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My professor mentioned a proper mapping theorem after the name of Remmert which says:

Let $X$ and $Y$ be complex manifolds, $f:X \to Y$ be a proper holomorphic map, and $V \subset X$ be a complex analytic subvariety of $ X$, then $f(V)$ is a subvariety of $Y$.

I know this is a deep result, and the proof is not easy, but is there any simple reason which can convince me $f(V)$ is a subvariety, at least intuitively?

Besides, what is the analog result in algebraic geometry? In Hartshorne, he can only say the image is constructible set.

EDIT: One of intuitive explanation which mix the language of complex and algebraic geoemtry may be following: suppose X,Y are projective analytic varieties over $\mathbb{C}$, than they are algebraic varieties. Because proper morphism are closed, and also because the image is a constructible set, it has to be a variety.

The problems of above explanation are: (1) The question is obviously local on Y, but I have to assume X,Y are projective analytic varieties in order to translate back to algebraic varities(GAGA).(2)I still use the result that the image is constructible which is not obvious intuitively´╝îand GAGA to connect analytic variety to algebraic variety. All in all, it is not a good idea to use algebraic geometry to explain complex geometry.

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It's Remmert actually. –  froggie Dec 16 '12 at 0:59
Thank you, I have edited the name. –  Li Zhan Dec 16 '12 at 1:13
There's a notion of "proper morphism" native to algebraic geometry -- a morphism is said to be "proper" if it's separated and every base extension of it is a closed map. In particular, it itself is a closed map, so of course the image of any subvariety of $X$ is a subvariety of $Y$. Of course, this doesn't really answer your question, since it's not clear why this is the correct generalization of the analytic "proper." But you can check that a map of projective varieties over $\mathbb{C}$ is analytically proper iff it is algebraically proper. –  Paul VanKoughnett Dec 16 '12 at 1:18
As for the intuitive reason, take $V$ to be a manifold and $f$ to have a Jacobian matrix of constant rank; then for every $x\in V$ there exists a neighborhood $W_x$ such that $f(W_x)$ is a manifold. By compactness, for every $y\in V$ there exist only finitely many $x\in f^{-1}(y)$, so we can find $W\subset\bigcap f(W_x)$ a neighborhood of $y$ which is a manifold. To deal first with the non-constant rank case and then with the singular case, you just need extension theorems (namely, Remmert-Stein), which don't have anything to do with properness of maps, and induction. –  wisefool Dec 16 '12 at 1:33

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