Is there a categorification of (infinite) ordinal arithmetic? Background for the curious reader:
An ordinal $\beta$ is a transitive set in the sense that $\alpha\in\beta$ implies $\alpha\subset\beta$. Any ordinal is naturally well-ordered under $\in$ (so any subset of it has a least element), and any well-order is isomorphic to an ordinal. In fact, any class (naively, collection) of ordinals is itself well-ordered under the relation $\in$. This fact allows for the usage of transfinite induction and transfinite recursion.
We have three types of ordinals: the empty set $0=\emptyset=\{\; \}$, successor ordinals $S(\alpha)=\alpha\cup\{\alpha\}$ where $\alpha$ is an ordinal, and limit ordinals, which are all the other ones. Finite ordinals are either $0$ or successors, the set $\omega=\{0,1,2,\dots\}$ is a limit ordinal. A limit ordinal $\alpha$ has the property that $\alpha=\sup_{\beta<\alpha}\{\beta\}=\bigcup_{\beta<\alpha}\beta$.
My question
Ordinal arithmetic can be defined recursively as follows:


*

*$\alpha+0=\alpha$, $\alpha+S(\beta)=S(\alpha+\beta)$, $\alpha+\sup_{\gamma<\beta}\{\gamma\}=\sup_{\gamma<\beta}\{\alpha+\gamma\}$; 

*$\alpha\cdot0=\alpha$, $\alpha\cdot S(\beta)=\alpha\cdot\beta+\alpha$, $\alpha\cdot\sup_{\gamma<\beta}\{\gamma\}=\sup_{\gamma<\beta}\{\alpha\cdot\gamma\}$;

*$\alpha^0=1$, $\alpha^{S(\beta)}=\alpha^\beta\cdot\alpha$, $\alpha^{\sup_{\gamma<\beta}\{\gamma\}}=\sup_{\gamma<\beta}\{\alpha^\gamma\}$.


Alternatively, one can define:


*

*$\alpha+\beta$ is the unique ordinal isomorphic to the disjoint union $\{0\}\times\alpha\cup\{1\}\times\beta$ given the lexicographic order.

*$\alpha\cdot\beta$ is the unique ordinal isomorphic to the Cartesian product $\beta\times\alpha$ given the lexicographic order.


As the disjoint union and Cartesian product are simply the categorical coproduct and the categorical product, I wonder if there is some way to actually categorify these alternate definitions. Additionally, I am not aware of any non-recursive version of exponentiation, so I would be curious if a categorical formulation of addition and product of ordinals also allows for a categorical (hence non-recursive) formulation of exponentiation.
 A: Your question is quite vague, therefore I don't know if this is what you want. But I think this is a neat "all-in-one" description of ordinal operations, which in fact come down to operations on well-orderings.
The category of well-orderings has no coproducts (basically because there is no way how to decide the relations between the summands). But there are "directed coproducts" (I don't know if this construction has a canonical name):
Let $S$ be a well-ordering. For every $s \in S$ let $X_s$ be a well-ordering. Consider their underlying sets and take their disjoint union $X = \coprod_{s \in S} X_s = \bigcup_{s \in S} \{s\} \times X_s$. Endow this set with the following well-order: We have $(s,x) < (s',x')$ when $s < s'$ or ($s=s'$ and $x < x'$ in $X_s$). We may write
$$X = \coprod\limits_{s \in S}^{\longrightarrow} X_s$$ 
for this well-ordering in order to indicate the direction induced by $S$. It is not a colimit in the usual sense (but perhaps the category theorists can tell you if it is a weighted colimit?). But it enjoys the obvious universal property within the category of well-orderings and increasing maps:  Morphisms $X \to Y$ correspond 1:1 to families of morphisms $f_s : X_s \to Y$ that satisfy the property
$\forall s,s' \in S (s < s' \Rightarrow \forall x \in X_s \forall x' \in X_{s'} \,(f_s(x) < f_{s'}(x'))).$
Here are two examples:


*

*If $S=\{0<1\}$, then we have just two well-orderings $X_0,X_1$, and we get $X_0 \stackrel{\longrightarrow}{\sqcup} X_1$. This categorifies the ordinal sum.

*If $X_s=Y$ is constant, then we get $S \times Y$ with the lexicographic well-ordering. This categorifies the ordinal product.
Similarily, one can define and construct "directed products", which categorifies ordinal exponentiation.
A: We can define $\alpha^\beta$ to be the order type of the set of functions from $\beta$ to $\alpha$ of finite support (finitely many non-$0$ function values), ordered in reverse lexicographic order--that is, if $f,g:\beta\to\alpha$ have finite support and are distinct functions, and if $\xi$ is the greatest ordinal in $\beta$ for which $f(\xi)\neq g(\xi)$, then we say $f<g$ if $f(\xi)<g(\xi)$.
I'm not well-versed in category theory, so I can't answer your other questions, but I can give you that much.
