Bourbaki

Proposition 4

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

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

(a) For all i∈Ii \in I, the mapping ϕi\phi_i is a homomorphism of MiM_i into M and ϕi∘hi=h\phi_i \circ h_i = h for all i∈Ii \in I. Further, M is generated by ⋃i∈Iϕi(Mi)\bigcup_{i \in I} \phi_i(M_i).

(b) Let M′M' be a monoid and f′:Mi→M′f' : M_i \to M' (for i∈Ii \in I) homomorphisms such that fi∘hif_i \circ h_i is independent of i∈Ii \in I. There exists one and only one homomorphism f:M→M′f : M \to M' such that fi=f∘ϕif_i = f \circ \phi_i for all i∈Ii \in I.

In what follows we shall make the following hypothesis:

(A) For all i∈Ii \in I, there exists a subset PiP_i of MiM_i containing eie_i such that the mapping (a,p)↦hi(a).p(a, p) \mapsto h_i(a).p of A×PiA \times P_i into MiM_i is bijective.

It implies that the homomorphisms hih_i are injective. Let x∈Mx \in M; every finite sequence σ=(a;i1,…,in;p1,…,pn)\sigma = (a; i_1, \ldots, i_n; p_1, \ldots, p_n) with a∈A,iα∈Ia \in A, i_\alpha \in I and pα∈Piαp_\alpha \in P_{i_\alpha} for 1≤α≤n1 \leq \alpha \leq n, satisfying

x=h(a)⋅∏α=1nϕiα(pα) x = h(a) \cdot \prod_{\alpha=1}^n \phi_{i_\alpha}(p_\alpha)

is called a decomposition of xx. The integer n≥0n \geq 0 is called the length of the decomposition σ\sigma and is denoted by l(σ)l(\sigma); the sequence (e)(e) is a decomposition of length 0 of the identity element of MM. The decomposition σ\sigma is called reduced if iα≠iα+1i_\alpha \neq i_{\alpha+1} for 1≤α<n1 \leq \alpha < n and pα≠eiαp_\alpha \neq e_{i_\alpha} for 1≤α≤n1 \leq \alpha \leq n.

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

Cited by

Cites

Nothing.

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