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A cogenerator in a category $\mathcal{C}$ is an object $\Omega$ such that for any pair of parallel arrows $f,g:X\rightrightarrows Y$ in $\mathcal{C}$ we have

$$ \forall h:Y\to\Omega\big(h\circ f=h\circ g\big)\implies f=g, $$

so morphisms into $\Omega$ are sufficient to delineate between distinct arrows. My question is

Are the Surreals ${\bf N_0}$ a cogenerator in the category ${\bf \leq-Field}$ of ordered fields and ordered field homomorphisms?

Obviously we have some size issues here; I will offer one remedy below, but feel free to use whatever tools you like to get the surreals into a category.

We will work in $ZFC+\text{there exist two}$ Grothendieck universes $U\subset V$, carying out our discourse in $V$ about the category ${\bf \leq-Field}_U$ of (possibly large in $U$) $U$-ordered fields. The surreals in $U$ are a weakly terminal object in this category; are they a cogenerator?


EDIT: To address some potentially confusing language that was initially pointed out by Pace Nielsen (thank you) and resolved in the comments below Kieth Kearnes' answer, by fields that are 'possibly large in $U$' I mean fields whose underlying classes are subclasses of $U$, not fields whose underlying classes are larger than all of $U$ or fields that are 'in $U$' in a membership sense.

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On page 21 of

Conway's Field of Surreal Numbers
Norman L. Alling
Transactions of the American Mathematical Society
Jan., 1985, Vol. 287, No. 1, pp. 365-386.

one finds the statement that, on page 43 of

J. H. Conway, On numbers and games, Academic Press, London, 1976,

Conway states that ``As an abstract Field, ${\bf N}_0$ is the unique universally embedding totally ordered Field.''

I don't have a copy of Conway's book, so don't know how he proved it. If the claim is true, then it is sufficient to answer your question, since given distinct $f, g\colon X\to Y$ one can take an embedding $h$ of $Y$ into ${\bf N}_0$ to separate them.

**Edit** I am editing my answer to respond to the comments.

Originally I thought the main question was in the last sentence of the question: The surreals in U are a weakly terminal object in this category; are they a cogenerator? The answer is affirmative, since in any category where all morphisms are monic, any weakly terminal object is a cogenerator.

Now the comments lead me to think that I trivialized the problem. It seems to be this, instead: Given that every $U$-small ordered field embeds in ${\bf N}_0(U)$, does it follow that every $U$-large ordered field in $U$ embeds in ${\bf N}_0(U)$?

I say Yes in NBG with Global Choice. Let $\mathbb G$ be a $U$-large ordered field. Use Global Choice to enumerate $\mathbb G$ by the ordinals in $U$. For each cardinal $\lambda$, let $\mathbb G_{\lambda}$ be the $U$-small ordered subfield of $\mathbb G$ generated by the elements enumerated by ordinals $<\lambda$. This yields a well-ordered (class size) filtration of $\mathbb G$. Starting with the embedding of $\mathbb G_0 = \mathbb Q$ into ${\bf N}_0(U)$, extend this embedding recursively to each ${\mathbb G}_{\alpha}$ using the following fact at successor stages:

Fact. Any embedding of $\mathbb G_{\alpha}$ into ${\bf N}_0$ can be extended to $\mathbb G_{\alpha+1}$. This follows from the fact that ${\bf N}_0$ is a $\kappa$-universally extending model for all $\kappa$ (see An alternative construction of Conway's ordered field ${\bf N}_0$ by Philip Ehrlich, Algebra Universalis, 25 (1988) 7-16). Here I am relying on the fact that $\mathbb G_{\alpha}$ and $\mathbb G_{\alpha+1}$ are $U$-small in order to reference the $\kappa$-universally extending model property.

At limit stages take unions. I believe that this process yields a $U$-large embedding of $\mathbb G$ into ${\mathbf N}_0(U)$. Hence ${\mathbf N}_0(U)$ is weakly terminal even with respect to $U$-large ordered fields in $U$, hence is a cogenerator even for the class of $U$-large ordered fields in $U$.

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  • $\begingroup$ The proof uses choice, correct? (probably in NBG?) $\endgroup$
    – Alec Rhea
    Commented Apr 16, 2021 at 20:23
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    $\begingroup$ See related question mathoverflow.net/q/227849/1946 concerning whether one can prove the universality results without global choice. Seems to be an open question, even just for the order structure. $\endgroup$ Commented Apr 17, 2021 at 20:39
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    $\begingroup$ For what it's worth, IIRC (I do have ONAG but haven't read it in a while) the result about $\mathbf{No}$ being the unique universally embedding totally ordered field is one of many in the book that's stated without proof. $\endgroup$ Commented Apr 18, 2021 at 4:06
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    $\begingroup$ @JoelDavidHamkins I realized yesterday that the next logical question was 'do we need choice/is this equivalent to global choice?', and then I remembered your question. Time for a bounty! $\endgroup$
    – Alec Rhea
    Commented Apr 18, 2021 at 4:31
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    $\begingroup$ @AlecRhea Thanks! $\endgroup$ Commented Apr 18, 2021 at 9:27

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