Continuous Function on Closed Interval is Bijective iff Strictly Monotone

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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:

\(\displaystyle \map f c\) \(=\) \(\displaystyle a\)
\(\displaystyle \map f d\) \(=\) \(\displaystyle b\)


It follows that for all $x \in \closedint c d$:

\(\displaystyle \map f x\) \(\ge\) \(\displaystyle \map f c\)
\(\displaystyle \) \(=\) \(\displaystyle a\)
\(\displaystyle \map f x\) \(\le\) \(\displaystyle \map f d\)
\(\displaystyle \) \(=\) \(\displaystyle b\)


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:

\(\displaystyle \map f c\) \(=\) \(\displaystyle b\)
\(\displaystyle \map f d\) \(=\) \(\displaystyle a\)


It follows that for all $x \in \closedint c d$:

\(\displaystyle \map f x\) \(\le\) \(\displaystyle \map f c\)
\(\displaystyle \) \(=\) \(\displaystyle b\)
\(\displaystyle \map f x\) \(\ge\) \(\displaystyle \map f d\)
\(\displaystyle \) \(=\) \(\displaystyle a\)


Thus the image of $f$ is again $\closedint a b$.

$\blacksquare$