g. for any $n \in N$ one has the map $m \mapsto m \otimes n$. In our situation though, $N = S$ is a ring, so the map $m \mapsto m \otimes 1_S$ seems to be canonical, right? $\endgroup$ – red_trumpet Jan 1 at 18:01
  • $\begingroup$ @red_trumpet Yes, good point. I'll edit my answer. $\endgroup$ – Alex Kruckman Jan 1 at 19:16
  • $\begingroup$ @AlexKruckman Indeed, this is more explict and well detailed. However I am not sure it is a constructive description of $f^*s$ especially for the sheafification step? Maybe we can't go more explicit than that and that we can only describe locally on an affine open $f^*s$ to be equal to $f^{\#}(s)$, and the sheafification step will remain forever unclear? Thank you for your lights. $\endgroup$ – PerelMan May 2 at 18:38
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    $\begingroup$ @PerelMan But the map from a presheaf to its sheafification is totally explicit. One way to construct the sheafification is this: Let $F$ be a presheaf on $X$. For open $U\subseteq X$, set $F^{\text{sh}}(U)$ be the set of all compatible families of germs of $F$ at all points of $U$ (meaning for all $p\in U$, we get a germ $g_p\in F_p$, and there exists a neighborhood $p\subseteq V\subseteq U$ such that there is a section $f\in F(V)$ and $g_q = f_q$ for all $q\in V$). Then the map $F(U)\to F^{\text{sh}}(U)$ takes a section $f\in F(U)$ to its family of germs $(f_p)_{q\in U}$. $\endgroup$ – Alex Kruckman May 2 at 21:05
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We treat pullbacks of global sections, nothing changes if we restrict to some other open set. We will be using the fact that tensor product of sheaves commutes with restriction to open sets.

Call $f: X \rightarrow Y$ our morphism of schemes. Let $\mathscr{G}$ be an $\mathcal{O}_Y$-module. There is a canonical morphism $\mathscr{G} \rightarrow f_* f^* \mathscr{G}$ of $\mathcal{O}_Y$-modules. This induces a morphism of global sections $\psi:\Gamma(Y,\mathscr{G}) \rightarrow \Gamma(X,f^*\mathscr{G})$. The question is how to describe concretely the image of $s \in \Gamma(Y,\mathscr{G})$ under this map.

It is instructive to begin with the special case $\mathscr{G}=\mathcal{O}_Y$. Consider the following rings $A=\Gamma(Y,\mathcal{O}_Y), \, B = \Gamma(X,f^{-1} \mathcal{O}_Y), \, C = \Gamma(X,\mathcal{O}_X)$. The data of $f$ include a morphism $\mathcal{O}_Y \rightarrow f_* \mathcal{O}_X$. This induces a ring homomorphism $\phi_1: A \rightarrow C$ on global sections. Now the canonical morphism $\mathcal{O}_Y \rightarrow f_* f^{*} \mathcal{O}_Y$ induces a ring homomorphism $\phi_1':A \rightarrow \Gamma(X,f^* \mathscr{G})$. The key is that $\phi_1$ and $\phi_1'$ have the same image. To see this note that we have a ring homomorphism $\phi_2: A \rightarrow B$ induced by $\mathcal{O}_Y \rightarrow f_* f^{-1} \mathcal{O}_Y$. We also have a $\phi_3 : B \rightarrow C$ induced by $f^{-1} \mathcal{O}_Y \rightarrow f^{-1} f_* \mathcal{O}_X \rightarrow \mathcal{O}_X$. Moreover $\phi_1 = \phi_3 \circ \phi_2$. Now $\Gamma(X,f^* \mathcal{O}_Y) = B \otimes_B C$. Finally by checking the definition of $\phi_1'$ we can see that $\phi_1'$ sends $s$ to $\phi_2(s) \otimes 1 \in B \otimes_B C$ which we can identify with $\phi_3(\phi_2(s))=\phi_1(s)$.

For general $\mathscr{G}$ the picture is analogous with the note that $\Gamma(X,f^*\mathscr{G}) = \Gamma(X,f^{-1}\mathscr{G}) \otimes_B C$. That is $\psi$ sends global sections of $\mathscr{G}$ to their canonical image in the $B$-module $\Gamma(X,f^{-1} \mathscr{G})$ and then to their image in the $C$-module $\Gamma(X,f^{-1} \mathscr{G}) \otimes_B C$, the latter being an extended module by scalars according to $\phi_3: B \rightarrow C$.

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