# Continuous Function on Closed Interval is Bijective iff Strictly Monotone It has been suggested that this article or section be renamed: specify that the function is real One may discuss this suggestion on the talk page.

## 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 $\closedint a b$.

$\blacksquare$