Exercise 8, § 1

This solution was written by hy3-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

The subsets $A$ and $B$ are ordered by the relations induced by the orderings of $E$ and $F$ respectively, by Example 2 of no. 1 (tag 03SI).

Let $\varphi$ denote the mapping $x \mapsto f(x)$ with domain $A$ and codomain $F$, and let $\psi$ denote the mapping $y \mapsto g(y)$ with domain $B$ and codomain $E$.

Since $f$ is an increasing mapping of $E$ into $F$ (Definition 1 of § 1, tag 03JC), for every $x \in A$ we have $g(f(x)) = x$ by the definition of $A$; applying $f$ to this equality gives $f(g(f(x))) = f(x)$, so $f(x)$ satisfies the defining condition of $B$ and therefore $f(x) \in B$. Hence $\varphi$ maps $A$ into $B$. Similarly, since $g$ is an increasing mapping of $F$ into $E$ (Definition 1 of § 1, tag 03JC), for every $y \in B$ the equality $f(g(y)) = y$ implies on applying $g$ that $g(f(g(y))) = g(y)$, so $g(y) \in A$; thus $\psi$ maps $B$ into $A$.

For $x \in A$, we have $\psi(\varphi(x)) = g(f(x)) = x$ by the definition of $A$, and for $y \in B$, $\varphi(\psi(y)) = f(g(y)) = y$ by the definition of $B$. Consequently $\psi \circ \varphi = \mathrm{id}_A$ and $\varphi \circ \psi = \mathrm{id}_B$, so $\varphi$ is a bijection of $A$ onto $B$ and $\psi$ is its inverse.

Let $x_1 \leqslant x_2$ in $A$, which means $x_1 \leqslant x_2$ in $E$. Because $f$ is increasing (Definition 1 of § 1, tag 03JC), $f(x_1) \leqslant f(x_2)$ in $F$; since $f(x_1), f(x_2) \in B$, this is precisely $\varphi(x_1) \leqslant \varphi(x_2)$ in the induced order on $B$. Thus $\varphi$ is increasing. Likewise, if $y_1 \leqslant y_2$ in $B$, then $g(y_1) \leqslant g(y_2)$ in $E$ because $g$ is increasing (Definition 1 of § 1, tag 03JC), so $\psi$ is increasing. By the standard definition of an isomorphism of ordered sets, a bijective increasing mapping whose inverse is also increasing is an isomorphism; therefore $\varphi$ is an isomorphism of the ordered set $A$ onto the ordered set $B$.

The isomorphism $\varphi$ and its inverse $\psi$ are obtained simply by restricting the given mappings $f$ and $g$, so this isomorphism is canonical. Hence the ordered sets $A$ and $B$ are canonically isomorphic.