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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeMay 31st 2010
• (edited May 31st 2010)

I was reading MacLane (like a good student) and when I saw the diagram for a universal arrow, it made me think of natural transformations.

It looked like we could describe a comma category $(d\downarrow F)$ as a natural transformation $\tau:\Delta d\to F$, where $\Delta d:C\to D$ is a constant functor.

If so, then the universal arrow, i.e. initial object of the comma category, would be like the “initial component” of a natural transformation.

Does that make any sense?

• CommentRowNumber2.
• CommentAuthorEric
• CommentTimeMay 31st 2010

Then, if I push my luck, it looks like if $C$ has an initial object $c$, then the component $\tau_c:d\to F(c)$ is a universal arrow. Is that correct?

• CommentRowNumber3.
• CommentAuthorFinnLawler
• CommentTimeMay 31st 2010

I’m in a bit of a rush, so I don’t have time to write more, but note that, as described at comma category, the category $d/F$ is the universal category with a transformation from $d/F \to * \overset{d}{\to} D$ (the constant functor) to $d/F \to C \overset{F}{\to} D$, where the arrows out of $d/F$ are the projections. So $d/F$ certainly depends on this transformation, but it’s also universal among squares-with-a-transformation-in over the cospan $C \overset{F}{\to} D \overset{d}{\leftarrow} *$.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeMay 31st 2010

To answer one question: a category isn’t a transformation, so you can’t describe a comma category like that. But as Finn already noted, there is a connection. If $p: d/F \to C$ is the functor that sends an object $(d \to F(c), c)$ to $c$, then a natural transformation $\tau: \Delta d \to F$ is essentially the same thing as a section $s: C \to d/F$ of $p$. For, such a section would have to take each object $c$ to some object of the form $(d \to F(c), c)$, where the first component would be your component $\tau_c: d \to F(c)$, and would take each morphism $f: c \to c'$ to the pair

$\array{ d & \stackrel{\tau_c}{\to} & F(c) & & & c \\ \tau_{c'} & \searrow & \downarrow F(f) & & & \downarrow f \\ & & F(c') & & & c' }$

where the triangle must commute by definition of $d/F$. The commutativity means the $\tau_c$ are components of a natural transformation $\tau: \Delta d \to F$.

As for #2: If $c$ is initial in $C$, there’s a universal cone $\Delta F(c) \to F$ whose component at an object $c'$ of $C$ is the arrow $F(!_{c'}): F(c) \to F(c')$, where $!_{c'}: c \to c'$ is the unique arrow to $c'$. Universality means that given any cone $\tau: \Delta d \to F(c)$, there’s a unique $g: d \to F(c)$ such that $\tau$ is the composite

$\Delta d \stackrel{\Delta g}{\to} \Delta F(c) \stackrel{univ}{\to} F$

and this $g$ is exactly $\tau_c$, as you surmise.

• CommentRowNumber5.
• CommentAuthorEric
• CommentTimeMay 31st 2010

Well, I made a little progress with Yoneda lemma and went back to Dugger hoping I’d get to the point I understand enough about sheaves so that I could get back to Baez-Hoffnung, but then realized that Dugger is STILL a bit over my head, so fell back to MacLane to study the basics again. I’m feeling things are less mysterious this time around, but it is not easy to penetrate my thick skull.

While I was reading the section in MacLane on universal arrows, the diagram

$\array{ c & \stackrel{u}{\to} & S r & & & r \\ f & \searrow & \downarrow S f' & & & \downarrow f' \\ & & S d & & & d }$

looked like the components of a natural transformation $\Delta c\to S$

$\array{ \Delta c(r) & \stackrel{u}{\to} & S(r) \\ \mathllap{\Delta c(f')\;\;}{\downarrow} & & \mathrlap{\quad S(f')}{\downarrow} \\ \Delta c(d) &\stackrel{f}\to & S(d) }\quad\quad\quad\quad\quad\array{ r \\ \downarrow f'\\ d }$

Hence my question. So you’re giving me a little too much credit, but I’m trying and making some progress :)

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeMay 31st 2010
• (edited May 31st 2010)

looked like the components of a natural transformation

Yes, indeedy. Exactly right.

So, there’s a running dictionary, that a cone $c \to S$ with vertex $c$ is the exact same thing as a natural transformation $\Delta c \to S$. And a morphism of cones from $\tau: c \to S$ to $\tau': d \to S$ is a map $f: c \to c'$ so that all the relevant triangles

$\array{ c & & \\ ^f \downarrow & \searrow ^{\tau_d} & \\ c' & \underset{\tau^{'}_d}{\to} & S(d) }$

commute for all objects $d$ in the domain of $S$, but that just translates to saying

$\array{ \Delta c & & \\ ^{\Delta f} \downarrow & \searrow ^\tau & \\ \Delta c' & \underset{\tau'}{\to} & S }$

commutes as a diagram of natural transformations.

(A next step would be to understand a universal cone in terms of a representing object, but please carry on.)

• CommentRowNumber7.
• CommentAuthorFinnLawler
• CommentTimeMay 31st 2010

I’ll just add to Todd’s explanation, in case you hadn’t spotted it, that the correspondence between your $\tau \colon \Delta d \to F$ and his section $s \colon C \to d/F$ of $p \colon d/F \to C$ (i.e. $p s = 1$) is precisely the universal property of the comma category:

$\array{ C & \to & \to & ! \\ \downarrow & \searrow s & & \searrow \\ \downarrow & & d/F & \to & * \\ 1 & \searrow & \downarrow & \swArrow & \downarrow \\ & & C & \to & D }$

That’s a bit messy, but your $\tau$ lives in the square ’behind’ the comma square. Then $s$ is unique making the rhombuses commute: the top one commutes trivially, because both morphisms land in the terminal object, and the left one says exactly that $s$ is a section of $p$.

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeMay 31st 2010

Ha ha – that was exactly the diagram I had in mind, but I knew I wouldn’t be able to render it nicely!

• CommentRowNumber9.
• CommentAuthorFinnLawler
• CommentTimeMay 31st 2010

I knew it was the one you were thinking of, but for my part I thought ’better drawn badly than not drawn at all.’ :)

• CommentRowNumber10.
• CommentAuthorEric
• CommentTimeJun 1st 2010

Thanks for the help, but just so you know, any diagram more complicated than a component of a natural transformation makes me nervous :)

It will take some time to try to parse what you’ve said, but I’ll try (when I get a chance).

• CommentRowNumber11.
• CommentAuthorEric
• CommentTimeJun 6th 2010
• (edited Jun 6th 2010)

Quick sanity check.

In Finn’s comment #7, the top path $C\to *\to D$ is the constant functor $\Delta d:C\to D$ if we label $d:* \to D$. Right? (Deja vu: I think I wanted to use the notation $d!$ for the constant functor at one point for this very reason.)

The bottom path is just $F:C\to D$.

Ok. I’m pretty sure I got that, but it felt good to say it out loud :)

So $d/F$ is the limit of that cospan in $Cat$, i.e. pullback. Right?

Ok. So we have the natural transformation $\tau:(C\to *\to D)\Rightarrow (C\to D)$ and the other one $\tau':(d/F\to *\to D)\Rightarrow (d/F\to C\to D)$.

So there is a 3-dimensional thingie mobob relating this two natural transformations. What is that thingie called? (Sorry to get so technical on you :))

• CommentRowNumber12.
• CommentAuthorFinnLawler
• CommentTimeJun 6th 2010

Not quite. You’re right about the top and bottom paths, though.

There is a cospan $F \colon C \to D \leftarrow * \colon d$. The category $d/F$ is not the pullback but the comma object over this. The pullback would be the universal commuting square over the cospan, whereas the comma object is the universal square containing a transformation. The transformation $\tau' \colon (d/F \to * \to D) \Rightarrow (d/F \to C \to D)$ is the transformation in the universal (comma) sqaure.

Now you also have the transformation $\tau \colon (C \to * \to D) \Rightarrow (C \to D)$, which also lives in a square over the cospan given by $F$ and $d$. So by the universality of the comma object $d/F$, there is a unique morphism (functor) $s \colon C \to d/F$ that makes the rhombuses commute. In particular, there’s nothing 3-dimensional going on.

The crappiness of my diagram is certainly no help. What I was trying to draw was two squares-with-transformations-in over the same cospan, together with a morphism between the objects (categories) in the top left corners of the square.

Does that help?

• CommentRowNumber13.
• CommentAuthorEric
• CommentTimeJun 6th 2010

The pullback would be the universal commuting square over the cospan, whereas the comma object is the universal square containing a transformation.

That is interesting. Is there any way this is an incarnation of a 2-limit or something? Not that I’d have much of a chance of understanding that.

In particular, there’s nothing 3-dimensional going on.

Well, a square with a natural transformation is 2-dimensional, so we have two solid squares connected by a morphism and everything in sight commutes, so I still think there is something 3-dimensional going on, but maybe that is semantics :)

For example, that section morphism $s:C\to d/F$ looks like some component of some 3-morphism. Or something…

The crappiness of my diagram is certainly no help.

Like you said, any diagram is better than no diagram :) I think the first time I looked at it, I thought $!$ and $1$ were objects. Once I knew they were labeling the morphisms, it made more sense :)

Does that help?

Yes, it does. Thanks! :)

• CommentRowNumber14.
• CommentAuthorFinnLawler
• CommentTimeJun 6th 2010

Gah! I typed out a reply, then accidentally hit ’Back’ and lost it all >:(

Is there any way this is an incarnation of a 2-limit or something?

Yes! In fact it’s a strict 2-limit, which is a Cat-weighted limit.

I still think there is something 3-dimensional going on, but maybe that is semantics :)

Certainly, the diagram is quite pleasing when thought of as 3-dimensional, but what I meant was that there are no 3-cells around. The only 3-morphisms I can think of here are the modifications in 2-Cat, but their components are 2-cells.