Is the function $x \mapsto \max\{f(x),g(x)\}$ always an automorphism if $f$ and $g$ are?

Question

Consider a totally ordered set $X$ and a pair of automorphisms (i.e., order-preserving bijections) $f,g$ of $X.$ Then the function $$h:x \in X \mapsto \max\{f(x),g(x)\} \in X$$

is an order-preserving surjection. So if $h$ is injective, then it must an automorphism.

Is $h$ necessarily injective? If not, what is a counterexample?

Answer 1

Suppose $h(x) = h(y)$. Then if $f(x) = h(x)$ and $f(y) = h(y)$, we have $x=y$ ; similarly if $h(x) = g(x)$ and $h(y) = g(y)$. So we are left with the case $f(x) = h(x)$ and $g(y) = h(y)$ with $f(x) \neq g(x)$ and $f(y) \neq g(y)$. Without loss of generality, suppose $x \le y$. If $x < y$, then $$ g(y) = h(y) = h(x) = f(x) < f(y), $$ a contradiction to $h(y) = \max \{f(y), g(y)\}$. Therefore $x = y$.

Hope that helps,

Answer 2

Suppose $x\leq y$ and $h(x)=h(y)$. Since $f$ and $g$ are injective, the cases where $f(x)=f(y)$ or $g(x)=g(y)$ are trivial. Hence we may assume WLOG that $h(x)=f(x)\geq g(x)$ and $h(y)=g(y)\geq f(y)$. But then $f(y)\leq g(y)=f(x)$, which implies $x=y$.