Paraphrased from Munkres' Topology:
Lemma 9.2. Given a collection $\mathcal{A}$ of nonempty sets, there
exists a choice function
\begin{equation*} f: \mathcal{A} \to \bigcup\limits_{A \in
\mathcal{A}}A \end{equation*} such that $f(A) \in A$ for each $A \in
\mathcal{A}$.
Now consider Proof 1 of the statement "$f$ is surjective iff it has a right inverse" as pictured below:
Instead of $S$, I am using $A$; and instead of $T$, I am using $B$.
Looking at the $f$ is surjective $\implies $ $f$ has a right inverse direction, we have this collection of preimages, $\{f^{-1}(\{y\})\}_{y \in B} = X$. The choice function given in the proof is $g: B \to A$. Shouldn't it technically be $g: X \to A$?
And is there not an abuse of notation throughout the proof as $y$ should be a singleton set? Especially in the last step?
Best Answer
I don't have a problem with their use of the phrase "using the axiom of choice" because it is not as binding as "by the axiom of choice". The word "using" suggests some details are hidden; here are those details.
To be absolutely precise, we need to introduce two operations on maps between sets. If $f:A\to B$ is any map, define the direct image map $f_*:P(A)\to P(B)$ by $f_*(U) = \{f(a) : a \in U\}$ and define the inverse image map $f^*:P(B)\to P(A)$ by $f^*(V) = \{a\in A:f(a)\in V\}$. With this notation, the preimage of a point $y$ under $f$, typically denoted $f^{-1}(y)$, is actually $f^*(\{y\})$.
The family $X = \{f^*(\{y\}) : y\in B\}$ is a collection of pairwise disjoint sets; none of them is empty because $f$ is surjective. By the axiom of choice (your lemma 9.2) there exists a map $$g_1:X\to\bigcup_{C\in X} C = A$$ such that $g_1(C) \in C$ for all $C$ in $X$. Note this means $g_1$ is injective, because if $g_1(C) = g_1(D)$ then they are both in $C\cap D$, which means $C\cap D\ne\varnothing$, which means $C = D$.
Since every element of $X$ is of the form $f^*(\{y\})$ for some $y$ in $B$, we are prompted to make the following definition. Define $$g_2:B\to X$$ by $g_2(y) = f^*(\{y\})$. Note that $g_2$ is also injective because if $g_2(x) = g_2(y)$ then applying $f_*$ to both sides yields $\{x\} = \{y\}$, hence $x = y$. For example, $$f_*(g_2(x)) = \{f(a):a\in f^*(\{x\})\} = \{f(a):a\in\{a\in A:f(a)\in\{x\}\}\} = \{f(a):a\in\{a\in A:f(a) = x\}\} = \{x\}.$$
Now let $g$ be the composition of the two injections $g_1\circ g_2:B\to A$, and the rest should be clear: $g(y)$ is an element of $f^*(\{y\})$, so $f(g(y))$ is an element of $\{y\}$, which means $f(g(y)) = y$.