Extending morphisms via Zorn's Lemma In the linked question here, the user demonstrates two examples of extension of morphisms using Zorn's Lemma arguments, and I've seen the same pattern to extend morphisms before in other sources.
However, none of them contain a verification of the condition of Zorn's lemma that every totally ordered subset has an upper bound. They jump straight from establishing the order relation to the existence of a maximal element.
Presumably this is because the verification follows a routine pattern that is obvious if you have seen it before. But supposing I haven't, how do we know that these posets meet the upper bound condition of Zorn's Lemma?
 A: You are correct that the verification is a necessary part of applying Zorn's lemma and it is often skipped because it follows a routine pattern.
The setup is generally that we have a chain $\{(A_i, f_i)\}_{i \in I}$ where the $A_i$ are sets and the $f_i\colon A_i \to X$ are morphisms satisfying $A_i \subseteq A_j$ and $f_i = f_j|_{A_i}$ whenever $(A_i, f_i) \leq (A_j, f_j)$.  Then let $A = \bigcup_iA_i$ and define $f\colon A \to X$ by the following rule:  For $a \in A$ choose $i$ such that $a \in A_i$ and define $f(a) = f_i(a)$.
Now you have to check two things, first our definition of $f$ required a choice and we need to prove $f$ does not depend on this choice.  This follows because of the restriction conditions on the $f_i$ which imply that $f_i(a)$ is the same for all $i$ such that $a \in A_i$.  Second, you have to check that $(A, f)$ is an upper bound for the chain $(A_i, f_i)$.  This amounts to showing that $f|_{A_i} = A_i$, which is true by definition of $f$ (because we can choose that $i$).
