I was reading the book Sets, Models and Proofs and I'm now stuck at the part where they prove that Zorn's Lemma implies AC. In particular, this part:
"Proof. We assume that Zorn’s Lemma is true. Suppose given a surjective function $f : X → Y$. A partial section of $f$ is a pair $(A, u)$ where $A$ is a subset of $Y$ and $u : A → X$ a function such that $f (u(y)) = y$ for each $y ∈ A$. Given two such partial sections $(A, u)$ and $(B, v)$, put $(A, u) ≤ (B, v)$ iff $A ⊆ B $ and u is the restriction of v to A. Let P be the set of partial sections (A, u) of f ; then with the relation ≤, P is a poset, as is easy to see.
P is nonempty; this is left to you. . ."
What I'm confused about is this: Doesn't the existence of (A, u) require the Axiom of Choice? Because you have to choose an u for which f(u(y)) = y. For each y ∈ A, you have to pick an u(y) = x. I honestly don't see how I can prove that P is nonempty without using the Axiom of Choice or by using Zorn's Lemma, which is defined in the book as:
"Definition 1.3.4 Zorn’s Lemma is the following assertion: if (P , ≤) is a poset with the property that every chain in P has an upper bound in P, then P has a maximal element. Note that if P satisfies the hypothesis of Zorn’s Lemma, then P is nonempty. This is so because the empty subset of P is always a chain. However, checking that every chain has an upper bound in P usually involves checking this for the empty chain separately; that is, checking that P is nonempty."
Any help would be much appreciated.
Edit: For anyone that comes here after me, Do We Need the Axiom of Choice for Finite Sets? was also really helpful.
No, in the proof, the author is just saying what means that a pair $(A, u)$ is a partial section of $f$, it doesn't say anything about if such a pair exists for any given $A\subseteq Y$, the important thing is that the set of ALL partial sections of $f$ is not empty, but this is easy to check, just take any $y\in Y$, since $f$ is surjective there exists $x\in X$ such that $f(x)= y$ and if you define $u : \{y\} \rightarrow X$ as $u(y):=x$ then the pair $(\{y\}, u)$ is a partial section of $f$.