Does the internal category forgetful functor have both adjoints?

Question

Let $\mathcal{E}$ be a category with finite limits, and let $Cat(\mathcal{E})$ be the category of internal categories in $\mathcal{E}$, with forgetful functor $Ob : Cat(\mathcal{E}) \to \mathcal{E}$, sending each internal category in $\mathcal{E}$ to its $\mathcal{E}$-object of objects. A colleague of mine has conjectured/stated that this functor $Ob$ has left and right adjoints - is there a simple conceptual argument for this, and/or an appropriate reference in the literature somewhere?

Answer 1

From our experience in the category of sets, we can see that the left adjoint should be a "discrete internal category" functor, and the right adjoint should be an "indiscrete internal category" functor. The discrete case is easier: we can define $L(X)$ to be the internal category $X\rightrightarrows X$ in which the source, target, identity-arrow, and composition morphisms are all the identity of $X$. One easily checks that this has the right universal property; this relies critically on the fact that the source and target morphisms have a common splitting, and wouldn't work at all for internal semicategories.

Now, what should the indiscrete internal category $R(X)$ be? If $X$ is a set, then the set of arrows in the indiscrete internal category is simply $X\times X$, so let's try defining $R(X)$ as $X\times X\rightrightarrows X$ in general. The source and target morphisms are simply the first two projections, while the identity-arrow morphism is the diagonal. The pullback $(X\times X)\times_X (X\times X)$ is just $X\times X\times X$, which the composition morphism the projection away from the middle factor.

Now given any morphism $f:A\to X$ in $\mathcal E$ and a category structure $(s,t):M\rightrightarrows A$ on $A$, we get a unique morphism $(f\circ s,f\circ t):M\to X\times X$ commuting with the source and target morphisms on either side, which one checks also respects the identity-arrow morphisms in both categories.

So both adjoints exist in the same manner as in the category of sets. It might be enlightening to observe that these adjoints are simply the adjoints to the evaluation-at-zero functor from the category of reflexive graphs internal to $\mathcal E$ to $\mathcal E$. Such evaluation functors always have adjoints on both sides, given by Kan extensions, if $\mathcal E$ admits enough limits and colimits, so one interesting thing that happens here is that the indexing diagram for reflexive graphs is so nice that no colimits are needed to construct the left adjoint to that evaluation functor, where normally some coproducts would be required. (This feels, morally, related to the fact that reflexive coequalizers are nicer than general coequalizers.)

The other interesting thing that happens is that the free and cofree reflexive graph on an object magically come equipped with canonical category structures. This is really a rather rare phenomenon, which is why you can't name any other algebraic structures for which the forgetful functor has a right adjoint.

Answer 2

Not 100% sure that this works, but I would argue as follows.

Your friend's claim amounts to say that every object defines a discrete and a codiscrete internal category; discreteness is the easiest: the object of morphisms $C_1$ is just the object of objects, and all structure is trivial.

Codiscreteness can be argued similarly, abstracting the request that in the "maximaly connected groupoid" over a set there is exactly one arrow between any two objects.

It might help to recall that an internal category in $\cal E$ is just a truncated simplicial object $\Delta^o \to \cal E$ satisfying a Segal condition, so $Cat(\cal E)$ is a suitable category of functors; $Ob$ now is the "evaluation at $[0]$" functor.