Let (X,e) be a Hopf space and m:X×X→X its composition law. For any loops c,c′ of X at e, one has:
c′∗c∼c∗c′∼m∘(c,c′)
where ∼ denotes the relation of strict homotopy. The law of composition of the group π1(X,e) is the homomorphism obtained by composing the canonical isomorphism π1(X,e)×π1(X,e)→π1(X×X,(e,e)) (III, p. 297, corollary) and the homomorphism π1(m,(e,e)) of π1(X×X,(e,e)) into π1(X,e). The group π1(X,e) is commutative.
Let μ:π1(X,e)×π1(X,e)→π1(X,e) be the group homomorphism obtained by composing the canonical isomorphism of π1(X,e)×π1(X,e) onto π1(X× X,(e,e)) and the homomorphism π1(m,(e,e)). It is to be proved that one has
μ(γ,γ′)=γγ′=γ′γ(1)
for all γ,γ′∈π1(X,e). Taking account of Remark 2 of III, p. 297, one has:
μ(γ,γ′)=μ(γ,εe)μ(εe,γ′)=μ(εe,γ′)μ(γ,εe)(2)
Let c∈Ωe(X) be a loop of class γ; let us denote by m1:X→X the mapping defined by m1(x)=m(x,e); then μ(γ,εe) is the class of m1∘c. By definition of a Hopf space, the mappings pointed at e, m1 and IdX, have the same pointed homotopy class at e. Hence the loops at e, m1∘c and c, are strictly homotopic (III, p. 296, Cor. 1 of Prop. 3) and one has μ(γ,εe)=γ. One proves analogously that one has μ(εe,γ′)=γ′. Relation (1) then follows from relation (2).
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.
Nothing.
Nothing.