Bourbaki

Corollary 2

Tag 01YR, permanent. Cite it as [Bourbaki, Tag 01YR].

Topologie algébrique, chapter III, § 3. Groupoïde de Poincaré

Let X and Y be topological spaces and let f:X→f: X\rightarrow Y be a homeomorphism. For every point xx of X, the homomorphism

π1(f,x):π1(X,x)→π1(Y,f(x)) \pi_1(f, x):\pi_1(X, x)\rightarrow \pi_1(Y, f(x))

is an isomorphism.

Let gg be a continuous mapping of Y into X, inverse to ff up to homotopy. Let xx be a point of X. It follows from Prop. 3 applied to the homotopic mappings IdX_X and g∘f:X→Xg\circ f: X\rightarrow X that the mapping π1(g∘f,x)\pi_1(g\circ f, x) is an isomorphism of the group π1(X,x)\pi_1(X, x) onto the group π1(X,g∘f(x))\pi_1(X, g\circ f(x)). Since

π1(g∘f,x)=π1(g,f(x))∘π1(f,x) \pi_1(g\circ f, x) =\pi_1(g, f(x))\circ \pi_1(f, x)

the homomorphism π1(f,x)\pi_1(f, x) is injective and the homomorphism π1(g,f(x))\pi_1(g, f(x)) is surjective. Since the mapping gg is also a homotopy equivalence, the homomorphism π1(g,f(x))\pi_1(g, f(x)) is injective; it is therefore an isomorphism. Hence π1(f,x)\pi_1(f, x) is an isomorphism.

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.