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

$\endgroup$
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 at 19:51

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .