Let $k$ be an algebraically closed field, $X,Y$ projective varieties (irreducible algebraic sets) and $f:X\to Y$ a morphism. Is $f(X)$ a projective variety? I think it is because the image of a morphism is closed and continuity preserves irreducibility. Is this correct?
I wonder because if $X$ and $Y$ are affine varieties, the statement is not true by this example: Image of a morphism of varieties.
Best Answer
Yes, this is correct. To be a bit more precise, if $X$ is a projective variety and $Y$ is any variety and $f:X\to Y$ is a morphism, then $f$ is a closed map (in particular its image is closed). And furthermore, the image of an irreducible set under any continuous map is irreducible.