Bourbaki

Lemma 1

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

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

Let Y be a topological space and let φ:I×I→Y\varphi :\mathbf{I}\times \mathbf{I}\rightarrow Y be a continuous mapping. For t∈It\in \mathbf{I}, put c0(t)=φ(t,0),c1(t)=φ(t,1)c_0(t) =\varphi (t,0),c_1(t) =\varphi (t,1), d0(t)=φ(0,t)d_0(t) =\varphi (0, t) and d1(t)=φ(1,t)d_1(t) =\varphi (1, t). The paths c0∗d1c_0*d_1 and d0∗c1d_0*c_1 are strictly homotopic.

Let cc denote the path in I×I\mathbf{I}\times \mathbf{I} obtained by juxtaposing the paths t↦(t,0)t\mapsto (t,0) and t↦(1,t)t\mapsto (1, t); let dd denote the path in I×I\mathbf{I}\times \mathbf{I} obtained by juxtaposing the paths t↦(0,t)t\mapsto (0, t) and t↦(t,1)t\mapsto (t,1). The paths cc and dd have the same origin (0,0)(0,0) and the same end (1,1)(1,1). The mapping (t,s)↦(t, s)\mapsto (1−s)c(t)+sd(t)(1-s)c(t)+sd(t) of I×I\mathbf{I}\times \mathbf{I} into I×I\mathbf{I}\times \mathbf{I} is a strict homotopy joining cc to dd. One has c0∗d1=φ∘cc_0*d_1=\varphi \circ c and d0∗c1=φ∘dd_0*c_1=\varphi \circ d; these two paths are therefore strictly homotopic.

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.

Cited by

Nothing.

Cites

Nothing.

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