TITLE: Composition of 2 involutions
QUESTION [7 upvotes]: How can we prove that any bijection on any set is  a composition of 2 involutions ?
Since involutions are bijections mapping elements of a set to elements of the same set, I find it weird that this applies to any bijection.
Thanks for your help !

REPLY [5 votes]: I assume that by "bijection" on a set $S$ you mean a bijection from $S$ to itself.  The question would not make sense for a bijection from $S$ to some other set.
The bijection decomposes $S$ into orbits.  It suffices to prove for a single orbit.
An orbit under the bijection is either a finite cycle $p_0 \to p_1 \to p_2\to \cdots \to p_n = p_0$ or a two-sided infinite sequence $\cdots \to p_{-2} \to p_{-1} \to p_0 \to p_{1} \to p_2 \to \cdots$.
In the infinite case, you can take the involutions $p_i \to p_{-i}$ and
$p_i \to p_{1-i}$.  In the finite case, $p_i \to p_{-i \pmod n}$ and $p_i \to p_{1-i \pmod n}$.