Zorn's lemma on proper class with chains bounded in cardinality

Question

Basically, I'm trying to understand a note in Jacob Lurie's paper On a conjecture of Conway (Illinois Journal of Mathematics 46.2, 2002).

Let $X\subseteq Y$ be sets, $\mathbf{U}$ a proper class, and $\varphi\colon X \to \mathbf{U}$ an injective function. Then in the paper, Zorn's lemma is used on all partial extensions of $\varphi$ to $Y$, i.e. it is used on the collection $$\Phi := \{\psi\colon Z\to \mathbf{U} \;:\; X\subseteq Z \subseteq Y,\; \psi|_X = \varphi\},$$ partially ordered by $\psi \le \psi'$ iff $Z\subseteq Z'$ and $\psi'|_Z = \psi$, for $\psi$, $\psi' \in \Phi$ with respective domains $Z$ and $Z'$.

I have no problem with seeing that every chain in $\Phi$ has an upper bound. The problem is however that $\Phi$ is a proper class (since $\mathbf{U}$ is a proper class), so from my understanding, to apply Zorn's lemma directly, the axiom of global choice is needed.

However, Lurie writes:

The fact that this partial order is actually a proper class introduces a technicality, but it is easy to sidestep since every chain in this partial order is bounded in size.

By that I think he means that global choice is not necessary, but by some "trick" that I do not understand, regular choice is sufficient. Can anyone help me whether this is correct, and if so, point out the trick to me?

(Note that in the paper, $X$, $Y$ and $\mathbf{U}$ are taken to be partially ordered abelian groups, and the considered maps are order-preserving group homomorphisms. I am rather certain however that this does not matter for my question at hand.)

Answer 1

Yes, that is the idea behind Lurie's point. This makes heavy use of the von Neumann hierarchy (and Replacement).

Suppose $\bf U$ is our partially ordered class, and suppose that we have some $x\in\bf U$ and we want to find a maximal element above it. And suppose that any chain has fewer than $\kappa$ members, for some regular cardinal $\kappa$. We define the following subclasses by recursion.

$U_0=\{x\}$, $U_{\alpha+1}$ is the collection of all minimal rank extensions of members of $U_\alpha$, and at limit stages take unions and minimally ranked upper bound to any chain which does not have an upper bound. Now consider $U_\kappa$. I claim that in this partial order every chain has an upper bound, and so by Zorn's Lemma there is a maximal element in $U_\kappa$, and it is going to be maximal in $\bf U$.

The idea here is simple. If $C$ is a chain in $U_\kappa$, it is certainly a chain in $\bf U$, so $|C|<\kappa$, and by the regularity of $\kappa$, $C\subseteq U_\alpha$ for some $\alpha<\kappa$. Therefore we added an upper bound to $C$ by $U_{\alpha+\omega}$.

Now, if $u$ is a maximal element in $U_\kappa$, then again, there is some $\alpha<\kappa$ such that $u\in U_\alpha$. If $u$ is not maximal in $\bf U$, then it has some extension, and therefore there is some minimally ranked extension, which we must have added in $U_{\alpha+1}$, which would contradict the maximality of $u$ in $U_\kappa$.