In



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



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> Carboni-Vitale, *Regular and exact completions*
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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?



Definition of infinitary regular category
**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.


