If $M$ is a (complex) affine algebraic manifold then it is intuitively obvious, that as a complex manifold, it has a finite number of connected components. This is strange, I can't find references for this. Can anybody enlighten me, where this is written?
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First recall the fact that if $X$ is a complex variety then $X(\mathbb{C})$ is connected (with the analytic topology) iff $X$ is connected (in the Zariski topology) (e.g. see [1, Theorem 6.1]).
Using this one easily deduces that $\pi_0(X)=\pi_0(X(\mathbb{C}))$ and thus it suffices to show that $\pi_0(X)$ is finite. But, this is well-known if $X$ is quasi-compact (e.g. see [2, Tag0BA8] and the fact that quasi-compact varieties are Noetherian schemes, and the fact that every connected component is a union of irreducible components).
EDIT: If you want a published reference, see [3, XII Proposition 2.4].
[1] Osserman, Brian. Complex varieties and the analytic topology. https://pdfs.semanticscholar.org/e72f/e3c8012daed6b9ae1ce1926a5c6c6bdeebba.pdf
[2] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/
[3] Grothendieck, A., 1971. Revêtement étales et groupe fondamental (SGA1). Lecture Note in Math., 224.
Alex Youcis
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Alex, thank you! – Sergei Akbarov Aug 21 '20 at 09:19
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Alex, this text by Osserman, it seems to be not published. Are there published books or articles for references? – Sergei Akbarov Aug 21 '20 at 09:53
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1@SergeiAkbarov Updated. – Alex Youcis Aug 21 '20 at 10:15
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Alex, I want to thank you in my paper. Probably it will be in Russian, so I need to know how your family name is pronounced: [ju:tsis] or maybe [ju:kis], or maybe else? – Sergei Akbarov Sep 04 '20 at 12:53
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@SergeiAkbarov Hey Sergei. There's no need for you to thank me! Just thank the writers of EGA :) – Alex Youcis Sep 04 '20 at 12:56
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Hm... OK.$\phantom{...}$ – Sergei Akbarov Sep 04 '20 at 13:14