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Let $X$ and $T$ be schemes and assume we have two coherent sheaves $\mathcal{F}$ and $\mathcal{G}$ on $X\times T$ which are flat over $T$, that is these are families of sheaves parametrized by $T$.

  • Assume we have a nontrivial morphism of sheaves $\phi: \mathcal{F}\rightarrow \mathcal{G}$ and that there is some $t_0\in T$ such that $\phi_{t_0}: \mathcal{F}_{t_0}\rightarrow G_{t_0}$ is surjective. Can we find an open subset $U\subseteq T$ containing $t_0$, such that for all $t\in U$ the map $\phi_t$ is surjective?

  • What can we say in general about the following set: $Z:=\{t\in T | \phi_t: \mathcal{F}_t\rightarrow \mathcal{G}_t \text{is surjective}\}$? Is it always an open subset in $T$? Or do wee need to put some restrictions to $\mathcal{F}$ and $\mathcal{G}$ to be true?

I read in Geometric Invariant Theory and Decorated Principal Bundles by Schmitt: "The condition that a morphism between vector bundles be surjective is an open condition in a suitable parameter space." But nothing more is said about what this "parameter space" should be.

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    $\begingroup$ I think you need the projection $X\times T\rightarrow T$ to be closed, e. g. $X$ proper over $k$ if you work over a field $k$. $\endgroup$ – abx Jun 24 '15 at 12:27
  • $\begingroup$ @abx : Can you elaborate a little bit? I don't see how this helps. My idea was to use the surjectivity of $\phi_{t_0}$ to get surjectivity of $\phi_{(x_0,t_0)}$ at some point $(x_0,t_0)$. This surjection would extend to some open subset $U$ in $X\times T$. Then my hope was that the image of $U$ under the projection could be an open subset in $T$. But then the projection had to be open. So maybe this is not a really good idea? $\endgroup$ – Bernie Jun 24 '15 at 14:29
  • $\begingroup$ Of course abx is correct, and it is easy to make counterexamples if you drop the properness hypothesis. The support of the cokernel of $\phi$ is closed in $X\times T$, but you are asking about the image of this set under $\phi$. $\endgroup$ – Jason Starr Jun 24 '15 at 15:31

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