Bourbaki

Proposition 3

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

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

Let X and Y be topological spaces, f0f_0 and f1f_1 continuous mappings of X into Y, and let σ:X×I→\sigma : X\times \mathbf{I}\rightarrow Y be a homotopy joining f0f_0 to f1f_1. Let xx be a point of X; put y0=f0(x)y_0=f_0(x), y1=f1(x)y_1=f_1(x), and denote by δ∈ϖy0,y1(Y)\delta \in \varpi_{y_0,y_1}(Y) the class of the path dd defined by d(t)=σ(x,t)d(t) =\sigma (x, t) for t∈It\in \mathbf{I}. For every γ∈π1(X,x)\gamma \in \pi_1(X, x), one has (f1)∗(γ)=(f_1)_*(\gamma ) = δ−1((f0)∗(γ))δ\delta^{-1}((f_0)_*(\gamma ))\delta.

Let cc be a loop in X at xx and let γ\gamma be its strict homotopy class. Put γ0=(f0)∗(γ)\gamma_0= (f_0)_*(\gamma ) and γ1=(f1)∗(γ)\gamma_1= (f_1)_*(\gamma ); these are the strict homotopy classes of f0∘cf_0\circ c and f1∘cf_1\circ c. For (t,s)∈I×I(t, s)\in \mathbf{I}\times \mathbf{I}, put φ(t,s)=σ(c(t),s)\varphi (t, s) =\sigma (c(t), s). For every t∈It\in \mathbf{I}, one has φ(t,0)=(f0∘c)(t),φ(t,1)=(f1∘c)(t)\varphi (t,0) = (f_0\circ c)(t),\varphi (t,1) = (f_1\circ c)(t) and φ(0,t)=φ(1,t)=d(t)\varphi (0, t) =\varphi (1, t) =d(t). The relation γ0δ=δγ1\gamma_0\delta =\delta \gamma_1 therefore follows from the following lemma.

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.