Lemma 5.6.2.label Let $G$ be a group acting on a set $X$. Suppose $Y\suf X$ satisfies $\abs{[x]\cap Y}=1$ for each $x\in X$ then $X/G\iso Y$. Addtionally
- (1)
If $X$ has a partial order $\leq_{X}$ such that for any $x_{1}\sim x_{1}',x_{2}\sim x_{2}'\in X$ we have
\begin{align*}x_{1}\leq x_{2}\iff x_{1}'\leq x_{2}'\end{align*}then $X/G\iso Y$ as posets.
- (2)
If $X,Y$ is a convex cone and for any $x_{1}\sim x_{1}',x_{2}\sim x_{2}'\in X,\lam\geq0$ we have
\begin{align*}x_{1}+x_{2}\sim x_{1}'+x_{2}'&&\lam x_{1}\sim \lam x_{1}'\end{align*}then $X/G\iso Y$ as convex cones.
Proof. Consider the map $\p:Y\to X/G$ defined by $y\mto [y]$. Let $y_{1},y_{2}\in Y$ then $\p(y_{1})=\p(y_{2})\iff [y_{1}]=[y_{2}]\iff y_{1}=y_{2}$ since $\curl{y_1,y_2}\suf [y_{1}]\cap Y$. Let $E\in X/G$ then $E=[x]$ for some $x\in X$. Since $\abs{[x]\cap Y}=1$ there exists $y\in Y$ such that $y\in [x]$ so $[y]=[x]=\p(x)$.
- (1)
By hypothesis $[x_{1}]\leq_{X/G}[x_{2}]\iff x_{1}\leq_{X} x_{2}$ is a well-defined partial order on $X/G$. The restriction of $\leq_{Y}\define \leq_{X}|_{Y}$ to $Y$ is a partial order on $Y$. Suppose $y_{1},y_{2}\in Y$ then
\begin{align*}y_{1}\leq_{Y} y_{2}\iff y_{1}\leq_{X} y_{2}\iff \braks{y_1}\leq_{X/G}\braks{y_2}\iff\p(y_{1})\leq_{X/G}\p(y_{2})\end{align*} - (2)
By hypothesis $X/G$ is a convex cone with
\begin{align*}\braks{x_1}+\braks{x_2}=\braks{x_1+x_2}&&\lam\braks{x}=\braks{\lam x_1}\end{align*}Suppose $y_{1},y_{2}\in Y,\lam\geq 0$ then
\begin{align*}\p(y_{1}+y_{2})=\braks{y_1+y_2}=\braks{y_1}+\braks{y_2}=\p(y_{1})+\p(y_{2})\\ \p(\lam y_{1})=\braks{\lam y_1}=\lam \braks{y_1}=\lam\p(y_{1})\end{align*}
$\square$
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