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• CommentRowNumber1.
• CommentAuthorYimingXu
• CommentTimeJul 24th 2022
• (edited Jul 25th 2022)

I am dealing with McLarty’s paper on axiomatising a category of categories here

I am confused by his description on Theorem 27. His description starts on page 1251, and my confusion is about the paragraph precisely before the statement of Theorem 27 on page 1252. I feel very messed-up by the following sentences:

Now consider any internal small category in which every “endomorphism” is an “identity arrow”, and suppose it has a skeleton. The skeleton is of the sort described in the last paragraph, so let $B$ be the corresponding actual category. Then an actual category corresponding to the pattern appears (in a number of ways) as a full subcategory of $B \times (Cl^{Ob})$.

I have the following concrete questions:

1. What does a “skeleton” refer to and why it has a corresponding category? From the construction in the paragraph before, I guess the skeleton is something whose corresponding actual category is the equalizer, but what is the thing?

2. Regarding:

Then an actual category corresponding to the pattern appears (in a number of ways) as a full subcategory of $B \times (Cl^{Ob})$.

What is “the pattern appears”?

1. In “then $Ar$, $Ob$ corresponds to an actual category”, is the witness of the existence of $B$ or some full subcategory of $B\times Cl^{Ob}$? If the later, is there any description of such full subcategory?

Sorry if it turns out to be my stupidness. But I am really confused now. I hope there is a rearranged clearer version of such a proof. Thank you for any hint ot comment which may help me understanding this.

• CommentRowNumber2.
• CommentAuthorGuest
• CommentTimeJul 24th 2022
The nForum is for issues concerning nLab. I suggest you repost your question on MathOverflow.

kirk sturtz
• CommentRowNumber3.
• CommentAuthorGuest
• CommentTimeJul 24th 2022
The nForum is for issues concerning nLab. I suggest you repost your question on MathOverflow.

kirk sturtz
• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2022

@Kirk

We discuss stuff like this all the time. As long as it’s relevant to the nLab, and it ends up being included on an nLab page at the end, I see no harm.

• CommentRowNumber5.
• CommentAuthorGuest
• CommentTimeJul 24th 2022

the nForum started out as a discussion forum about higher category theory. The wiki came later.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeJul 25th 2022
• (edited Jul 25th 2022)

Re #5:

No, the nForum was started in order to provide a place for discussing nLab edits. Before we had the nForum we would leave those (now deprecated) query boxes in nLab entries and log edits on dedicated nLab pages (see here).

$\,$

Re #1:

By “skeleton” the text must mean the usual notion of skeleton: equivalent full subcategories all whose isomorphisms are endomorphisms.

A moment of reflection shows that if a category is both skeletal as well as satisfying the condition that every endomorphism is an identity, then it satisfies McLarty’s condition on p. 1251 that anti-parallel morphisms are equal (namely both equal to an identity, in this case).

If we believe the conclusion on p. 1252 that this “does give $\mathbf{Ar}$, $\mathbf{Ob}$ as internal small category” [sic], it follows that skeletal such categories exist as actual categories in McLarty’s scheme.

Now what does he mean by “pattern”? From the line (on p. 1251)

Any finite pattern of objects and arrows that intuitively forms a category

I gather he means that a “pattern” is a set of consistent composition rules defining a category. Saying that the “pattern appears (in a number of ways) as a full subcategory” seems to mean “the prescribed category is found (in a number of ways) as a full subcategory”.

So, finally, with $\mathbf{B}$ the “actual category” being the skeleton of the prescribed category, that full prescribed category differs from $\mathbf{B}$ by having more isomorphisms to further objects.

Concretely, any morphism in a category with a chosen skeleton is the composite of

1. the chosen morphism identifying the domain with its representative in the skeleton,

2. a morphism in the skeleton,

3. the chosen morphism identifying the codomain with its representative in the skeleton.

Since $\mathbf{Cl}$ is the walking isomorphism between a pair of distinct objects (by top of p. 1249), this shows that/how any prescribed category with objects $\mathbf{Ob}$ may be regarded as a full subcategory of its skeleton times $\mathbf{Cl}^{\mathbf{Ob}}$.

That’s my understanding of the text, anyway. I haven’t tried to verify the veracity of the claims based on this reading, but it seems rather plausible.

• CommentRowNumber7.
• CommentAuthorYimingXu
• CommentTimeJul 25th 2022
• (edited Jul 25th 2022)

Thanks @Urs for the explanation, and also thanks @DavidRoberts for supporting such discussion.

Now it makes some sense to me, thanks a lot! I read it as follows, and some confusion persists.

The $S$ and $T$ represents the objects and arrows of the skeleton of the category prescribed by $Ar$ and $Ob$. As I see such $S$ and $T$ correspond to a skeleton satisfiy the condition that “anti-parallel arrows are equal”, the construction on page 1251 applies, and give an actual category, which is the equalizer in the paper, with $S$ and $T$ form a internal small category of such an actual category.

Now $B$ will be the skeleton of the prescribed category, and the remaining issue is:

Given a skeleton $B$ of a category $C$ and the discrete category $Ob$ of objects of $C$, and given the information that every endomorphism of $C$ is an identity, how to recover the category $C$ with internal small category $Ar$, $Ob$ as a subcategory of $B\times Cl^{Ob}$?

I understand given the skeleton $B$ of $C$, every map in $C$ is of form $i\circ b\circ j$, where $b$ is an arrow in $B$, and $i$ and $j$ are isomorphisms. If $i$ is not an identity, then its codomain cannot live in $B$. And similarly, if $j$ is not an identity, then its domain cannot live in $B$, since $B$ is the skeleton and isomorphic objects are cut out of it. Finding out $C$ in $B\times Cl^{Ob}$ as a full subcategory amounts to finding out a suitable collection of objects of $B\times Cl^{Ob}$, such a collection will uniquely determine a map $B\times Cl^{Ob}\to Cl$, and the pullback along $1\mapsto Cl$ will be $C$. But I still have the question how concretely can we be pick out such a collection of objects. The intuition comes to my is to use the fact that each object $b:1\to B$ determines an object of $Cl^{Ob}$, which is a map $Ob\to Cl$, by: $c\mapsto$ if $c = b$ then the $id_1$ in $Cl$, if $c$ has a non-identity arrow from $b$ then $i_1:1\to 0$, and if $c$ has an non-identity arrow to $b$ then $0\to 1$, if there is no map between $c$ and $b$ then $id_0$, and the collection we pick will have some relation to the collection of pairs $(b,f_b)$, where $f_b$ is the map determined by $b$. But it seems to me that as we have not use the information about whole $Ar$, we cannot recover the information of parallel arrows in this way.

Thank you for pointing it out if I have already misunderstand something. And I appreciate for any further hint on how to construct one of “a number of ways” of the embedding of the category in $B\times Cl^{Ob}$.

EDIT: I realise that I am wrong with defining the map $Ob\to Cl$! I was thinking about sending objects of $Ob$ to arrows in $Cl$, which is totalling wrong. But Given $Ob$ is discrete, since $Ar$, $Ob$ forms an internal small category, a map $Ob\to Cl$ is completely determined by where the objects goes to, so a map $Ob\to Cl$ is same as a map $Ob\to 1+1$. Therefore, why do we use $Cl$ instead of $1+1$?

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeJul 26th 2022

$Ob$ is discrete, since $Ar$, $Ob$ forms an internal small category, a map $Ob\to Cl$ is completely determined by where the objects goes to, so a map $Ob\to Cl$ is same as a map $Ob\to 1+1$. Therefore, why do we use $Cl$ instead of $1+1$?

Notice that $\mathbf{Cl}^\mathbf{Ob}$ is not just the set of maps, but is the full functor category, whose objects are such maps, and whose morphisms are the natural transformations between these.

• CommentRowNumber9.
• CommentAuthorYimingXu
• CommentTimeJul 26th 2022
• (edited Jul 29th 2022)

For #8: Thank you for the hint! A natural transformation from functors $Ob\to Cl$ is an assignment of objects in $Ob$ to arrows in $Cl$, which is relevant to what I thought yesterday…

Then an arrow $b:2\to B$ naturally corresponds to an assignment sending an object to an arrow in $Cl$. I think I see I can identify pairs of objects which has isomorphism to domain of $B$ or out of codomain of $B$ in this way, by collecting the three pieces of information as in #6. But if we find out the arrows just as:

1.the chosen morphism identifying the domain with its representative in the skeleton, 2.a morphism in the skeleton, 3.the chosen morphism identifying the codomain with its representative in the skeleton.

Then I think the resultant category will have $(Ob,T)$ instead of $(Ob,Ar)$ as its internal small category, since all information about parallel maps are missing. Therefore, does the claim:

$Ar,Ob$ corresponds to an actual category”

not in the sense that “there exists a category $C$ such that $(Ar,Ob)$ is the internal small category of $C$”, but modulo some equivalence?

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeJul 28th 2022
• (edited Jul 28th 2022)

In case the question still concerns this point:

I appreciate for any further hint on how to construct one of “a number of ways” of the embedding of the category in $B \times Cl^{Ob}$.

Here is how I imagined doing it:

So we have a category $C$ with objects $Ob$ and skeleton

$B \overset{i}{\hookrightarrow} C \overset{r}{\longrightarrow} B$

and we are asking for a full inclusion

$C \overset{j}{\hookrightarrow} B \times Cl^{Ob} \,,$

where $Cl = \{ 0 \overset{\sim}{\leftrightarrow} 1\}$.

To achieve this, declare $j$ on objects $c \in C$ by

$j(c) \coloneqq \left( r(c), \; c' \mapsto \left\{ \begin{array}{ll} 1 & c' = c \\ 0 & else \end{array} \right. \right)$

On morphisms declare $j$ to be $r$ in the first component and the unique possible choice in the second.

This should do the trick.

The same formula with “$0$” and “$1$” exchanged gives a second way of doing it.

• CommentRowNumber11.
• CommentAuthorYimingXu
• CommentTimeJul 29th 2022

Thank you for writing out this solution! This definition is pretty clear.

I was careless when I read the final statement of the statement: The condition given there is “if there is a monic $t:T\to Ar$ selectinng one from each family of parallel arrows between objects in $S$, not objects in the whole $Ob$”, therefore arrows which has its source or target which is not in $S$ is not possible to be selected. I think I have missed something here. From where can we recover the information “the chosen morphism identifying the domain with its representative in the skeleton”?

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeAug 4th 2022
• (edited Aug 4th 2022)

From where can we recover the information “the chosen morphism identifying the domain with its representative in the skeleton”?

$B \overset{i}{\hookrightarrow} C$

is in particular an equivalence of categories, hence there is a functor

$C \overset{r}{\to} B$

and, in particular, a natural isomorphism

$\eta \;\colon\; i \circ r \Rightarrow id_c \,.$

For any object $x \in C$, the component

$\eta_x \;\colon\; i\big(r(x)\big) \overset{\sim}{\longrightarrow} x$

connects $x$ to its representative

$r(x) \in B \overset{i}{\hookrightarrow} C$

in the skeleton.

• CommentRowNumber13.
• CommentAuthorYimingXu
• CommentTimeAug 4th 2022
• (edited Aug 4th 2022)

Thank you!

This seems that we need a $C$ to work with, but the whole task is to prove the category $C$ exists! I agree that as for #10, once we have a category and a skeleton, then we can use the equivalence of category to define an embedding(I was thinking about turning your description of the inclusion $C\to B\times Cl^{Ob}$ into a map $Cl^{Ob}\to 2$ to avoid $C$ these days, and also avoid the usage of $r$, but have not made any progress yet). But throughout the proof, the fact that “$B$ is the skeleton of some category $C$” is just a thing to keep in mind, we just construct $B$ on its own, not as a skeleton of any category yet, therefore, I think we cannot explicitly use $C$, $i$ and $r$ here, because this is what we want to show to exist.

• CommentRowNumber14.
• CommentAuthorYimingXu
• CommentTimeAug 11th 2022
• (edited Aug 11th 2022)

I have emailed the author (professor McLarty) of this paper, according to the reply, this is a mistake:this version of the proof should not be published. I will try to write up a complete proof here after receiving the corrected proof from the author. Thank you for keep helping!

• CommentRowNumber15.
• CommentAuthorDavidRoberts
• CommentTimeAug 13th 2022

Excellent, thanks!