= Solution
An <inverse function> of $f:X\to Y$ is a function $h:Y\to X$ satisfying both
$$
h\circ f=\operatorname{id}_X,\qquad f\circ h=\operatorname{id}_Y.
$$
These are two-sided inverse identities. If $f(x)=f(x')$, apply $h$ to get $x=x'$, 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 \b[a function has a two-sided inverse exactly when it is a bijection]. This also makes its <inverse function> unique.
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