Solution
= Solution
Let $h:B\to A$ be a two-sided <inverse function> of $f$. If $f(a)=f(a')$, apply $h$ and use $h\circ f=\operatorname{id}_A$ to obtain $a=a'$, proving <injectivity>. For any $b\in B$, the other identity gives $b=f(h(b))$, proving <surjectivity>. Therefore \b[$f$ is a <bijection>]. Both inverse identities matter: a one-sided inverse alone would establish only the corresponding one of these two properties.