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Mar 26, 2019 at 4:27 comment added abx 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.
Mar 25, 2019 at 22:50 review First posts
Mar 25, 2019 at 23:03
Mar 25, 2019 at 22:48 history asked user137493 CC BY-SA 4.0