Use the choice axiom to show that if \(f: A \to B\) is surjective, then \(f\) has a right inverse \(h: B \to A\) .
Show that if \(f: A \to B\) is injective and \(A\) is not empty, then \(f\) has a left inverse. Is the axiom of choice needed?
Solution
Since \(f:A\to B\) is surjective, every fiber \[ f^{-1}(\{b\})\qquad(b\in B) \] is nonempty. The axiom of choice applied to this family of nonempty sets gives a choice function selecting \[ h(b)\in f^{-1}(\{b\}) \] for every \(b\in B\). Then \[ f(h(b))=b \] for every \(b\), hence \[ f\circ h=i_B. \] Thus \(h\) is a right inverse of \(f\).
Suppose \(f:A\to B\) is injective and \(A\ne\varnothing\). Choose one fixed point \(a_0\in A\). Define \[ g:B\to A, \qquad g(b)= \begin{cases} f^{-1}(b),&b\in f(A),\\a_0,&b\notin f(A). \end{cases} \] The inverse value in the first case is unique because \(f\) is injective. Hence, for every \(a\in A\), \[ g(f(a))=a, \] so \(g\circ f=i_A\). Thus \(g\) is a left inverse.
No axiom of choice is needed in (b): only one element \(a_0\) is chosen from the single nonempty set \(A\), and all remaining values are uniquely determined.