Definition of infinitary regular category

Question

In

Makkai, A theorem on Barr-exact categories, with an infinitary generalization

a definition of infinitary regular category is given: a complete regular category $C$ with the additional requirement

(DC) for every diagram $F : \alpha^{\text{op}} \to C$, $\alpha$ an ordinal, such that each $F(\beta+1) \to F(\beta)$ is regular epi and $F(\lambda) = \lim_{\beta < \lambda} F(\beta)$ for each limit ordinal $\lambda$, then each projection $\lim F \to F(\beta)$ must be regular epi.

In

Carboni-Vitale, Regular and exact completions

another definition (called "completely regular") is given: a complete regular category such that products of regular epis are regular epi.

The first definition implies the second. Are the two definitions equivalent?

Answer 1

The two definitions are not equivalent: The following counter-example might not be the simplest, and I mostly learned it from Christian Espindola. It is a nice counterexample to quite a lot of similar questions...

Let $I$ be the poset of rational number $0 \leqslant q \leqslant 1$, seen as a category with a unique morphism $q \rightarrow q'$ when $q \leqslant q'$, and put the topology on $I$ so that the only non-trivial covering sieve on $q$ is the set of all $q'<q$.

Let $T$ be the topos of sheaves over $I$ for this topology.

A notable properties of $T$ is that its sheafification functors commutes to all limits (and not just finite ones). Indeed, due to the fact that each object has a minimal cover, the explicit expression of the $+$ construction involve no colimits (only limits) in particular the $+$ construction (hence the sheafification functor) commutes to all limits.

It follows from this that in $T$ any product of epimorphisms is an epimorphisms (assuming choice in the base topos). Here is an explicit proof:

Indeed if $(X_i \rightarrow Y_i)_{i \in I}$ is a collection of epimorphisms for each $i$, then given of section $(y_i)_{i \in I}$ of $\prod Y_i$ on $q$, and a $q'<q$, the fact that $X_i \rightarrow Y_i$ is an epi means that you have for each $i$ a section of $x_i$ of $X_i$ on $q'$ whose image in $Y_i$ is the restriction of $y_i$ to $q'$, take the $(x_i)_{i \in I}$ as a section of $\prod X_i$ on $q'$ is image in $\prod Y_i$ is the restriction of $(y_i)$ to $q'$, as we can do that for all $q' <q$ it shows that $\prod X_i \rightarrow \prod Y_i$ is an epimorphisms.

A similar arguement can be used to show that a limit of a countable tower of epimorphisms is again an epimorphisms, i.e. $T$ satisifes $DC$ for countable ordinals, but it fails for uncountable tower as the following example will show:

One construct a tower $V_{\alpha}$ by induction as folows:

$V_0=1$ is the terminal objects For each $\alpha$, $V_\alpha$ is a coproduct of representable. At limit ordinal $V_\alpha$ is definded as the limit of the tower below it.

$V_{\alpha^+}$ is constructed as follow: for each representable $U_q \subset V_{\alpha}$ appearing in the decomposition, on replace $U_q$ by:

$$ \coprod_{q' < q} U_{q'} \twoheadrightarrow U_q $$

and $V_{\alpha^+} \rightarrow V_{\alpha}$ is the proproduct of all these maps in the decomposition of $V_{\alpha}$.

The $V_{\alpha}$ forms a tower as in your definition, but one can check that $V_{\alpha}$ is the initial object for all uncountable ordinals.