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Let $\varphi:X\to Y$ be a surjective morphism of schemes which are connected and of finite type.

Let $A$ be an abelian group, $\mathscr{F}$ be the constant sheaf on $X$ with fibers $A$ and $\mathscr{G}$ the constant sheaf on $Y$ with fibers $A$.

Is it true, then, that we have an isomorphism $$ \varphi^* \mathscr{G} \cong \mathscr{F} \quad ?$$

Or does this hold only under some more restrictive hypothesis?

In case it is true, could you give me a reference for a proof of it or give me some hints to prove it by myself?

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    $\begingroup$ First of all, one should write $f^{-1}$, the inverse image sheaf. $f^*$ is suitable for sheaves of $\mathcal{O}_Y$-modules. Next, it has nothing to do with schemes: it works in the category of topological spaces. As for the question, I suggest you read carefully any introductory text on sheaf theory. The proof is obvious once you understand all the definitions. $\endgroup$
    – Anton Fonarev
    Commented Dec 6, 2013 at 10:23
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    $\begingroup$ Thanks for the suggestion Anton. I already read an introductory text on sheaves, but I'm still not very experienced about the theory. Give me some more time to practice! ;-) $\endgroup$
    – Abramo
    Commented Dec 6, 2013 at 10:33
  • $\begingroup$ Now I see it, thinking in terms of the inverse image this is really a triviality. I was making my life difficult using the pullback when it was not needed. $\endgroup$
    – Abramo
    Commented Dec 6, 2013 at 11:00
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    $\begingroup$ The constant sheaf on $X$ with fibre $A$ is the pullback of sheaf $A$ on a point $pt$ by the unique map $\pi_X:X\to pt$. In particular, for any map $f:Y\to X$, its composition with $\pi_X$ is $\pi_Y$. In particular, $f^*A_X=f^*\pi_X^*A=\pi_Y^*A=A_Y$. This argument works just as well in the category of topological spaces as in schemes. $\endgroup$
    – Pulcinella
    Commented Mar 14, 2021 at 16:03

2 Answers 2

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Yes, it is true. Maybe the easiest way to see it is to think of your sheaves as "étalé space" (see Wikipedia): $\mathcal{F}$ is the sheaf of (local) sections of the projection $X\times A\rightarrow X$, and similarly for $\mathcal{G}$. Now the pull back $\varphi ^*\mathcal{G}$ corresponds to the pull back on $X$ of $Y\times A$, which is obviously $X\times A$.

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If $\mathscr{F} = g^{-1}\underline{A}$ is the constant sheaf on $Y$, $f^{-1}\mathscr{F} = (g \circ f)^{-1}\underline{A}$ is constant on $X$, where $f: X \to Y$.

$g$ is the map to the final object of the category $X$ and $Y$ live in.

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  • $\begingroup$ Could you make explicit what is the map $g$ you are using? Thanks! $\endgroup$
    – Abramo
    Commented Dec 6, 2013 at 12:25
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    $\begingroup$ Unless the "final object of the category" is a point, you are invoking the fact that the pullback of a constant sheaf is constant in full generality, which is exactly what the question asks. $\endgroup$
    – Ryan Reich
    Commented Dec 6, 2013 at 15:25
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    $\begingroup$ I would say it also holds if the final object is an irreducible topological space (so the constant presheaf is flasque and a sheaf), e.g. $\mathrm{Spec}\mathbf{Z}$. $\endgroup$
    – user19475
    Commented Dec 6, 2013 at 15:52

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