Can a nonmonotone function $ f : \mathbb R \to \mathbb R $ have an inverse?

Question

I'm asked to give an example of a non-monotone function $ f : \mathbb R \to \mathbb R $ which has an inverse function. As far as I know, a function has an inverse if $ f ( x ) $ is one-to-one and strict monotonicity is required for invertibility.

I know that $ f ( x ) = \sin x $ is non-monotonic, but it doesn't pass the horizontal line test.

My conclusion is that providing such example is not possible because it doesn't exist.

Is that right or am I missing something? It's my first time with abstract algebra and I'm struggling to grasp some concepts.

Thank you very much :)

Answer 1

There's a reason why it is difficult to come up with an example of an invertible function that is not monotone. If $f$ is continuous and one-to-one on $\Bbb R$, then $f$ must be monotone on $\Bbb R$. (For a proof of this fact, see Theorem 2 of Chapter 12 of Michael Spivak's Calculus.)

Therefore, if $f:\Bbb R\to\Bbb R$ is not monotone, but is still one-to-one, then $f$ must be discontinuous. Here is one example: $$ f(x)=\begin{cases} \arctan(1/x)&\text{if $x\neq0$,} \\ 0 &\text{if $x=0$.} \end{cases} $$

Answer 2

Yes, it is possible if your function is not continuous.

For example $$f(x)=x+1, x\in Q $$ $$ f(x)=x-1, x\notin Q$$ has an inverse but it is not monotonic.