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Let $f: X \to Y$ be a morphism of varieties, with $X$ projective. Suppose I tell you that for any two points $p_1 \neq p_2 $ in $X$, $f(p_1) \neq f(p_2)$ in $Y$. Does it necessarily follow that $f$ is a homeomorphism onto its image?

I can see that this is true in the analytic category: a continuous map from a compact space to a Hausdorff space is a homeomorphism onto its image. Perhaps there is an analogue of this statement in the algebraic category with "compact" $\mapsto $ "proper" and "Hausdorff" $\mapsto $ "separated"? Or am I just missing something really obvious?

[I ask this question because I'm trying to understand Hartshorne's proof that a linear system separating points and tangent vectors defines a projective embedding, II 7.3. Here, the "homeomorphism onto the image" property is needed but Hartshorne appears only to prove "injectivity on points".]

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    I assume you are working over an algebraically closed field, otherwise things can go wrong. If so, since $X$ is projective, $f$ is proper and replacing $Y$ with image of $f$ (which is allowed, since the image is closed in $Y$) we have a proper injective and surjective map. Since proper implies closed and coupled with injectivity implies open, we get a homeomorphism. – 2017-02-27
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    I see - $f$ itself is proper here. If $g: Y \to {\rm Spec \ } k$ is the natural projection, then $g \circ f$ is proper and $g$ is separated, hence f is proper. That's very useful, Mohan! – 2017-02-27

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