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Let $k$ be either the field $\Bbb C$ of complex numbers or the field $\Bbb R$ of real numbers. Let $X$ be an algebraic variety over $k$, say, quasi-projective and smooth (but not necessarily projective). We consider the set of $k$-points $X(k)$ with the usual topology.

Question. Is $X(k)$ homotopically equivalent to a CW complex?

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    $\begingroup$ Yes - see for example Triangulations of semi-algebraic sets by Hironaka. (I learned of that from this question mathoverflow.net/questions/26927/…) $\endgroup$
    – Will Sawin
    Commented Apr 25, 2021 at 15:22
  • $\begingroup$ Many thanks indeed! $\endgroup$
    – Mikhail Borovoi
    Commented Apr 25, 2021 at 15:33

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This is same answer I left at the question linked above in the comments. The one benefit it offers is that it directly addresses the class of varieties in question.

In Triangulation of Locally Semi-Algebraic Spaces. by K.R. Hofmann, necessary and sufficient conditions are given for a locally semi-algebraic space to be homeomorphic to a simplicial complex, with a corollary that any abstract algebraic variety over $\mathbb{R}$ or $\mathbb{C}$ admits a triangulation.

Here is an excerpt from the abstract:

"We give necessary and sufficient conditions for a locally semi-algebraic space to be homeomorphic to a simplicial complex. Our proof does not require the space to be embedded anywhere, and it requires neither compactness nor projectivity of the space. A corollary is that every real or complex algebraic variety is triangulable, a result which does not seem to be available in the literature when the variety is neither projective nor real and compact."

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