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Lifting property for proper morphism

MathOverflow Asked by Simon Parker on November 3, 2021

Let $X subseteq mathbb{C}^n$ be a complex affine variety and $tilde{X} to X$ a surjective proper morphism where $tilde{X}$ is smooth. Is it true that every morphism $mathbb{C} to X$ can be lifted to $mathbb{C} to tilde{X}$?

I’m particularly interested in the case where $X$ has an isolated singularity and $tilde{X} to X$ is a resolution.

One Answer

Regarding your general question, the answer is no.

Take any hyperbolic projective variety $Y$ (for instance, a ball quotient) of dimension $n$, and project it generically onto $mathbb{P}^n$. Removing a hyperplane from $mathbb{P}^n$ and its preimage from $Y$, we get a surjective finite morphism $f colon Y^{circ} to mathbb{C}^n$. There are plenty of morphisms $mathbb{C} to mathbb{C}^n$, but none of them lifts to $Y^{circ}$, since by the hyperbolicity assumption there are no non-constant entire curves in $Y^{circ}$.

As a toy model, you can consider the case where $Y$ is a genus $2$ curve mapping $2:1$ to $mathbb{P}^1$, and then remove a point from $mathbb{P}^1$ and the two points over it from $Y$.

Regarding your question about the isolated singularity, the answer is yes. In fact, the resolution is an isomorphism away from the singular point $p in X$. Thus, you can take the image of your entire curve in $X-{p}$, lift it to the resolution via the resolution map and then consider the Zariski closure in $tilde{X}$.

Answered by Francesco Polizzi on November 3, 2021

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