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Hello i have a little question.Let $f:X\rightarrow Y$ where $X$ is the normalization of $Y$ (projective algebraic variety). Is $$ Hom(\mathbb{P}^{1},X) \rightarrow Hom(\mathbb{P}^{1},Y)$$ finite and surjective ?

I know that $X \rightarrow Y$ is finite and surjective by the fact that $X$ is normalization and $ Hom(\mathbb{P}^{1},X) \rightarrow Hom(\mathbb{P}^{1},Y)$ is surjective by the universal property of normalization. Anyone could help me with this thank u.

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    $\begingroup$ I think the map is not surjective in general. That is, if the image of $\mathbf{P}^1$ lands in the singular locus of $Y$, there is no reason why the map can be lifted to $X$. One can find an example for surfaces, and I think one such example is a surface $x^2 u = y^2 v$ in coordinates $[x,y,z] \in \mathbf{P}^2$, $[u,v] \in \mathbf{P}^1$. Its singular locus is isomorphic to a projective line, and taking normalization yields a connected $2:1$ cover on the exceptional locus, hence there is no lifting for the identity map on the exceptional locus. $\endgroup$
    – Evgeny Shinder
    Commented Mar 20, 2020 at 22:34
  • $\begingroup$ As to why your argument using the universal property of normalization doesn't go through, see the answer at mathoverflow.net/questions/46/…: the issue is that the universal property only allows you to lift maps $Z\to X$ up to $Y$ if they send associated points of $Z$ to those of $X$; if $\dim X\geq 2$, this won't be the case for maps $\mathbb P^1\to X$. $\endgroup$
    – Devlin Mallory
    Commented Mar 21, 2020 at 0:43

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