Let $X$ be an irreducible normal projective scheme over $\mathbb{C}$. Let $U$ be the open subscheme of smooth points of $X$. Consider the closed subscheme $Z = X \setminus U$. Suppose that the codimension of $Z$ in $X$ is at least $2$. Is it true that the fundamental group of $U$ and $X$ are isomorphic?

Edit: Is it true for $X$ an integral normal projective scheme over $\mathbb{C}$?

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    $\begingroup$ In codimension $2$, is it not a cone over an elliptic curve a counterexample? $\endgroup$ Nov 19, 2018 at 16:05
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    $\begingroup$ @FrancescoPolizzi Is cone over an elliptic curve reduced? (sorry for asking a stupid question). Actually, I want to know the result for X normal projective integral scheme over $\mathbb{C}$. $\endgroup$
    – user124771
    Nov 19, 2018 at 16:53
  • $\begingroup$ Of course it is reduced $\endgroup$ Nov 19, 2018 at 17:02

3 Answers 3


Let me expand my comment into an answer.

Take as $X$ the cone of vertex $v$ over an elliptic curve $E$. Then $X$ is simply connected (this is a general property of projective cones). However, $U = X-\{v\}$ is not simply connected: in fact, the projection $\pi \colon U \to E$ onto the basis gives $X$ the structure of a topological fibration with fiber homeomorphic to $\mathbb{C}$, so the corresponding long exact sequence of homotopy groups yields $$\pi_1(U) = \pi_1(E) = \mathbb{Z} \oplus \mathbb{Z}.$$

  • $\begingroup$ Am I mistaken or is the fiber homeomorphic to $\mathbb{C}$? (not that it makes a difference for the conclusion). $\endgroup$
    – cgodfrey
    Apr 23, 2019 at 18:00
  • $\begingroup$ @cgodfrey: yes, thanks. I will correct it $\endgroup$ Apr 23, 2019 at 18:35

In fact, quite the opposite tends to be true. Mumford [1] showed that for $(X,0)$ the germ of a normal surface singularity (over $\mathbf{C}$), $U=X\setminus 0$, one has $\pi_1(U)=\{1\}$ if and only if $X$ is smooth. At the same time, $\pi_1(X) = \{1\}$ since $0\to X$ is a homotopy equivalence.

If $X$ is smooth, this is true (the etale variant is called "Zariski-Nagata purity").

EDIT. To address Francesco's comment: of course the example is not projective. The easiest projective example was given by Francesco in his comment: $X$ is the (projective) cone over an elliptic curve $E$ and $U$ the complement of the vertex. Then $\pi_1(U)= \pi_1(E) = \mathbb{Z}^2$ and $\pi_1(X) = \{1\}$.

[1] Mumford, D., The topology of normal singularities of an algebraic surface and a criterion for simplicity, Publ. Math., Inst. Hautes Étud. Sci. 9, 5-22 (1961). ZBL0108.16801.

  • $\begingroup$ Well, strictly speaking, $X$ is not projective in your example. $\endgroup$ Nov 19, 2018 at 17:04

Although $X$ and $U$ need not have isomorphic fundamental group (as the accepted answer shows), the induced map from the inclusion $U\hookrightarrow X$ is generally $\pi_1$-surjective.

Precisely, if $X$ is a normal projective variety, and $A\subset X$ is a proper closed subvariety, then the natural homomorphism $\pi_1(X-A) \to \pi_1(X)$ is surjective.

You can find this result in On the fundamental groups of normal varieties by Donu Arapura, Alexandru Dimca, Richard Hain.

Interestingly, although natural inclusions only give $\pi_1$-surjections, it turns out that natural projections do give $\pi_1$-isomorphisms.

Precisely, let $G$ be a connected reductive algebraic affine group over an algebraically closed field $k$ (arbitrary characteristic). Assume $G$ is acting on a smooth connected projective variety $M$ (and there is an appropriate ample line bundle $\mathcal{L}$). Then the homomorphism (induced by the GIT projection) of algebraic fundamental groups $\pi_1(M)\to \pi_1(M/\! /_{\mathcal{L}}G)$ is an isomorphism. If $k = \mathbb{C}$, then there is also an isomorphism between the topological fundamental groups.

You can find this result in Fundamental group of a geometric invariant theoretic quotient by Indranil Biswas, Amit Hogadi, A. J. Parameswaran.


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