Supposed that $X \subset K^n$ is the zero set of some polynomials, where $K = \mathbb{R}$ or $\mathbb{C}$. Is it the case that for every point $x \in X$ there is an $\epsilon > 0$ so that $B = B_{\epsilon}(x) \cap X$ is contractible around $x$? $(B_{\epsilon}$ is a ball of radius $\epsilon$ in the ambient Euclidean space.)

Seems so but I don't know how to prove it.

My guess would be to show that there is a neighborhood where all of the homotopy groups vanish, and this presumably follows from some finiteness about the topology of an algebraic variety.

But I don't know how to show that $X$ is locally simply connected to begin with, though the intuition that there are no arbitrarily small holes seems reasonable for an algebraic variety.

Just wondering.

Follow up: Is there a reasonable sense in which local rings are contractible?

  • $\begingroup$ They are all simplicial complexes hence locally contractible. $\endgroup$ Sep 11, 2016 at 19:51

1 Answer 1


In the spirit of studiosus's comment: an algebraic variety can be triangulated (in $\mathbb{R}$ or $\mathbb{C}$; an algebraic set is a particular example of a semi-algebraic set, where inequalities are allowed in the definition). A triangulated space is locally contractible, which is for example proved in the appendix of Hatcher's book on algebraic topology (recalling that a triangulated space is a CW-complex). Hence yes, an algebraic variety is locally contractible. This proof is probably killing a fly with a bazooka though.


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