Bourbaki

Remark

Tag 056Y, permanent. Cite it as [Bourbaki, Tag 056Y].

Algebra, chapter I, § 2. Identity element; cancellable elements; invertible elements

Let E be a monoid and x an element of E. If x is left invertible it is left cancellable; for, if x′x' is a left inverse of xx, then

γx′∘γx=γx′⊤x=γe=IdE \gamma_{x'} \circ \gamma_x = \gamma_{x' \top x} = \gamma_e = \mathrm{Id}_E

and γx\gamma_x is injective. In particular, if x is left invertible, the left and right translations by x are bijective. Conversely, suppose that γx\gamma_x is bijective; there exists x′∈Ex' \in E such that xx′=γx(x′)=exx' = \gamma_x(x') = e; then γx(x′x)=(xx′)x=x=γx(e)\gamma_x(x'x) = (xx')x = x = \gamma_x(e) and hence x′x=ex'x = e, so that x is invertible. We see similarly that if δx\delta_x is bijective, x is invertible.

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

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Cites

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