Proposition 3

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

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

Let E be a monoid and x an element of E.

(i) For x to be left (resp. right) invertible it is necessary and sufficient that the right (resp. left) translation by x be surjective.

(ii) For x to be invertible it is necessary and sufficient that it be left and right invertible. In that case x has a unique inverse, which is also its unique left (resp. right) inverse.

If x' is a left inverse of x then (Proposition 1)

\delta_x \circ \delta_{x'} = \delta_{x' \top x} = \delta_e = \mathrm{Id}_E

and $\delta_x$ is surjective. Conversely, if $\delta_x$ is surjective, there exists an element $x'$ of E such that $\delta_x(x') = e$ and $x'$ is the left inverse of $x$ . The other assertions of (i) follow similarly.

If $x'$ (resp. $x''$ ) is a left (resp. right) inverse of $x$ , then

x' = x' \top e = x' \top (x \top x'') = (x' \top x) \top x'' = e \top x'' = x'',

whence (ii).

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

Cited by

Cites

Proposition 1