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Let $X$ be a connected scheme, $X\rightarrow \mathrm{Spec}\,\mathbb{C}$ be a morphism of finite type and relative dimension 1. Assume that there is a $\mathbb{C}$-morphism $f:X\rightarrow \mathbb{P}^2$ which is injective on the underlying topological spaces.

By Chevalley's theorem we know that $f(X)$ is a finite disjoint union of locally closed subsets of $\mathbb{P}^2$. Is it true that $f(X)$ is a locally closed subset of $\mathbb{P}^2$?

I think one way to show this would be to show that $f$ is a topological embedding. Then if it is also true that $\overline{f(X)}\backslash f(X)$ is finite, it is game over (because for a connected one-dimensional scheme of finite type over $\mathrm{Spec}\,\mathbb{C}$ the topology on the set of closed points is cofinite).

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    $\begingroup$ I think this is true even if you don't assume $f$ injective. Then passing to the normalization you can assume that $X$ is a smooth connected curve, that is, $X=\hat{X}\smallsetminus F$ where $\hat{X}$ is a smooth projective curve and $F$ a finite set. Now $f$ extends to a morphism $\hat{f}:\hat{X}\rightarrow \mathbb{P}^2$, and $f(X)$ is equal to $\hat{f}(\hat{X})$ minus a finite set, hence is locally closed. $\endgroup$
    – abx
    Commented Mar 26, 2019 at 4:27

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