Bourbaki

Proposition 2

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

Topologie algébrique, chapter III, § 2. Homotopie et chemins

Let X be a topological space. The map which associates with a path cc the opposite path c‾\overline{c} is a homeomorphism of Λ(X) onto itself. The juxtaposition of paths (c,d)↦c∗d(c, d)\mapsto c*d is a homeomorphism of the space (Λ(X),e)×X(Λ(X),o)(\Lambda (X), e)\times_X(\Lambda (X), o) onto Λ(X).

The mapping t↦1−tt\mapsto 1-t is a homeomorphism of the space I\mathbf{I} onto itself; the first assertion follows.

Let C denote the fibered product (Λ(X),e)×X(Λ(X),o)(\Lambda (X), e)\times_X(\Lambda (X), o). Let γ:C→\gamma : C\rightarrow Λ(X) be the mapping (c,d)↦c∗d(c, d)\mapsto c*d and δ:C×I→\delta : C\times \mathbf{I}\rightarrow X the mapping ((c,d),t)↦(c∗d)(t)((c, d), t)\mapsto (c*d)(t). The restrictions of the mapping δ\delta to C×[0,12]C\times [0,\frac{1}{2}] and C×[12,1]C\times [\frac{1}{2},1] are continuous by virtue of formula (1) and of prop. 1. The mapping δ\delta is therefore continuous (TG, I, p. 19, prop. 4), as is the mapping γ\gamma (prop. 1). We have c(t)=(c∗d)(t2)c(t) = (c*d)(\frac{t}{2}) and d(t)=(c∗d)(1+t2)d(t) = (c*d)(\frac{1+t}{2}), hence γ\gamma is injective.

Let gg be a path in X; for t∈It\in \mathbf{I}, put

cg(t)=g(t2)c_g(t) =g(\frac{t}{2}) and dg(t)=g(1+t2)d_g(t) =g(\frac{1 + t}{2}).

The mappings cgc_g and dgd_g thus defined are juxtaposed paths in X and we have cg∗dg=gc_g*d_g=g. Moreover, the mappings g↦cgg\mapsto c_g and g↦dgg\mapsto d_g are continuous mappings of the space Λ(X) into itself (prop. 1). It follows that the mapping γ\gamma is a homeomorphism.

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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Cites

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Machine translation of the English, by gpt-5-6-mini, gpt-5-mini, gpt-5-6, gpt-5.4, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.