Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 1E / Solution

An inverse function of $f:X\to Y$ is a function $h:Y\to X$ satisfying bothThese are two-sided inverse identities. If $f(x)=f(x^{\prime })$, apply $h$ to get $x=x^{\prime }$, so $f$ is injective. For every $y\in Y$, the identity $f(h(y))=y$ supplies a preimage, so $f$ is surjective. Thus $f$ is a bijection.

Conversely, if $f$ is a bijection, each $y\in Y$ has exactly one preimage. Define $h(y)$ to be that preimage. Then $f(h(y))=y$ and $h(f(x))=x$, giving both identities. Hence a function has a two-sided inverse exactly when it is a bijection. This also makes its inverse function unique.