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I want to show the following statement:

Given a small regular category $\mathcal C$ there exists a topos $\widetilde{\mathcal C}$ and a regular functor $F\colon\mathcal C\to\widetilde{\mathcal{C}}$ such that any regular functor $G\colon\mathcal{C}\to\mathcal E$ from $\mathcal C$ to a cocomplete topos $\mathcal E$ there exists an essentially unique functor $\widetilde G\colon\widetilde{\mathcal C}\to\mathcal E$ which preserves finite limits and all colimits such that $\widetilde G\circ F\simeq G$.

This seems very similar to the universal property of the Yoneda embedding, and since regular functors preserve finite limits and regular epis, my hunch is to define a Grothendieck topology on $\mathcal C$ so that Yoneda followed by sheafification preserves regular epis. I think this can be done as follows: for regular epi $f\colon B\to C$ in $\mathcal C$ coequalizing $A\rightrightarrows B$, we let $X$ be the coequalizer of $y_A\rightrightarrows y_B$ in $\widehat{\mathcal C}$ and choose the least Grothendieck topology $j_f$ such that sheafification sends the unique morphism $X\to y_C$ to an isomorphism. Taking the union of all such Grothendieck topologies over all regular epis we get the least Grothendieck topology $j$ such that the composition $\mathcal C\to\widehat{\mathcal C}\to\mathsf{Sh}(C, j)$ is regular.

From here I'd like to show that if $F\colon\mathcal C\to\mathcal E$ is regular, then it is flat and continuous for $j$, hence factors through $\mathsf{Sh}(C, j)$, but I am not sure how to do that.

  1. Am I on the right track?
  2. What would be the next step? (if I missing some important theorem I'd appreciate a reference)
14159
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  • The Grothendieck topology you want has a much simpler description: it is the one where a sieve is covering iff it contains a regular epi. Then apply standard results about morphisms of sites. – Zhen Lin Jun 22 '23 at 22:10
  • Hmm, my guess would have been: define a family to be a covering if it's an epimorphic family, i.e. $g \circ f_i = h \circ f_i$ for all $f_i$ in the family implies $g = h$. – Daniel Schepler Jun 23 '23 at 17:04
  • @ZhenLin That's a much nicer description. I think I understand why it would make the composition $\mathcal C\to\widehat{\mathcal C}\to\mathsf{Sh}(\mathcal C)$ regular, but I'm can't quite see why $F$ would be continuous with respect to this topology – 14159 Jun 24 '23 at 14:33
  • @DanielSchepler I don't see how this would work. Would you mind explaining why you'd choose this Grothendieck topology? – 14159 Jun 24 '23 at 14:34

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