Collection of surjective functions implies axiom of choice

Question

if I have this:

(a) If $\left \{ f_i:A_i\rightarrow B_i|i\epsilon I \right \}$ is a collection of surjective functions then $\prod_{i\epsilon I} f_i: \prod_{i\epsilon I} A_i\rightarrow \prod_{i\epsilon I} B_i$ with $(a_i)_{i\epsilon I} \mapsto (f_i(a_i))_{i\epsilon I} $ is a surjective function.

(b) Every nonempty set has a choice function.

How can I prove that (a) implies (b)?

Answer 1

We will prove the contraposition. It is known that, the axiom of choice is equivalent to following statement:

If $\{A_i\}_{i\in I}$ is a family of nonempty sets, then $\prod_{i\in I}A_i$ is nonempty.

Assume that the choice fails, then there is a family of nonempty sets $\{A_i\}_{i\in I}$ with $\prod_{i\in I}A_i=\varnothing$. For each $i$, define $f_i:A_i\to \{0\}$ such that $f_i(x)=0$. We can check that each $f_i$ is surjective. However, the function $\prod f_i:\prod A_i\to\{0\}^I$ has empty domain and nonempty codomain so is not surjective.