Bijective function $f:[a,b] \to [a,b]$ with $f(a) = a$ and $f(b) = b$ is monotonic increasing.

Question

If I have a bijective real function $f:[a,b] \to [a,b]$ with $f(a) = a$ and $f(b) = b,$ I want to show it is monotonic increasing or find a counterexample.

Notice that I'm not assuming continuity, for the intermediate value theorem and the condition on endpoints would easily take care of the problem in this case, and that is why I'm stuck. I can't think o a counterexample nor prove it. By removing the restrictions on the range (i.e. making it $\mathbb R$, forgetting about the endpoints condition and only asking injectivity) I can find a counterexample to the affirmation, so I think I must use $f(a) = a$ and $f(b) = b$ in some way...

Any help is appreciated, thanks.

Answer 1

Define $f:[0,1]\to[0,1]$ by $f(0)=0$, $f(1)=1$, and $f(x)=1-x$ otherwise.

Answer 2

Here's a counterexample (assuming $a<b$):

$f(x)=x$ except $f\left(a+\dfrac {b-a}3\right)=a+\dfrac {2(b-a)}3$ and $f\left(a+\dfrac {2(b-a)}3\right)=a+\dfrac {b-a}3;$

$f$ is a bijective real function with $f(a)=a$ and $f(b)=b$,

but $f$ is not monotonic increasing because $a+\dfrac {b-a}3<a+\dfrac {2(b-a)}3$

but $f\left(a+\dfrac {b-a}3\right)>f\left(a+\dfrac {2(b-a)}3\right)$.

Answer 3

If you want nowhere monotonic function, so take $f:[0,1]\rightarrow[0,1]$ such that $f(x)=x$ for rational $x$ and $f(x)=1-x$ for irrational $x$.