Solution
= Solution
\b[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 $g\ne\operatorname{id}_X$. 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.