Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 5D / iii / Solution

It can be false even on a two-element finite set. The same constant maps $f=g=0$ on $\{0,1\}$ satisfy $fg=f$, while $g\ne {\mathrm{id}}_X$. Indeed, a constant $f$ makes $fg=f$ for every $g$. If $f$ were an injection, one could cancel it pointwise to conclude $g(x)=x$, but finiteness alone supplies no such hypothesis. As before, the assertion is trivial on the empty set or a singleton.