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1. • CommentRowNumber1.
   • CommentAuthorYimingXu
   • CommentTimeMar 6th 2022
   • (edited Mar 6th 2022)
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   Author: YimingXu
   Format: MarkdownItexHi all, I am dealing with CCAF by McLarty, which can be found [here](https://www.jstor.org/stable/2275472?casa_token=hi4gsrP5DHMAAAAA:l5929tBx2JWhyJS9B1aoR8iB3AaD1o8NByGOmoj2MHTMoI-LrnB4EzfRSa-si01RyxIz4dHbuldXesD-x9LvW-TmmojxoTMS7XbUrNWEsykCsx0p) . I am confused by the definition of $Cl$ on page 1248, the construction is: > Coequalize $n \circ_P m$ with $m \circ 0$, then do the same with $m$ and $n$ reversed. Call this coequalizer $Cl$. What is **this equalizer $Cl$**? There are two equalizers here, does **this coequalizer** actually mean the pushout of the coequalizer of $n \circ_P m$ with $m \circ 0$ and the coequalizer of $m \circ_P n$ with $n \circ 0$? I guess a bit like the author want to start with the category with exactly two objects, and a pair of arrows between them which are not necessarily inverses of each other, call them $m$ and $n$, then any arrow of such category will be a combination of $m$ and $n$, and taking the coequalizer quotients out the relation $m\circ n = id$ and $n\circ m = id$, but then the category will just be the category containing precisely two objects and a pair of isomorphisms between them, so there will be no point of mentioning "every arrow parallel to $m$" in the proof. Could someone please explain what does the quoted sentence mean here, or how does this category look like? Or some hint of reading this proof? Since I am very unsure about his construction, I had a hard time guess what does he mean after this sentence.

   Hi all, I am dealing with CCAF by McLarty, which can be found here .

   I am confused by the definition of $Cl$ on page 1248, the construction is:

   Coequalize $n \circ_P m$ with $m \circ 0$, then do the same with $m$ and $n$ reversed. Call this coequalizer $Cl$.

   What is this equalizer $Cl$? There are two equalizers here, does this coequalizer actually mean the pushout of the coequalizer of $n \circ_P m$ with $m \circ 0$ and the coequalizer of $m \circ_P n$ with $n \circ 0$?

   I guess a bit like the author want to start with the category with exactly two objects, and a pair of arrows between them which are not necessarily inverses of each other, call them $m$ and $n$, then any arrow of such category will be a combination of $m$ and $n$, and taking the coequalizer quotients out the relation $m\circ n = id$ and $n\circ m = id$, but then the category will just be the category containing precisely two objects and a pair of isomorphisms between them, so there will be no point of mentioning “every arrow parallel to $m$” in the proof.

   Could someone please explain what does the quoted sentence mean here, or how does this category look like? Or some hint of reading this proof? Since I am very unsure about his construction, I had a hard time guess what does he mean after this sentence.

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 7th 2022
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   Author: DavidRoberts
   Format: MarkdownItexNote that it's the full subcategory classifier, and a full subcategory is determined by which objects it contains. So figuring out what the full subcategory classifier should be in the ordinary description of categories would be a good start. I don't really understand why the category $\mathbf{T}$ from earlier won't do the job. But it might be that it's difficult to prove the universal property of the classifier without cooking up a careful construction that in the end doesn't do much, but bakes in the universal property.

   Note that it’s the full subcategory classifier, and a full subcategory is determined by which objects it contains. So figuring out what the full subcategory classifier should be in the ordinary description of categories would be a good start. I don’t really understand why the category $\mathbf{T}$ from earlier won’t do the job. But it might be that it’s difficult to prove the universal property of the classifier without cooking up a careful construction that in the end doesn’t do much, but bakes in the universal property.

3. • CommentRowNumber3.
   • CommentAuthorYimingXu
   • CommentTimeMar 8th 2022
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   Author: YimingXu
   Format: MarkdownItexThanks for replying! >I don’t really understand why the category T from earlier won’t do the job. I think this is because we only know that $T$ contains such pair of objects but $T$ may contain some thing extra and far from being merely "two objects with a pair of isomorphisms". So the morphism that makes the pullback will not be unique. >Note that it’s the full subcategory classifier, and a full subcategory is determined by which objects it contains. So figuring out what the full subcategory classifier should be in the ordinary description of categories would be a good start. I thought about $1 + 1$ the first moment I see this hint from you. Then I realise that if the category looks like $A \to B \to C$ and we want the full subcategory to be $A \to B$ then $1 + 1$ cannot do the job. But the category with precisely two objects and a pair isomorphism can do this job, by sending the arrow $A\to B$ on the identity of one of the object, and $B \to C$ to the non-identity arrow. I will try to check it in more detail tomorrow and update if I figure out more details.

   Thanks for replying!

   I don’t really understand why the category T from earlier won’t do the job.

   I think this is because we only know that $T$ contains such pair of objects but $T$ may contain some thing extra and far from being merely “two objects with a pair of isomorphisms”. So the morphism that makes the pullback will not be unique.

   Note that it’s the full subcategory classifier, and a full subcategory is determined by which objects it contains. So figuring out what the full subcategory classifier should be in the ordinary description of categories would be a good start.

   I thought about $1 + 1$ the first moment I see this hint from you. Then I realise that if the category looks like $A \to B \to C$ and we want the full subcategory to be $A \to B$ then $1 + 1$ cannot do the job. But the category with precisely two objects and a pair isomorphism can do this job, by sending the arrow $A\to B$ on the identity of one of the object, and $B \to C$ to the non-identity arrow.

   I will try to check it in more detail tomorrow and update if I figure out more details.

4. • CommentRowNumber4.
   • CommentAuthorHurkyl
   • CommentTimeMar 8th 2022
   • (edited Mar 8th 2022)
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   Author: Hurkyl
   Format: MarkdownItexIn the ordinary setting this can be packaged up neatly by observing that the "indiscrete category" functor is right adjoint to the "set of objects" functor. This correspondence identifies subsets of objects, which correspond to maps from the object set to the two-point set, to functors to the two-object indiscrete category.

   In the ordinary setting this can be packaged up neatly by observing that the “indiscrete category” functor is right adjoint to the “set of objects” functor. This correspondence identifies subsets of objects, which correspond to maps from the object set to the two-point set, to functors to the two-object indiscrete category.

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 9th 2022
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   Author: DavidRoberts
   Format: MarkdownItex@Hurkyl I was thinking that the indiscrete 2-object category was the classifier, just didn't sit down and check. But I was thinking that $T$ was this category, though, again, without checking it!

   @Hurkyl

   I was thinking that the indiscrete 2-object category was the classifier, just didn’t sit down and check. But I was thinking that $T$ was this category, though, again, without checking it!

6. • CommentRowNumber6.
   • CommentAuthorYimingXu
   • CommentTimeMar 10th 2022
   • (edited Mar 10th 2022)
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   Author: YimingXu
   Format: MarkdownItexThe $T$ as in Theorem 15 is "**a** (not **the**) category with a pair of isomorphic objects", can certainly contain many other things! He used the category $A^2$, where $A$ is the category exists by CC6, and then hides the construction and call it $T$. I see the adjunction, thanks for pointing out this! (It is a bit unfortunate here since they are a pair of meta-functor in this formulation, hence a pair of "meta-adjunction". Let me try if I can get a nice way to express this.)

   The $T$ as in Theorem 15 is “a (not the) category with a pair of isomorphic objects”, can certainly contain many other things! He used the category $A^2$, where $A$ is the category exists by CC6, and then hides the construction and call it $T$.

   I see the adjunction, thanks for pointing out this! (It is a bit unfortunate here since they are a pair of meta-functor in this formulation, hence a pair of “meta-adjunction”. Let me try if I can get a nice way to express this.)

7. • CommentRowNumber7.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 10th 2022
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   Author: DavidRoberts
   Format: MarkdownItexAh, I see. Thanks for clarifying.

   Ah, I see. Thanks for clarifying.