Bourbaki

Lemma 2

Tag 05FO, permanent. Cite it as [Bourbaki, Tag 05FO].

Algebra, chapter I, § 7. Free monoids, free groups

Let MM be the monoidal sum of the family (Mx)x∈X(M_x)_{x \in X} defined by Mx=NM_x = \mathbf{N} for all x∈Xx \in X and let ϕx\phi_x denote the canonical homomorphism of MxM_x into MM. The mapping x↦ϕx(1)x \mapsto \phi_x(1) of XX into MM extends to an isomorphism hh of Mo(X)\mathrm{Mo}(X) onto MM.

Let hh be the homomorphism of Mo(X)\mathrm{Mo}(X) into MM characterized by h(x)=ϕx(1)h(x) = \phi_x(1). For every integer n≥0n \geq 0, ϕx(n)=ϕx(1)n=h(x)n\phi_x(n) = \phi_x(1)^n = h(x)^n and as MM is generated by ⋃x∈Xϕx(N)\bigcup_{x \in X} \phi_x(\mathbf{N}), it is also generated by h(X)h(X). Hence hh is surjective. Moreover, for all xx in XX, the mapping n↦xnn \mapsto x^n is a homomorphism of N=Mx\mathbf{N} = M_x into Mo(X)\mathrm{Mo}(X); there thus exists (no. 3, Proposition 4) a homomorphism h′h' of MM into Mo(X)\mathrm{Mo}(X) such that h′(ϕx(n))=xnh'(\phi_x(n)) = x^n for x∈Xx \in X and n∈Nn \in \mathbf{N}; in particular, h′(h(x))=xh'(h(x)) = x for x∈Xx \in X and hence h′∘hh' \circ h is the identity homomorphism of Mo(X)\mathrm{Mo}(X). Therefore hh is injective. It has thus been proved that hh is bijective.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

Transcribed from the printed edition, 1998, Springer. Not translated.