Eric Forgy Cograph

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Warning: This page is just a note for me in an attempt to understand cograph of a functor. Comments welcome!

Attempt #3

After agreeing to use cograph of a functor as the definition of cograph, let’s try to walk through what a cograph is in detail for some cases.

Case 1: (0,0)(0,0)-Functor Between (0,0)(0,0)-Categories

In this case, we have (0,0)(0,0)-categories, or sets, XX and YY and we consider a (0,0)(0,0)-functor F:XYF:X\to Y.

If we reinterpret this as a concrete (0,1)(0,1)-category, or poset, with 2 objects and 1 morphism F:XYF:X\to Y, then cograph(F)cograph(F) is the category of elements El F¯(2)El_{\bar{F}}(\mathbf{2}), where

2F¯Set\mathbf{2}\stackrel{\bar{F}}{\to} Set

with F¯(0)=X\bar{F}(0) = X, F¯(1)=Y\bar{F}(1) = Y, and F¯()=F\bar{F}(\to) = F.

This means that cograph(F)cograph(F) is a (0,1)(0,1)-category, or poset.

Case 2: (0,1)(0,1)-Functor Between (0,1)(0,1)-Categories

Under Construction

Eric: Help!!

In this case, we have (0,1)(0,1)-categories, or posets, XX and YY and we consider a (0,1)(0,1)-functor F:XYF:X\to Y.

Note that the diagram \bullet \to \bullet is just a poset, no matter what concretely XX and YY might be. —Toby

If we reinterpret this as a concrete (0,1)(0,1)-category, or poset, with 2 objects and 1 morphism F:XYF:X\to Y, then cograph(F)cograph(F) is the category of elements El F¯(2)El_{\bar{F}}(\mathbf{2}), where

2F¯Pos\mathbf{2}\stackrel{\bar{F}}{\to} Pos

with F¯(0)=X\bar{F}(0) = X, F¯(1)=Y\bar{F}(1) = Y, and F¯()=F\bar{F}(\to) = F.

This means that cograph(F)cograph(F) is a (0,1)(0,1)-category, or poset.


Diagrams

Graph

The following diagram describes a graph:

X graph(F) X×Y F Y\array{ X & \stackrel{graph(F)}{\rightleftarrows} & X\times Y \\ {} & \mathllap{\scriptsize{F}}\searrow & \darr \\ {} & {} & Y }

If you’re anything like me, you like to dig in and see what a diagram means in good old fashion equations. To do this, we need to stair at the diagram and find a pair of paths that start and end at the same object. The fact that the diagram commutes means that the two paths are “equal”. One such pair starts and ends at XX, i.e.

This means that

π Xgraph(F)=Id X.\pi_X\circ graph(F) = Id_X.

Another pair starts at XX and ends at YY. These are given by

This means that

π Ygraph(F)=F.\pi_Y\circ graph(F) = F.

Putting those together, we see that

graph(F)=(Id X,F).graph(F) = (Id_X,F).

Cograph

Y cograph(F) XY F X\array{ Y & \stackrel{cograph(F)}{\leftrightarrows} & X\sqcup Y \\ {} & \mathllap{\scriptsize{F}}\nwarrow & \uparr \\ {} & {} & X }

Putting these together, we see that

cograph(F)={FId Y}.cograph(F) = \left\{{F \atop Id_Y}\right\}.

Toby: (This works for Lawvere's cograph of a function between sets; the cograph of a functor is not in general isomorphic to YY.)

However, these diagrams are dangerous in that the top row does not commute if you compose in the other direction.

Eric: Thanks! I think I should probably focus on trying to understand the diagram

cograph(f) * I f¯ Set \array{ cograph(f) &\to& {*} \\ \downarrow && \downarrow \\ I &\stackrel{\bar f}{\to}& Set }

then.


Attempt #2

I have a question…

When I look at

Cographs of functors between 0-categories

In the case that C,DC, D are 0-categories, i.e. sets, a functor f:CDf : C \to D is just a function between sets. The cograph 2-pullback

\array{
cograph(f) &\to& {*}
\\
\downarrow && \downarrow
\\
I &\stackrel{\bar f}{\to}& Set

}

>iscomputedbytheordinaryaclass=existingWikiWordhref=/nlab/show/pullback>pullback/a> is computed by the ordinary a class=existingWikiWord href=/nlab/show/pullbackpullback/a

\array{ cograph(f) &\to& Set_{} \ \downarrow && \downarrow \ I &\stackrel{\bar f}{\to}& Set }

>andidentifies >and identifies

\array{ cograph(f) &\to& {} \ \downarrow && \downarrow \ I &\stackrel{\bar f}{\to}& nCat }

Then $cograph(f)$ should probably be an $(n+1)Cat$. But what if instead of sticking $$0\to 1

in the bottom left corner, we stick in

Then Then

This is kind of the primordial non-degenerate 2-category. Any smaller 2-category can be obtained form this one by changing 1-arrows to identity arrows (or so I think).

Then, I THINK cograph(f)cograph(f) would be an (n+2)(n+2)-category.

We could then introduce a primordial non-degenerate 3-category, which would be a (directed) 3-cube. Any smaller 3-category can be obtained from this one by setting some morphisms to identity morphisms. Putting that into the bottom left corner, cograph(f)cograph(f) would be an (n+3)(n+3)-category. Etc.

Does that make any sense?

Urs: Right, so the answer to your question about higher cographs is basically: Yes.

The cograph of an nn-functor is defined almost literally like that of a functor, only that everything is taking place in the corresponding higher context. So you regard your nn-functor F:CDF: C \to D as an nn-functor InCatI \to nCat and then look at the weak pullback of the point along this morphism.

But are we ready for this already? I thought we were still waiting to hear back from you about the degree to wich you are happy now with talking about ordinary 1-functors in terms of cographs. So that is clear?

Eric: Well, I think I understand how, given a functor F:CDF:C\to D, you get a cograph, but what I would like to see is a way to define functors in terms of cographs, i.e. I understand cographs in terms of functors, but now I want to understand functors in terms of cographs.

I also do not yet quite understand the pullback diagram.

cograph(f) * I f¯ nCat \array{ cograph(f) &\to& {*} \\ \downarrow && \downarrow \\ I &\stackrel{\bar f}{\to}& nCat }

Concerning the other bit of your question:

you may of course consider pulling back along more general maps DnCatD \to nCat. As DD becomes larger, it encodes not just a single functor but a whole system of such. And if you want you can call the corresponding weak pullback of the point the cograph of that system of functors. Just beware that such systems of nn-functors are then more traditionally called (n+1)(n+1)-presheaves on D opD^{op} and that the corresponding cograph would be called their nn-Grothendieck construction or something like that. But that’s just words.

Eric: I must be confused. I was trying to draw a link between the difference in degree between an nn-functor and its cograph by the degree of the shape that goes in the bottom left corner of the pullback diagram. Since an interval II has degree 1, I thought that might be related to why the cograph is one degree higher than the functor. If you use a “2-interval” or something, I thought the pullback might be an (n+2)(n+2)-category or something.

PS: Another thing I’m thinking about is how you can go down in degree. For example, given categories CC and DD, you get a functor category [C,D][C,D], which I’d think of as some special kind of 2-category. Can we go down in degrees? Given 0-categories CC and DD, couldn’t we form a “morphism category” [C,D][C,D], which is a 1-category? How about given (1)(-1)-categories CC and DD, can we form a “truth category” [C,D][C,D], which is a 0-category? Etc etc.

Eric: Thanks Toby. So I’m even more confused than I thought :) Is there any sense in which a functor category is a 2-category with

  • 0-morphisms: Obj(C)Obj(D)Obj(C)\sqcup Obj(D)
  • 1-morphisms: Mor(C)Mor(D)Mor(C)\sqcup Mor(D)
  • 2-morphisms: blah!!

Ok! Maybe what I’m thinking is a “exploded” version of [C,D][C,D]?

Eric: Ok, but please tell me at least that given a functor F:CDF:C\to D, that cograph(F)cograph(F) is a 2-category :o

Eric: If you look at Section 2.3.1 in Lurie’s paper, he defines a “category” specifying object and morphisms, but (I think!) there should also be 2-morphisms

Hom(Hom 𝒞(X,Y),Hom 𝒟(F(X),F(Y)))Hom(Hom_{\mathcal{C}}(X,Y),Hom_{\mathcal{D}}(F(X),F(Y)))

or something.

Once I understand that (if there is any truth to it, then I’d like to go up the ladder again start at (2)(-2). In other words, I hope to be able to say something like

That is probably totally wrong, but in the process of discovering why it is wrong, I will have learned something (I hope!).

Attempt #1


From cograph of a functor

The category cograph(L)cograph(L) is the category with Obj(cograph(L))=Obj(C)⨿Obj(D)Obj(cograph(L)) = Obj(C)\amalg Obj(D) and with

Hom cograph(L)(x,y)={Hom C(x,y) ifx,yC Hom D(x,y) ifx,yD Hom D(L(x),y) ifxC,yD ifxD,yC Hom_{cograph(L)}(x,y) = \left\{ \begin{aligned} Hom_C(x,y) & if x,y \in C \\ Hom_D(x,y) & if x,y \in D \\ Hom_D(L(x),y) & if x \in C ,y \in D \\ \emptyset & if x \in D ,y \in C \end{aligned} \right.

Eric: If that is a cograph, is there a name for the category similarly defined as

The category C?DC?D is the category with Obj(C?D)=Obj(C)⨿Obj(D)Obj(C?D) = Obj(C) \amalg Obj(D) and with

Hom_{C?D}(x,y) = \left\{
\begin{aligned}
  Hom_C(x,y) & if x,y \in C
  \\
  Hom_D(x,y) & if x,y \in D
  \\
  \emptyset & otherwise
\end{aligned}

\right.

>? The question mark in "$C?D$" really is a question mark because I don't know what should go there. It seems to me that this category should always exist (at least for small categories), which makes me think it is not a coproduct. _Toby_: Yes, this is called the 'disjoint union', the missing symbol is just ‘$\amalg$’ again, and it *is* a coproduct in $Cat$. (Technical note: the small midfix symbol is `\amalg` ‘$\amalg$’, while `\coprod` ‘$\coprod$’ is the large prefix symbol. That is, ‘$C \amalg D$’ but ‘$\coprod_i C_i$’.) <a class='existingWikiWord' href='/ericforgy/published/Eric'>Eric</a>: Thanks! Ok. I'm a little confused about the notation though. For the disjoint union of categories, would you write it $C\coprod D$ or $C\amalg D$? Edit: Oh. I think `\amalg` is not rendering. _Toby_: The disjoint union of $C$ and $D$ is $C \amalg D$, while the disjoint union of the family $(C_i)_i$ is $\coprod_i C_i$. Compare: the direct sum of $C$ and $D$ is $C \oplus D$, while the direct sum of the family $(C_i)_i$ is $\bigoplus_i C_i$. Also, the disjoint union of $C$ and $D$ is $C + D$, while the disjoint union of the family $(C_i)_i$ is $\sum_i C_i$ (since there are many notations for <a class='existingWikiWord' href='/nlab/show/disjoint+union'>disjoint union</a>, and more generally for <a class='existingWikiWord' href='/nlab/show/coproduct'>coproduct</a>). But is ‘$\amalg$’ not showing up in your fonts? Is that you what you mean by 'not rendering'? <a class='existingWikiWord' href='/ericforgy/published/Eric'>Eric</a>: Right. `\amalg` is not showing up. All I see is ‘’ (using Firefox). _Toby_: That is a problem. Urs has been using `\sqcup` ‘$\sqcup$’ sometimes, which is not technically the correct symbol but almost looks right. (It should be just like ‘$\coprod$’, except smaller, very much like ‘$\otimes$’ and ‘$\bigotimes$’ or ‘$\cup$’ and ‘$\bigcup$’.) Do you know which fonts your browser is using?
Revised on November 17, 2009 at 21:40:31 by Eric Forgy