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Aug 29, 2022 at 5:43 history edited user127776 CC BY-SA 4.0
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Aug 29, 2022 at 5:39 comment added user127776 @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.
Aug 28, 2022 at 20:43 history edited user127776 CC BY-SA 4.0
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Aug 28, 2022 at 20:37 comment added user127776 @KhashF Thanks, yes this is not true in general I should add what $Y$ and $G$ are supposed to be.
Aug 28, 2022 at 20:28 comment added KhashF 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 $X\rightarrow Y$.
Aug 26, 2022 at 19:10 history asked user127776 CC BY-SA 4.0