Can you build the surreal numbers as a simple direct limit of ordered fields? The surreal numbers are sometimes called the "universally embedding" ordered field, in that every ordered field embeds into them. What "universally embedding" means seems to be somewhat complicated, so I am curious if a simpler method of building them will do.
The idea is, suppose we take "all of" the ordered fields, with homomorphisms as embeddings between them. We can then build the direct limit, which "all of" the ordered fields embed into. Is the result equal to the surreal numbers?
I put "all of" in quotes because there are clearly a few quirks that are needed to avoid set-theoretic paradoxes. So for instance, instead of looking at "all of" the ordered fields, we can look at only those of cardinality less than, suppose, some strongly inaccessible cardinal $\kappa$. Then we can build the direct limit without any problem. Is the result an "initial subfield" of the surreal numbers with birthday up to that inaccessible cardinal. (Or do we get something strictly larger?)
I am sure there are other ways to deal with the set theory issues posed by this question, so I will kind of leave it open to whatever makes for the best answer (as is pretty common when asking questions about surreal numbers in general).
EDIT: a few clarifications

*

*Clearly sometimes the choice of embedding from one field into another is non-unique. To phrase the question differently: suppose we have any arbitrary directed system in which the objects are all of the ordered fields. Is the direct limit of this directed system always isomorphic to the surreal numbers?


*Whatever the direct limit is, every ordered field embeds into it. Also, every ordered field embeds into the surreal numbers. Does this tell us anything about the relationship between the two fields?
 A: Here is one way to get a positive answer to the title question.
Theorem. There is a definable class $\mathcal{F}$ of ordered fields, containing isomorphic copies of any given field, and a directed order $\unlhd$
on them, with a definable commutative system of embeddings between them $\pi_{F,K}:F\to K$ for $F\unlhd K$, such that the direct limit of the system is the surreal field $\newcommand\No{\text{No}}\No$.
Proof.
The surreal field $\No$ itself is definable. Let $\mathcal{F}$ be the class of set-sized subfields $F\subseteq\No$. Define $F\unlhd K$ if and only if $F\subseteq K$, and let $\pi_{F,K}:F\to K$ be the inclusion map. This is a definable, directed, commutative system of embeddings, and the direct limit is clearly the surreal field $\No$ itself, since every surreal number is an element of some set-sized subfield. Every ordered field is isomorphic to a subfield of $\No$ by the universality property, and so $\mathcal{F}$ contains copies of any given ordered field. $\Box$
I realize that this answer is not achieving the goal you had wanted, which was an alternative route to universality, since I am using universality to prove that the construction has the property of containing copies of all fields.
