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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,and 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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