Axiom of choice need help with functions If $S$ is a set of $4$ arbitrary sets. And $B$ is a set of Union of those $4$ arbitrary sets. Let $$C = \{f:S \to B \mid f\text{ is a choice function}\}\;.$$ Show there is a bijection $q:C \to \text{set}_1\times\text{set}_2\times\text{set}_3\times\text{set}_4$.
 A: Let $S=\{S_1,S_2,S_3,S_4\}$, and let $B=S_1\cup S_2\cup S_3\cup S_4$. Then $C$ is the set of all functions $f:S\to B$ such that $f(S_k)\in S_k$ for $k=1,2,3,4$. We want to match up that function $f$ with some member of the Cartesian product $S_1\times S_2\times S_3\times S_4$.
Suppose that $f(S_1)=s_1\in S_1$, $f(S_2)=s_2\in S_2$, $f(S_3)=s_3\in S_3$, and $f(S_4)=s_4\in S_4$; then $\langle s_1,s_2,s_3,s_4\rangle\in S_1\times S_2\times S_3\times S_4$. More generally, for any $f\in C$ we can look at the $4$-tuple $\langle f(S_1),f(S_2),f(S_3),f(S_4)\rangle$ and see that it’s in $S_1\times S_2\times S_3\times S_4$. That suggests that we should let
$$q(f)=\langle f(S_1),f(S_2),f(S_3),f(S_4)\rangle\;.$$
To finish the problem, you have to show that this map $q$ really is a bijection: if $f,g\in C$, then $q(f)=q(g)$ if and only if $f=g$, and for each $\langle s_1,s_2,s_3,s_4\rangle\in S_1\times S_2\times S_3\times S_4$ there is an $f\in C$ such that $q(f)=\langle s_1,s_2,s_3,s_4\rangle$. Neither of these is very hard to do.
A: For $f\in C$, define $q(f)$ to be $(f(set_1), f(set_2), f(set_3), f(set_4))$.
Then $q(f) = q(g) \implies (f(set_1), f(set_2), f(set_3), f(set_4)) = (g(set_1), g(set_2), g(set_3), g(set_4))$ so $f = g$, therefore, q is injective.
For any point $(a,b,c,d) \in set_1\times set_2 \times set_3 \times set_4$, define $h$ by
$$h(set_1) = a,\, h(set_2) = b, h(set_3) = c,\,  h(set_4) = d$$
Then, $q(h) = (a,b,c,d)$ so q is surjective.
Therefore, q is a bijection.
