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Stream: learning: questions

Topic: structure of regular subobjects

Naso (Mar 13 2022 at 02:53):

If $C$ is 1. a cateogory or 2. a preorder, what kind of structure does its set of regular subobjects have? I know it is a poset and i think it is a meet-semilattice, but is it also something more?

Morgan Rogers (he/him) (Mar 13 2022 at 06:59):

In a preorder the structure is pretty boring, since the only regular subobjects are the isomorphisms

Naso (Mar 13 2022 at 08:28):

oh, i thought they were subsets with the induced ordering

Oscar Cunningham (Mar 13 2022 at 08:31):

Ah, it sounds like you wanted the regular subobjects of $C$, whereas Morgan was talking about the regular subobjects of $A$, for $A\in C$.

Morgan Rogers (he/him) (Mar 13 2022 at 08:52):

Ah right, I see that I misread the question. Do you mean regular subobjects in the 1-category of categories / the 1-category of preorders, respectively?

Naso (Mar 13 2022 at 08:56):

Yep that's what I meant, sorry

Morgan Rogers (he/him) (Mar 13 2022 at 09:46):

The hard part is identifying which the regular subobjects are. Given an inclusion of a subcategory $F: \mathcal{B} \hookrightarrow \mathcal{C}$, consider its cokernel pair pushout along itself. This is obtained essentially by taking two copies of $\mathcal{C}$ and identifying all of the objects and morphisms in the respective copies coming from $\mathcal{B}$ (and freely generating the new composites of what remains).
I think that this pushout is also a pullback, which amounts to saying that $\mathcal{B} \xrightarrow{\Delta} \mathcal{B} \times \mathcal{B} \hookrightarrow \mathcal{C} \times \mathcal{C}$ is precisely the collection of all morphisms which are identified by the functor to the cokernel. Is that the case?

Morgan Rogers (he/him) (Mar 13 2022 at 12:11):

In fact, the above isn't true. Consider the "walking isomorphism": the category with two objects $A$ and $B$ and a unique object between each pair of objects. If we take the subcategory omitting the morphism $A \to B$, then performing the pushout I described, we must also identify the respective copies of the inverse morphisms $B \to A$, so the equalizer of the coproduct inclusions includes it.
More generally, we know that the image under any functor of an isomorphism is an isomorphism, so any regular subcategory of a category $\mathcal{C}$ must be closed under any inverses of its morphisms which exist.

Morgan Rogers (he/him) (Mar 13 2022 at 12:15):

It's also worth noting that the regular subobjects of a preorder in the respective categories you were asking about are different. If I take the two-element preorder $x < y$, the two element discrete order $x$ $y$ (with two incomparable elements) is not a regular subobject in the category of preorders but it is one in the 1-category of categories.

Morgan Rogers (he/him) (Mar 13 2022 at 14:39):

Morgan Rogers (he/him) said:

More generally, we know that the image under any functor of an isomorphism is an isomorphism, so any regular subcategory of a category $\mathcal{C}$ must be closed under any inverses of its morphisms which exist.

I strengthened this property in this SE answer I just wrote about categorical generalizations of normal subgroups :+1: