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Let $X$ and $Y$ be complex projective varieties. Assume there is a finite group $G$ acting on $Y$ and we denote the quotient projective variety by $Y/G$. We have a morphism of $\mathcal{Hom}$-schemes induced by the quotient of the group action in the following form: $$f: \mathcal{Hom}(X,Y) \rightarrow \mathcal{Hom}(X, Y/G)$$ A morphism that can be lifted to the left side is called a liftable morphism. For a connected quasi-projective variety $U$, we say a morphism $\tilde{g}: U\rightarrow Y/G$ is a restriction of a morphism $g: X\rightarrow Y/G$ if there is an etale map $h:U\rightarrow X$ such $ \tilde{g}$ is the restriction of $g$ along $h$. Note that if we replace $X$ by the algebraic closure of its function field i.e. $\overline{K(X)}$, then all morphisms become liftable. This implies that every morphism on the right is liftable after some restriction.

  • For every morphism on the right is there a neighborhood (with the topology on the $\mathcal{Hom}(X, Y/G)$ explained below) such that it becomes liftable after a restriction?

Edit: The topology on $\mathcal{Hom}(X, Y/G)$ can be considered to be the analytic topology of complex points on the $\text{Hom}$-scheme.

For the purpose of this problem you can assume $Y=(\mathbb{P}^d)^{\times n}$ for arbitrarily large $n$ and $d$, $G = S_n$ and $Y/G = \text{Sym}^{n}(\mathbb{P}^d)$

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    $\begingroup$ What if we take $X$ and $Y$ to be compact Riemann surfaces with $g(X)<g(Y)$, and $Y$ admitting a Galois ramified cover $Y\rightarrow Y/G=\Bbb{CP}^1$? By the Riemann Existence Theorem, there exists a non-constant morphism $X\rightarrow \Bbb{CP}^1$, but there is no non-constant morphism from a non-empty Zariski open subset of $X$ to $Y$ since such a thing can be extended to a non-constant morphism $X\rightarrow Y$. $\endgroup$
    – KhashF
    Commented Aug 28, 2022 at 20:28
  • $\begingroup$ @KhashF Thanks, yes this is not true in general I should add what $Y$ and $G$ are supposed to be. $\endgroup$
    – user127776
    Commented Aug 28, 2022 at 20:37
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    $\begingroup$ @KhashF BTW I am not sure if that example contradicted anything since the $U$ mentioned above is an etale open not a Zariski open. Meaning if we consider etale covers it should be possible to make your map from $X$ to $\mathbb{CP}^1$ liftable. I think my mentioning of Zariski or analytical open has confused many people, I should remove that. What I meant, is that there are two types of topologies on the hom space which comes from the Hilbert scheme one is Zariski and the other one is analytic. $\endgroup$
    – user127776
    Commented Aug 29, 2022 at 5:39

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