Let $n \in \omega$. Suppose $f:n \to A$ is onto $A$. Prove that $A$ is finite.
I have: Let $I_a = \{i \in n:f(i)=a\}$ for $a \in A$. Since $f$ is onto $A$, $I_a$ is nonempty, and by the well-ordering principle, it has a least element $l$. I know I'm to prove by induction, by I'm a bit stuck. Any help is appreciated.
Since $f $ is surjective, it has a section $g:A\to n $, that's $f\circ g=\mathrm {is} $ Then $g $ is injective hence induce a bijection onto its image, that's $A\cong g [A] $. Since $g [A]\subseteq n$, it is finite, hence $A$ is finite as well.