Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone
Theorem
Let $\closedint a b$ and $\closedint c d$ be closed real intervals.
Let $f: \closedint c d \to \closedint a b$ be a continuous real function.
Let $\map f c, \map f d \in \set {a, b}$.
Then $f$ is bijective if and only if $f$ is strictly monotone.
Proof
Necessary condition
Let $f$ be a bijection.
From Continuous Injection of Interval is Strictly Monotone, it follows that $f$ is strictly monotone.
$\Box$
Sufficient condition
Let $f$ be strictly monotone.
From Strictly Monotone Real Function is Bijective, it follows that $f$ is a bijection on its image.
From Image of Interval by Continuous Function is Interval, it follows that the image of $f$ is a real interval.
Suppose that $f$ is strictly increasing.
Then by definition $\map f c < \map f d$.
It follows that:
\(\ds \map f c\) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds \map f d\) | \(=\) | \(\ds b\) |
It follows that for all $x \in \closedint c d$:
\(\ds \map f x\) | \(\ge\) | \(\ds \map f c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds \map f x\) | \(\le\) | \(\ds \map f d\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b\) |
Thus, the image of $f$ is $\closedint a b$.
$\Box$
Suppose instead that $f$ is strictly decreasing.
Then by definition $\map f c > \map f d$.
It follows that:
\(\ds \map f c\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \map f d\) | \(=\) | \(\ds a\) |
It follows that for all $x \in \closedint c d$:
\(\ds \map f x\) | \(\le\) | \(\ds \map f c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \map f x\) | \(\ge\) | \(\ds \map f d\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a\) |
Thus the image of $f$ is again $\closedint a b$.
$\blacksquare$