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Let $X$ be a (singular) projective variety, in other words something given by a collection of polynomial equations in $\mathbb CP^n$ or $\mathbb RP^n$. How one proves that it is a finite $CW$ complex?

Similar question. Suppose that $X$ affine (i.e. given by polynomial equations in $\mathbb C^n$, or $\mathbb R^n$). How one proves that its one point compactification is a finite $CW$ complex?

These questions are sequel to the discussions here:

For which classes of topological spaces Euler characteristics is defined?

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Particularly my answer :) – David Speyer Jun 3 '10 at 14:41
You may also want to look at the paper by Hironaka: "Triangulations of Algebraic Sets", p. 165--185, in the proceedings from the 1974 AMS Arcata conference in Algebraic Algebraic geometry. The purpose of the article is exactly to give a simple demonstration of a fact which "everyone knows", but which is reputed to be difficult. – Mike Roth Jun 3 '10 at 14:51
up vote 6 down vote accepted

The Lojasiewicz theorem says that every semi-algebraic subset of $\mathbf{R}^n$ can be triangulated. Moreover, there is a similar statement for pairs of the form (a semi-algebraic set, a closed subset). See e.g. Hironaka, Triangulations of algebraic sets, Arcata proceedings 1974 and references therein (including the original paper by Lojasiewicz).

The case of an arbitrary (not necessarily quasi-projective) complex algebraic variety follows from Nagata's theorem (every variety can be completed) and Chow's lemma (every complete variety can be blown up to a projective one).

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Algori, thanks! What about one-point compactification? How do you show that this is a finit CW-complex? – Dmitri Jun 3 '10 at 17:02
This is not too difficult - take a completion and then trianguate the resulting couple. If $K$ is a subpolyhedron of a polyhedron $K$, then $K/L$ can also be triangulated. – algori Jun 3 '10 at 17:33
That is, if $L$ is a subpolyhedron... – algori Jun 3 '10 at 17:34

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