Proposition 1

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

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

If the law $\mathsf{T}$ is associative, then for all $x \in E$ and $y \in E$

\gamma_{x \mathsf{T} y} = \gamma_x \circ \gamma_y, \qquad \delta_{x \mathsf{T} y} = \delta_y \circ \delta_x.

For all $z \in E$ :

\gamma_{x \mathsf{T} y}(z) = (x \mathsf{T} y) \mathsf{T} z = x \mathsf{T} (y \mathsf{T} z) = \gamma_x(\gamma_y(z))

\delta_{x \mathsf{T} y}(z) = z \mathsf{T} (x \mathsf{T} y) = (z \mathsf{T} x) \mathsf{T} y = \delta_y(\delta_x(z))

In other words, the mapping $x \mapsto \gamma_x$ is a homomorphism from the magma $E$ to the set $E^E$ of mappings of $E$ into itself with the law $(f, g) \mapsto f \circ g$ ; the mapping $x \mapsto \delta_x$ is a homomorphism of $E$ into the set $E^E$ with the opposite law. If $E$ is a monoid, these homomorphisms are unital.

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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