If $f\circ f\circ f=id$, then $f=id$

Question

Let $f$ a continuous function on all $\mathbb R$. How can I prove that if $f\circ f\circ f=id$, then $f=id$ ? I really have no idea.

Answer 1

By studying the domain, you have $f : \mathbb{R} \rightarrow \mathbb{R}$.

$\forall x \in \mathbb{R}$ the image by $f$ of $f(f(x))$ is $x$ so $f$ is onto. Also if $f(a)=f(b)$, then $a=f(f(f(a)))=f(f(f(b)))=b$, so $f$ is injective. So $f$ is one to one. If $f$ is decreasing $f\circ f$ is increasing and $f\circ f\circ f$ is decrasing, but $id$ is increasing, it is absurd, so $f$ is strictly increasing.

Suppose $f(x)>x$, since $f$ is strictly increasing $f(f(x))>f(x)>x$, so $x=f(f(f(x))>f(f(x))>f(x)>x$, it is impossible. In the same way $f(x)<x$ is impossible, so $\forall x, f(x)=x$. So $f=id$.