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Oct 10, 2018 at 2:28 vote accept Harry Gindi
Oct 9, 2018 at 7:55 answer added Harry Gindi timeline score: 1
Jul 23, 2018 at 22:44 history edited Harry Gindi CC BY-SA 4.0
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Jul 23, 2018 at 22:22 comment added Tim Campion Good catch. That simplifies things (and makes your question nontrivial!)-- it just says that all the morphisms in $\mathcal A_+$ are monomorphisms. So all told, we have a Reedy category $\mathcal A$ where the morphisms in $\mathcal A_+$ are monomorphisms, the morphisms in $\mathcal A_-$ are split epimorphisms, and a morphism in $\mathcal A_-$ is determined by its sections.
Jul 23, 2018 at 22:00 comment added Harry Gindi Also, for the category to be (pre)regular, that regularity property only needs to hold for representables, not all presheaves.
Jul 23, 2018 at 21:58 comment added Tim Campion Drat! There's a typo in my first comment. In the second-to-last sentence, it should say "normale" rather than "reguliere".
Jul 23, 2018 at 21:54 comment added Tim Campion Note that a categorie skeletique normale is the same thing as a Cisinski-generalized Reedy category which happens to be an ordinary Reedy category. So all told, a regular skeletal Reedy category is a Reedy category which satisfies the above condition (every nondegenerate map $A \to X$ in $\hat{\mathcal A}$ with $A$ representable is a monomorphism), as well as the condition from Definition 2.4 from the nlab page linked to above (to ensure that it is a categorie skeletique).
Jul 23, 2018 at 21:51 comment added Tim Campion Let's see. "Regular skeletal Reedy category" is Cisinski's categorie skeletique reguliere. A categorie skeletique reguliere is a categorie skeletique normale $\mathcal A$ such that every nondegenerate map $A \to X$ in $\hat{\mathcal A}$ with $A$ representable is a monomorphism. A categorie skeletique reguliere in turn is a categorie skeletique with no nonidentity isomorphisms. And a categorie skeletique is what the nlab calls a Cisinski-generalized Reedy category.
Jul 23, 2018 at 21:19 history asked Harry Gindi CC BY-SA 4.0