Homotopy Theory

user131753

4:34 AM

@DenisNardin Unfortunately I don't know anything about $(2,1)$-categories. What if I define the dual of a category $\mathbf{A}$ as follows: "A category $\mathbf{A}^{\text{op}}$ is said to the dual of a category $\mathbf{A}$ if for any category $\mathbf{B}$ and for all functors $\mathscr{F},\mathscr{G}$ such that $\mathscr{F}:\mathbf{A}\to\mathbf{B}$ and $\mathscr{G}:\mathbf{A}^{\text{op}}\to\mathbf{B}$ there exists an unique isomorphism from $\mathbf{A}$ to $\mathbf{A}^{\text{op}}$."

Jonathan Beardsley

@user170039 in general, there's not going to be an isomorphism from $A$ to $A^{op}$ unless $A$ is a groupoid, so that's probably not going to be a super helpful notion of dual.

user131753

Since I don't know about the natural transformations yet, I don't know whether what you said is essentially the same as what I said.

Jonathan Beardsley

Also, it looks like you're saying that there's such an isomorphism for every single category $B$ and every single pair of functors $F:A\to B$ and $G:A^{op}\to B$. So that's a whole lot of isomorphisms! Are you sure you want that many isomorphisms?

@user170039 I think that, looking at your original question, it might be helpful to think more about what's meant by talking about the set of morphisms between two objects in a category.

user131753

@JonathanBeardsley Actually I wanted the existence and uniqueness of the isomorphism independent of $\mathbf{B}$ and the functors $\mathscr{F}$ and $\mathscr{G}$. Not sure how I would phrase that.

Jonathan Beardsley

@user170039 Well, I mean, I would just not mention $B$ or the other functors at all.

(sorry for not doing the same typesetting as you, I find it takes me a lot of work and I end up making typos anyway!)

As for your original question let $A$ be a category. So that means for any two objects $a,b\in A$, we have a set of morphisms $Hom_{A}(a,b)$. Right now we don't know anything about this category. It's just got objects and then between pairs of objects it has morphisms, right?

user131753

4:43 AM

@JonathanBeardsley Yes sure.

Jonathan Beardsley

Now, I'm going to define a new category called $A^{op}$. Its objects are the same objects as $A$, but given any two objects in $A^{op}$, e.g. $a,b\in A^{op}$, I simply DEFINE the set of morphisms from $a$ to $b$, $Hom_{A^{op}}(a,b)$ to be the set $Hom_A(b,a)$.

user131753

@JonathanBeardsley You mean exactly equal?

Jonathan Beardsley

And then I check that this category I've defined actually satisfies all the necessary conditions, i.e. everything has an identity and I can compose morphisms associatively.

Yes. I mean exactly equal.

user131753

@JonathanBeardsley Only after that we define it to be the dual category of $\mathbf{A}$ right?

Jonathan Beardsley

I mean, what I just described defines a functor $Cat\to Cat$.

You can call it dual, or op, or whatever.

Sorry, maybe this is not answering your question, I'm not sure.

I guess you were asking if this functor has a universal property?

I think I was responding to your original question about $A\times B$ versus $B\times A$, which maybe you resolved.

user131753

4:49 AM

@JonathanBeardsley Yes, that was resolved (my pathethic confusion created it).

user131753

Actually my main motivation of asking the question was to obtain a categorical explanation of uniqueness of $\mathbf{A}^{\text{op}}$ (just like we do in case of products).

Jonathan Beardsley

I see.

I suppose, if one asks for uniqueness of something, one needs to have a way of defining that thing which isn't just explicitly describing it.

In other words, if I say "Prove that the smallest positive number is unique," then that's meaningful, but if I say "Prove that the number 1 is unique," well... I'm not sure how to understand that second question.

user131753

@JonathanBeardsley Exactly!! That's what I wanted to do.

Jonathan Beardsley

Right. Yeah I don't know. I mean, I guess you could obtain the opposite of a category by doing what Denis suggested. That is, first prove that, up to isomorphism, there is a unique non-trivial automorphism of $Cat$, and then prove that the target of this automorphism will always be $A^{op}$, as defined above.

user131753

@JonathanBeardsley Or simply define the target of the only non-trivial automorphism from $\mathbf{Cat}$ to $\mathbf{Cat}$ to be the dual.

Jonathan Beardsley

4:58 AM

Right, that's what I'm saying basically. Prove that there's one non-trivial automorphism, call it $\phi:Cat\to Cat$. Then say "When I write $A^{op}$ I mean $\phi(A)$, even though I have no idea what $\phi$ does."

Or something along these lines.

But why would you want to do that?

user131753

@JonathanBeardsley I think this definition is much more simpler than that of the "standard" one.

Jonathan Beardsley

Well I don't know. I mean first you need to prove that there's a unique non-trivial automorphism.

user131753

@JonathanBeardsley Because I want to take an "external" viewpoint of the objects of $\mathbf{Cat}$.

Jonathan Beardsley

Sure.

user131753

@JonathanBeardsley That's the non-trivial part I suppose. But at least I am happy so long as it is purely categorical. (Psychological bias, sorry!)

Jonathan Beardsley

5:01 AM

I mean, so maybe define $A^{op}$ in the classical way, and then try to prove that, if $\phi:Cat\to Cat$ is an automorphism, and isn't the identity, then $\phi(A)\simeq A^{op}$ or something.

user131753

@JonathanBeardsley Yes that approach, I think, would answer my original question categorically.

user131753

@JonathanBeardsley What I actually wanted (now I see that it is not an isomorphism) the unique morphsim to be is the "contravariant version" of the identity functor from. Is there any special name for it? Otherwise, I would need to replace "morphism" in place of "isomorphism" here.

Jonathan Beardsley

Well, unfortunately that's not a functor at all.

user131753

@JonathanBeardsley What? I meant that the functor would take objects to the same objects and morphisms to the same morphism but only reversing their direction.

Jonathan Beardsley

So it's the identity on objects. Given a morphism $f:a\to b$ in a category $A$, we need to choose a morphism that goes from $a$ to $b$ in $A^{op}$. Which one do we pick?

Recall that the morphisms from $a$ to $b$ in $A^{op}$ are $Hom_A(b,a)$. So which map from $b$ to $a$ do we send $f$ to?

user131753

5:16 AM

@JonathanBeardsley I don't understand. Why would we need to choose a morphism that goes from $a$ to $b$ in $\mathbf{A}^{\text{op}}$? I need to choose a morphism from $b$ to $a$ in $\mathbf{A}^{\text{op}}$.

Jonathan Beardsley

A functor $F$ from $C$ to $D$ is a function on objects, and for every map $f:a\to b$ in $C$, a map $F(f):F(a)\to F(b)$ in $D$.

So you're telling me that you want a functor $F:A\to A^{op}$ such that $F(a)=a$ for all $a$ in $A$.

user131753

Actually I want the functor to act on an $\mathbf{A}$-morphism as follows: $$\mathscr{F}(f:a\to b)=f:b\to a$$

Jonathan Beardsley

That's not how one defines a functor.

user131753

@JonathanBeardsley Now I understand. You are right, it isn't a functor.

Jonathan Beardsley

Great! :-)

user131753

5:21 AM

@JonathanBeardsley Thanks for helping.

Jonathan Beardsley

Sure thing.

I remember having the same exact difficulty with $op$ at some point.

Probably more recently than my high falootin' talk might imply.

Eric Peterson

1:42 PM

check out these cool animations, due to barnes-poduska-shick, of the unstable adams spectral sequences stabilizing: one for p = 2 chromotopy.org/images/UASS2.gif & one for p = 3 chromotopy.org/images/UASS3.gif

Cup Fever

2:55 PM

how goes your work on this?

Nov 1 '17 at 4:41, by Eric Peterson

https://github.com/ecpeterson/FormalGeomNotes/raw/master/main.pdf this got sent off to the publisher today 🎉🎉 but there's still lots of time to take advice, should any of you like to try to read it

Eric Peterson

4:05 PM

i'm reviewing the copyeditor's proofs now, and CUP has a website for the book. should be purchasable in october :)

Cup Fever

4:24 PM

cool

:)

Praphulla Koushik

4:42 PM

I am trying to read Differentiable stacks and gerbes by Behrend and Ping Xu in which there is a discussion of sheaves over stacks and their cohomology.. Are there other places where there is a discussion of sheaves over stacks and their cohomology?

Denis Nardin

6:05 PM

@PraphullaKoushik How well do you know the theory of sheaves over sites? I would recommend SGA 4 ½ (github.com/dkmiller/sga4.5/blob/master/sga4.5.pdf) and Vistoli's notes (homepage.sns.it/vistoli/descent.pdf) if you need an introduction

Praphulla Koushik

6:26 PM

@DenisNardin I only know the definitions and some basic properties of sheaves over a site, I read it from Vistoli’s notes some time back.. I do not read French :(

Vistoli’s notes does not say more than definition of sheaf on a site, right? Did I miss something }

Denis Nardin

They also contain the definition of stack and examples

Praphulla Koushik

No no I mean anything more than sheaf on site.. I do see that they discuss stacks.. I read from there about stacks...

Denis Nardin

Sheaves on a stack are sheaves on a site associated to the stack

Praphulla Koushik

Yes that’s true.. given a stack, fixing a site on category of smooth manifolds, this induce a grothendieck topology on stack $\mathcal{D}\rightarrow Man$ and sheaf on stack is sheaf on Category $\mathcal{D}$ with grothendieck topology said just now...

that would be definition... anything else you want to say...

Denis Nardin

6:52 PM

Not really... Are you ok with the examples in Vistoli's notes? They are a bit geared towards algebraic geometry but it's important have examples in mind in this subject

Praphulla Koushik

Only examples I am ok with is $B\mathcal{G}$ for a Lie groupoid $\mathcal{G}$.. do you have any suggestion for differentiable stacks.. there are so much material for algebraic stacks but not so much (in beginner level) for differentiable stacks...

Denis Nardin

So, usually stack arise when you need to study moduli problems. Are you familiar with this: webusers.imj-prg.fr/~gregory.ginot/papers/DiffStacksIGG2013.pdf ?

At page 12 there's the usual example of the moduli stack of triangles which is always nice to have in mind

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