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Let $f: X\to Y$ be a regular map of projective varieties that is closed (in the sense that it takes Zariski closed sets to Zariski closed sets). Let $V\subset X$ be a quasiprojective subvariety (i.e. locally closed and irreducible). Is $f(V)$ a quasiprojective subvariety of $Y$?

(I'm aware that under arbitrary regular maps, the image of quasiprojective need only be constructible, but I am assuming the map is closed).

(I doubt that this question is really at the level of Mathoverflow, but I asked it at math.stackexchange and got a small number of upvotes but no answer, so I thought I might try to slyly provoke one of y'all into answering before you close it.)

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    $\begingroup$ No, the map from the blow-up of $\mathbb{P}^2$ at a point $x$ back to $\mathbb{P}^2$ is closed(e.g. because this is morphism is projective, hence proper), but the image of a complement to the strict transform of a line passing through $x$ is $\mathbb{P}^2\setminus \mathbb{P}^1\cup\{x\}$ which is not locally closed. $\endgroup$ – SashaP Jul 5 at 15:57
  • $\begingroup$ Thanks so much! $\endgroup$ – birdiehoweblazer Jul 5 at 19:34
  • $\begingroup$ Do you think there's a counterexample among projective space bundles (i.e. $X$ is the total space, $Y$ the base, and $f$ the projection)? $\endgroup$ – birdiehoweblazer Jul 5 at 19:42
  • $\begingroup$ That restriction doesn't really change anything: given a projective morphism $f:X\to Y$ which comes with an embedding $i:X\to \mathbb{P}^n_Y$ of $Y$-schemes, a locally closed subscheme $V\subset X$ gives a locally closed subscheme $i(V)\subset \mathbb{P}^n_Y$ whose image under the projection $\mathbb{P}^n_Y\to Y$ coincides with $f(V)$, so any counterexample with a projective morphism(for instance, the one above) gives rise to a counterexample for a projection of the sort $\mathbb{P}^n\times Y\to Y$. $\endgroup$ – SashaP Jul 5 at 20:25
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