Past exam of the logic and set theory course of the University of Oxford / 2011 / B / 4 / a / ii

Suppose $h:\omega \hookrightarrow X$ is an injection. Shift its image by definingThe two pieces have disjoint images, so $g$ is injective, but $h(0)$ is omitted. Thus $g$ is not surjective.

Conversely, let $g:X\to X$ be an injection which omits some $a\in X$. The natural-number recursion theorem defines $h(0)=a$, $h(n+1)=g(h(n))$. If $h(m)=h(n)$ with $m<n$, repeated injectivity cancels $m$ iterations and gives $a=g^{n-m}(a)$, putting $a$ in the image of $g$, a contradiction. Hence $h$ is injective. ThereforeChoosing one witness $a$ from a single known nonempty set uses no axiom of choice. This is the Dedekind-infinite characterization.