Proposition 2

Tag 03JE, permanent. Cite it as [Bourbaki, Tag 03JE].

Theory of Sets, chapter III, § 1. Order relations. Ordered sets

Let E, E' be two ordered sets, and let $u : E \to E'$ and $v : E' \to E$ be two decreasing mappings such that for all $x \in E$ and all $x' \in E'$ we have $v(u(x)) \geqslant x$ and $u(v(x')) \geqslant x'$ . Then

u \circ v \circ u = u \quad \textit{and} \quad v \circ u \circ v = v.

For the relation $v(u(x)) \geqslant x$ implies $(u(v(u(x))) \leqslant u(x)$ because $u$ is decreasing; on the other hand, we have $u(v(u(x))) \geqslant u(x)$ by replacing $x'$ by $u(x)$ in the inequality $u(v(x')) \geqslant x'$ . Hence the first equality; the proof of the second is similar.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

§ 4. Modules over integrally closed Noetherian domains
§ 1. Algebras

Cites

Nothing.