IMC 1997 · Problem 5

Let $X$ be an arbitrary set, let $f$ be an one-to-one function mapping $X$ onto itself. Prove that there exist mappings $g_1, g_2 : X \to X$ such that $f = g_1 \circ g_2$ and $g_1 \circ g_1 = \mathrm{id} = g_2 \circ g_2$, where $\mathrm{id}$ denotes the identity mapping on $X$.