I always thought that strictly increasing functions always have inverses. However, now I am given a task to find a strictly increasing function $f(x)$ that is bounded between [0,1] and doesn't have an inverse $f(x)^{-1} : [f(0), f(1)] \rightarrow [0,1]$

I can't seem to come up with any examples that satisfy the criteria. Even discontinuous functions seem to have inverses. Any tips on this problem?

  • $\begingroup$ Maybe try making a discontinuous function that is not surjective onto $[f(0),f(1)]$. $\endgroup$ – Antonios-Alexandros Robotis Feb 23 '17 at 2:36

You seek a strictly increasing function $f\colon [0,1]\to [f(0),f(1)]$ that does not have an inverse. If $x<y$ then $f(x)<f(y)$ so $f$ is injective. If $f$ is surjective then it would therefore be a bijection and have an inverse. So to construct a counterexample, we need $f$ to not be surjective. By the intermediate value theorem, any continuous function will be surjective onto $[f(0),f(1)]$, so our counterexample must be discontinuous.

This leads us to the following counterexample: $$ f(x)=\begin{cases} \frac{x}{2},& 0\leq x\leq \frac{1}{2}\\ \frac{x+1}{2},&\frac{1}{2}<x\leq 1. \end{cases} $$ It is a strictly increasing discontinuous function that maps $[0,1]$ to $$\left[0,\frac{1}{4}\right]\cup \left(\frac{3}{4},1\right].$$


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