# Fundamental group of an open subscheme of a normal scheme

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 $$\mathbb{C}$$?

• In codimension $2$, is it not a cone over an elliptic curve a counterexample? – Francesco Polizzi Nov 19 '18 at 16:05
• @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 $\mathbb{C}$. – user124771 Nov 19 '18 at 16:53
• Of course it is reduced – Francesco Polizzi Nov 19 '18 at 17:02

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 $$\pi_1(U) = \pi_1(E) = \mathbb{Z} \oplus \mathbb{Z}.$$

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

In fact, quite the opposite tends to be true. Mumford  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 $$\pi_1(X) = \{1\}$$.

 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.

• Well, strictly speaking, $X$ is not projective in your example. – Francesco Polizzi Nov 19 '18 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.