Michael Shulman 2-pretopos

Definition

Let $n$ be 2, (2,1), (1,2), or 1. That is, $1\le n\le 2$ and $n$ is directed; see n-prefix.

Definition

An $n$-pretopos is an $n$-exact $n$-category which is also extensive. An infinitary $n$-pretopos is an $n$-pretopos which is infinitary-extensive.

As remarked here, regularity plus extensivity implies coherency. Thus an $n$-pretopos is, in particular, a coherent $n$-category. Conversely, we have:

Theorem

An $n$-category is an $n$-pretopos if and only if it is coherent and every (finitary) $n$-polycongruence is a kernel.

Examples

• $Cat$ is a 2-pretopos. Likewise, $Gpd$ is a (2,1)-pretopos and $Pos$ is a (1,2)-pretopos.

• A 1-category is a 1-pretopos precisely when it is a pretopos in the usual sense. Note that, as remarked for exactness, a 1-category is unlikely to be an $n$-pretopos for any $n\gt 1$.

• Since no nontrivial (0,1)-categories are extensive, the definition as phrased above is not reasonable for $n=(0,1)$. However, for some purposes (such as the n-Giraud theorem), it is convenient to define an (infinitary) (0,1)-pretopos to simply be an (infinitary) coherent (0,1)-category (exactness being automatic).

Colimits

An $n$-pretopos has coproducts and quotients of $n$-congruences, which are an important class of colimits. However, it can fail to admit all finite colimits, for essentially the same reason as when $n=1$: namely, some ostensibly “finite” colimits secretly involve infinitary processes. In a 1-category, this manifests in the construction of arbitrary coequalizers and pushouts, where we must first generate an equivalence relation by an infinitary process and then take its quotient.

For 2-categories it is even easier to find counterexamples: the 1-pretopos $FinSet$ does in fact have all finite colimits, but the 2-pretopos $FinCat$ of finite categories (that is, finitely many objects and finitely many morphisms) does not have coinserters, coinverters, or coequifiers. (The category $FPCat$ of finitely presented categories does have finite colimits, but fails to have finite limits.)

However, I conjecture that just as in the case $n=1$, once an $n$-pretopos is also countably-coherent, it does become finitely cocomplete. See colimits in an n-pretopos.