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: $(0,0)$-Functor Between $(0,0)$-Categories

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

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

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

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

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

Under Construction

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

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

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

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

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

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Diagrams

Graph

The following diagram describes a graph:

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 $X$, i.e.

• $\pi_X\circ graph(F)$
• $Id_X$

This means that

Another pair starts at $X$ and ends at $Y$. These are given by

• $\pi_Y\circ graph(F)$
• $F$

This means that

Putting those together, we see that

Cograph

• $cograph(F)\circ i_X = F$
• $cograph(F)\circ i_Y = Id_Y$

Putting these together, we see that

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

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

then.

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Attempt #2

I have a question…

When I look at

Cographs of functors between 0-categories

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

\array{

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

}

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

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

in the bottom left corner, we stick in

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 $cograph(f)$ would be an $(n+2)$-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, $cograph(f)$ would be an $(n+3)$-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 $n$-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 $n$-functor $F: C \to D$ as an $n$-functor $I \to nCat$ 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 $F:C\to D$, 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.

Concerning the other bit of your question:

you may of course consider pulling back along more general maps $D \to nCat$. As $D$ 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 $n$-functors are then more traditionally called $(n+1)$-presheaves on $D^{op}$ and that the corresponding cograph would be called their $n$-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 $n$-functor and its cograph by the degree of the shape that goes in the bottom left corner of the pullback diagram. Since an interval $I$ 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 $(n+2)$-category or something.

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

• Toby: You're not going far enough down! If $C$ and $D$ are categories, then the functor category $[C,D]$ is simply a category, not a $2$-category (except in the sense that any category can be interpreted as a $2$-category whose hom-categories are all discrete). Similarly, if $C$ and $D$ are sets, then the function set $[C,D]$ is simply a set; and if $C$ and $D$ are truth values, then the conditional $[C,D]$ is a truth value (of the proposition $C \Rightarrow D$).

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

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

• Not that I can see. A functor category certainly ‘is’ not that; its has functors (from $C$ to $D$) as objects and natural transformations as morphisms. And I don't see how its category of elements would be that either (although that needs a functor from $[C,D]$ to $Set$ to even make sense). Maybe there is some $2$-category like that, but I don't know what it would be; the only $2$-category with those objects and morphisms that I can think of offhand is simply $C \sqcup D$, which is also really just a category (so its only $2$-morphisms are identities).

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

• I'm not entirely sure what Urs is after; he jumps quickly to the $(\infty,1)$-case, where the cograph is of course an $(\infty,1)$-category. But his cograph of a functor is, if not literally a category, at least a $2$-category equivalent to a category. I would just call it a category: no nontrivial $2$-morphisms. Am I wrong, Urs?

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

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

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 $(-2)$. In other words, I hope to be able to say something like

• “A $(-1)$-category is a cograph of a $(-2)$-functor.”
• “A $0$-category is a cograph of a $(-1)$-functor.”
• “A $1$-category is a cograph of a $0$-functor.”
• $\dots$
• “An $n$-category is a cograph of an $(n-1)$-functor.”

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

Attempt #1

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From cograph of a functor…

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

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

The category $C?D$ is the category with $Obj(C?D) = Obj(C) \amalg Obj(D)$ 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.