Bourbaki

Theorem 1

Tag 021A, permanent. Cite it as [Bourbaki, Tag 021A].

Topologie algébrique, chapter IV, § 2. Groupes de Poincaré des espaces délaçables

Let X be a compact contractible topological space and let xx be a point of X. The Poincaré group π1(X,x)\pi_1(X, x) has a finite presentation.

The arcwise connected component of xx in X is open, closed and contractible; this makes it possible to suppose that the space X is connected. Since the space X is contractible, the X-space E=λx(X)E =\lambda_x(X), endowed with the terminal mapping, is a nonempty arcwise simply connected covering (IV, p. 342, th. 1). The group G = AutX(E)_X(E) is isomorphic to π1(X,x)\pi_1(X, x); it therefore suffices to prove that the group G has a finite presentation.

Since X is compact, every point xx of X possesses a compact neighbourhood KxK_x over which the covering E is trivializable (TG, I, p. 65, corollary). Since X is locally connected, every x∈Xx\in X possesses an open connected neighbourhood WxW_x, contained in KxK_x. Let F be a finite subset of X such that the WxW_x, for x∈Fx\in F, cover X. Let nn be the cardinal of F.

We shall prove by induction that, for every integer kk such that 1 ⩽k⩽n\leqslant k\leqslant n, there exist a subset A of cardinal kk contained in F and, for x∈Ax\in A, a section sxs_x of pp over KxK_x, such that the union of the sx(Wx)s_x(W_x), for x∈Ax\in A, is a connected subset of E. The assertion is true for k=1k= 1. Suppose it true for an integer kk such that 1⩽k<n1\leqslant k < n and let us prove that it is true for k+1k+ 1. Let A be a subset of F of cardinal kk and, for every x∈Ax\in A, let sxs_x be a section of E over KxK_x, such that ⋃x∈Asx(Wx)\bigcup_{x\in A}s_x(W_x) is connected. The open sets ⋃x∈AWx\bigcup_{x\in A}W_x and ⋃x∈F−AWx\bigcup_{x\in F-A}W_x are nonempty and cover X; their intersection is therefore nonempty, since X is connected. There thus exist x∈A,y∈F−Ax\in A,y\in F-A and z∈Wx∩Wyz\in W_x\cap W_y. Put A′=A∪{y}A'= A\cup \{y\} and choose a section sys_y of E over KyK_y such that sy(z)=sx(z)s_y(z) =s_x(z). The connected open sets ⋃p∈Asp(Wp)\bigcup_{p\in A}s_p(W_p) and sy(Wy)s_y(W_y) are nonempty and have a common point; their union is therefore connected. This proves the assertion for k+1k+ 1. By induction, it is therefore true for every integer k∈{1,...,n}k\in \{1, . . . , n\}.

Applying the preceding to k=nk=n, there then exists, for every x∈Fx\in F, a section sxs_x of E over KxK_x, in such a way that U=⋃x∈Fsx(Wx)U =\bigcup_{x\in F}s_x(W_x) is a connected open subset of E. We have p(U)=Xp(U) = X. Since the group G operates transitively in each fibre of the covering E, we have GU = E.

The closure of U is contained in ⋃x∈Fsx(Kx)\bigcup_{x\in F}s_x(K_x) and therefore is compact. The operation of G in E is proper (I, p. 96, cor. 1), hence the set of pairs (g,x)∈G×E(g, x)\in G\times E such that x∈U‾x\in \overline{U} and gx∈U‾gx\in \overline{U} is a compact subset of G×EG\times E (TG, I, p. 77, prop. 6). It follows that the set of g∈Gg\in G such that U‾∩gU‾=∅̸\overline{U}\cap g\overline{U}=\not\emptyset is compact. Since it is discrete, it is finite. A fortiori, the set of g∈Gg\in G such that U∩gU=∅̸U\cap gU=\not\emptyset is finite. The theorem thus follows from prop. 10 of I, p. 136.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, against glossary version 34. Not checked by a person.