Continuous Injection of Interval is Strictly Monotone

Theorem
Let $I$ be a real interval.

Let $f: I \to \R$ be a bijective continuous real function.

Then $f$ is strictly monotone.

Proof
Suppose that $f$ is not strictly monotone.

That is, there exist $x, y, z \in I$ with $x < y < z$ such that either $f \left({x}\right) \le f \left({y}\right) \ge f \left({z}\right)$, or $f \left({x}\right) \ge f \left({y}\right) \le f \left({z}\right)$.

Suppose $f \left({x}\right) \le f \left({y}\right) \ge f \left({z}\right)$.

If $f \left({x}\right) = f \left({y}\right)$, or $f \left({y}\right) = f \left({z}\right)$, or $f \left({x}\right) = f \left({z}\right)$, $f$ is not bijective, which is a contradiction.

Thus, $f \left({x}\right) < f \left({y}\right) > f \left({z}\right)$.

Suppose $f \left({x}\right) < f \left({z}\right)$.

That is:
 * $f \left({x}\right) < f \left({z}\right) < f \left({y}\right)$

As $f$ is continuous on $I$, the Intermediate Value Theorem can be applied.

Hence there exists $c \in \left({x \,.\,.\, y}\right)$ such that $f \left({c}\right) = f \left({z}\right)$.

As $z \notin \left({x \,.\,.\, y}\right)$, we have $c \ne z$.

So $f$ is not bijective, which is a contradiction.

Suppose instead $f \left({x}\right) > f \left({z}\right)$.

That is:
 * $f \left({z}\right) < f \left({x}\right) < f \left({y}\right)$

Again, as $f$ is continuous on $I$, the Intermediate Value Theorem can be applied.

Then, there exists $c \in \left({y \,.\,.\, z}\right)$ such that $f \left({c}\right) = f \left({x}\right)$.

So $f$ is not bijective, which is a contradiction.

If we suppose $f \left({x}\right) \ge f \left({y}\right) \le f \left({z}\right)$, we reach a similar contradiction.

By Axiom:Proof by Contradiction, $f$ is strictly monotone.