Let $X$ be a compact complex manifold and $V$ an analytic subset of $X$ of codimension $\ge 2$. By intuition, il might be true that the fundamental group of $X$ equals that of $X \backslash V.$ I would like to ask if the last claim is true or not and in the affirmative case, how do we prove it? Thank you in advance. 



The fundamental group of the complement of an analytic subset of codimension at least $2$
The answer is *yes*.


Just triangulate $X$ in such a way that $V$ is a subpolyhedron (this can be done by Lojasiewicz's theorem). Now it is well-known that removing a subpolyhedron of (real) codimension at least $3$ does not affect the fundamental group for transversality reasons.


**Edit.** Actually, Lojasiewicz's theorem is not necessary. In fact, the analytic subset $V$ provides a closed real submanifold of codimension at least $4$ of the underlying real manifold of $X$, so we can use the following result, whose proof can be found for instance in


C. Godbillon, *Eléments de Topologie Algebrique*, Théorème 2.3 page 146:



> 
> **Theorem.** Let $X$ be a smooth, connected real manifold without boundary, $V \subset X$ a closed submanifold, $x$ a point of $X\setminus V$ and $i \colon X \setminus V \longrightarrow X$ the inclusion map. Then
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> * if the codimension of $V$ is at least $2$, the group homomorphism $$i\_\* \colon \, \pi\_1(X \setminus V, \, x) \longrightarrow \pi\_1(X, \, x)$$ is surjective;
> * if the codimension of $V$ is at least $3$, the group homomorphism $$i\_\* \colon \, \pi\_1(X \setminus V, \, x) \longrightarrow \pi\_1(X, \, x)$$ is an isomorphism.
> 
> 
> 


