Let X be a topological space, let x,y,z,u be points of X, and let γ1∈ϖx,y(X),γ2∈ϖy,z(X),γ3∈ϖz,u(X) be path classes. One has
εxγ1=γ1εy=γ1(1)
γ1γ1=εx,γ1γ1=εy(2)
(γ1γ2)γ3=γ1(γ2γ3)(3)
Let c1∈Λx,y(X),c2∈Λy,z(X),c3∈Λz,u(X) be representatives of γ1,γ2 and γ3 respectively. Let φ:I→I be the function defined by
t/2 for 0⩽t⩽1/2,
(4) φ(t)=t−1/4 for 1/2⩽t⩽3/4,
2t−1 for 3/4⩽t⩽1.
The function φ is affine on each of the three intervals [0,1/2], [1/2,3/4] and [3/4,1]; it is therefore continuous. One has φ(0)=0 and φ(1)=1. It follows from formula (1) of III, p. 256 defining the juxtaposition of paths that
c1∗(c2∗c3)=((c1∗c2)∗c3)∘φ
By the example of III, p. 289, the paths c1∗(c2∗c3) and (c1∗c2)∗c3 are strictly homotopic, whence equality (3).
Analogously, the function ψ:I→I defined by
2t for 0⩽t⩽1/2,
ψ(t)=(5)
1 for 1/2⩽t⩽1
is continuous and satisfies ψ(0)=0,ψ(1)=1. The equality γ1εy=γ1 then follows from the fact that
c1∗ey=c1∘ψ
and the equality εxγ1=γ1 is proved analogously, whence (1).
The mapping σ:I×I→X defined by
c1(2ts) for 0⩽t⩽1/2,
σ(t,s)=(6)
c1(2(1−t)s) for 1/2⩽t⩽1
is continuous; it is a strict homotopy joining the path ex to the path c1∗c1, whence the first equality in (2). The second results from the first and from the fact that, for every path c, one has c=c.
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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