Exercise 2, § 1

This solution was written by laguna-s-2.1-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

(a) Let $E/S$ denote the quotient set of an equivalence relation $S$ on a preordered set $E$. The relation $R(X,Y)$ on $E/S$ is defined by requiring that for each $x \in X \in E/S$, there exists $y \in Y \in E/S$ such that $x \leqslant y$. To verify $R$ is a preorder relation:

Reflexivity: For any $X \in E/S$, take $x \in X$. Since $x \leqslant x$, there exists $y = x \in X$ such that $x \leqslant y$. Thus, $R(X,X)$.

Transitivity: Suppose $R(X,Y)$ and $R(Y,Z)$. For each $x \in X$, there exists $y \in Y$ with $x \leqslant y$, and for each such $y$, there exists $z \in Z$ with $y \leqslant z$. Hence $x \leqslant z$ for each $x \in X$, so $R(X,Z)$. Thus, $R$ is a preorder. This $R$ is the quotient of $E$ by $S$.

(b) Let $\varphi: E \to E/S$ be the canonical map. Suppose $g : E/S \to F$ is a mapping such that $g \circ \varphi$ is increasing. For $X \leqslant Y \in E/S$ (i.e., $R(X,Y)$), each $x \in X$ satisfies $x \leqslant y$ for some $y \in Y$. Since $g \circ \varphi$ is increasing, $\varphi(x) \leqslant \varphi(y)$. But $\varphi(x) = X$ and $\varphi(y) = Y$, so $g(X) \leqslant g(Y)$. Thus, $g$ is increasing.

We now prove $\varphi$ is increasing if and only if $S$ satisfies condition (C).

$(\Rightarrow)$ Suppose $\varphi$ is increasing and assume $x \leqslant y$ and $x \equiv x' \pmod{S}$. Then $\varphi(x) = \varphi(x')$, and $\varphi(x) \leqslant \varphi(y)$ since $\varphi$ is increasing. By the definition of $R$, the relation $R(\varphi(x), \varphi(y)) = R([x],[y])$ means: for every $x' \in [x]$, there exists $y' \in [y]$ such that $x' \leqslant y'$. Here $x' \in [x] = [x']$, so there exists $y' \in [y]$ with $x' \leqslant y'$. This is exactly condition (C).

$(\Leftarrow)$ Suppose (C) holds. To show $\varphi$ is increasing, assume $x \leqslant y$ in $E$. We must show $R([x],[y])$, i.e., for each $x' \in [x]$, there exists $y' \in [y]$ with $x' \leqslant y'$. Given $x' \in [x]$, we have $x \equiv x' \pmod{S}$. Applying (C) to $x \leqslant y$ and $x \equiv x' \pmod{S}$, there exists $y' \equiv y \pmod{S}$ (so $y' \in [y]$) such that $x' \leqslant y'$. Hence $\varphi$ is increasing.

Therefore, $\varphi$ is increasing if and only if $S$ is weakly compatible.

(c) Consider $E_1 \times E_2$ with order $z \leqslant t$ component-wise. Let $S_1$ group elements by their first component. For $z \equiv z' \pmod{S_1}$ (same first component) and $z \leqslant t$, condition (C) requires a $t' \equiv t \pmod{S_1}$ (same first component as $t$) such that $z' \leqslant t'$. Since $z'$ shares $z$'s first component and $t'$ shares $t$'s, $z'_1 = z_1 \leqslant t_1 = t'_1$, so $z' \leqslant t'$ holds. Thus, $S_1$ is weakly compatible. The projection $\mathrm{pr}_1$ factors through $\varphi_1$, and $f_1$ maps $\varphi_1(z) \mapsto \mathrm{pr}_1(z)$, yielding an isomorphism with $E_1$.

(d) Assume $E$ is ordered and (C') holds. For $R(X,Y)$ and $R(Y,X)$ in $E/S$, each $x \in X$ has $y \in Y$ with $x \leqslant y$, and each $y \in Y$ has $x' \in X$ with $y \leqslant x'$. Then $x \leqslant y \leqslant x'$. Since $x \equiv x' \pmod{S}$, (C') implies $x \equiv y \pmod{S}$, hence $X = Y$. Thus, $R$ is antisymmetric, forming an order.

(e) Let $E = \{a, b, c, d\}$ ordered $a < b < c < d$, with $S$ defined as $\{a, c\}, \{b, d\}$. In $E/S$, classes $[a,c]$ and $[b,d]$ are ordered by $[a,c] \leqslant [b,d]$, making $E/S$ a poset. Condition (C) fails for $x = b \equiv x' = d$, $y = c$, since no $y' \equiv y$ exists with $d \leqslant y'$. Condition (C') fails for $x = a, y = b, z = c$, as $a \equiv c \pmod{S}$ but $a \not\equiv b$. Yet $E/S$ is ordered.

(f) For $f$ increasing, (C') holds since $f(x) \leqslant f(y) \leqslant f(z)$ and $f(x) = f(z)$ implies $f(y) = f(x)$. Condition (C) requires for $x \equiv x' \pmod{S}$, $x \leqslant y$, finding $y' \equiv y \pmod{S}$ with $x' \leqslant y'$, aligning with the problem statement. The canonical decomposition $f = g \circ \varphi$ has $g$ injective and increasing. For $g$ to be an iso onto $f(E)$, the converse must hold via the given order conditions.