Exercise 9.5

Proof. Suppose that $f:A\to B$ is surjective. Now, by the choice axiom, the collection $A=P\left(A\right)-\left\{\text{∅}\right\}$ is a collection of nonempty sets and thus has a choice function $c$. Consider any $b\in B$ and the set $A_b=\left\{x\in A\mid f(x)=b\right\}$. Then $A_b\ne \varnothing$ since $f$ is surjective, and hence $A_b\in A$ since clearly also $A_b\subset A$ so that $A_b\in P\left(A\right)$. So set $h(b)=c(A_b)\in A_b$ so that $h(b)\in A$ since $A_b\subset A$. Hence $h$ is a function from $B$ to $A$.

Recall that, by definition, $h$ is a right inverse if and only if $f\circ h=i_B$, which we show presently. So consider any $b\in B$ and let $a=h(b)=c(A_b)\in A_b$ so that $f(a)=b$. Then clearly

which shows that $f\circ h=i_B$ since $b$ was arbitrary. Hence $h$ is the right inverse of $f$. □

Proof. Suppose that $f:A\to B$ is injective and $A\ne \varnothing$. Then $f$ is a bijection from $A$ to its image $f(A)\subset B$ and hence its inverse $f^{-1}$ is a function from $f(A)$ to $A$. Now, since $A$ is nonempty, there is an $a_0\in A$. So define $h:B\to A$ by

for any $b\in B$. Recall that $h$ is a left inverse of $f$ if and only if $h\circ f=i_A$ by definition, which we show now.

So consider any $a\in A$ and let $b=f(a)$ so that clearly $b\in f(A)$. Hence by definition $h(b)=f^{-1}(b)=f^{-1}(f(a))=a$. Finally, we have

This shows that $h\circ f=i_A$ since $a$ was arbitrary. Therefore $h$ is a left inverse of $f$ as desired. □