4
$\begingroup$

Consider a morphism $f: Y \to X$ between two varieties and consider the stacks parametrizing coherent sheaves on them $\mathcal{M}_X, \mathcal{M}_Y$.

Does one have for free an induced pullback morphism $f^*: \mathcal{M}_X \to \mathcal{M}_Y$?

I guess the question reduces to: if $S$ is a base scheme and $E$ is a family of sheaves flat over $S$ on $X$, and if $f_S: Y_S \to X_S$ is the induced morphism, then is $f_S^* E$ still a flat family of coherent sheaves? (as usual, for ugly $S$ coherent $E$ should be replaced by quasi-coherent (locally) of finite presentation).

Thanks

$\endgroup$

1 Answer 1

2
$\begingroup$

(I had to delete my earlier answer; the following answer is based on a comment by ulrich which vanished when I deleted the answer.)

The answer to your question is no.

Counterexample: Let $X$ be the plane and $Y$ the blow-up of $X$ at the origin. Consider the tautological family of length 1 skyscraper sheaves on $X$, parametrized by $X$. This means $S=X$, and $E=\Delta_\ast\mathscr{O}_X$, where $\Delta:X\to X\times X$ is the diagonal. Then $f_S^\ast E=(\Gamma_f)_\ast\mathcal{O}_Y$, where $\Gamma_f:Y\to Y\times X$ is the graph of $f$. This is not flat over $X$.

This makes sense, geometrically: there is no way to pull back the skyscraper sheaf at the origin in $X$ to a skyscraper sheaf in $Y$, in a way which is compatible with families of skyscraper sheaves, for example along lines through the origin.

$\endgroup$
1
  • $\begingroup$ What is $f$ is a flat morphism? $\endgroup$
    – Raffaele C
    Sep 1, 2021 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.