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Either I miss the point here, or reflexive globular sets are not describeable as presheaves on the reflexive globular category.
To see what I mean, let G be the reflexive globular category, that is objects are the [n]th and morphisms are generated by maps sn:[n]→[n+1], tn:[n]→[n+1] and in:[n+1]→[n] devoted to the structure equations
s∘s=t∘s, t∘t=s∘t and i∘s=id as well as i∘t=id.
Now from the last two equations we see, that both s and t have to be injective, but then the first two equations say that they have to be equal. ?? That’s not what the globular shape is all about! (presheaves from here to set are not what one has in mind with globular sets, since source and targets can be different in many situations)
Maybe its better to not just copy the experience from simplicial sets, but to think of them covariant?
but then the first two equations say that they have to be equal.
I don’t understand what you mean. Think of the cellular structure on a circle with two 0-cells, two 1-cells and a 2-cell, as a model for a 2-cell in G. (I think something may be wrong with the composition order in what you wrote)
The composition order has to be reversed, if a globular set is seen as a presheaf G→Set. I mean that IF this is true, then a globular set is not the same as the globular sets we know, because in G there is necessarily s=t, since otherwise the structure equations can’t be satisfied.
Can you write down the presheaf that gives your example? Or give me s0:[0]→[1] and t0:[0]→[1] and s1:[1]→[2] and t1:[1]→[2] and some i’s such that the structure equations are satisfied.
I think David, in fact you supported my argument, since you see globular sets as covariant functors G′→Set, too? Am I right? Where G′ still has to be defined…
Where did.you get your definition of G from? And do you mean G^op -> Set?
Ok maybe I was a little to sloppy,posting this. Sry. Here is the thing in more detail:
When one looks at the nLab entry http://ncatlab.org/nlab/show/globular+set, it states that globular sets can be seen as presheaves. I think this is wrong. To be more precise:
Let G be the category, which elements are the ordinal numbers [n]:=(0,....,n) for any n∈{0,1,2,...} and morphism are generated by the following maps
s:[n]→[n+1]
t:[n]→[n+1]
i:[n+1]→[n]
subject to the conditions:
1.) ss=ts
2.) tt=st
3.) is=Id
4.) it=Id
Then it is said, that a globular set, is a functor
F:Gop→Set
===================
On a first guess, one would say, ok this is similar to the situation of simplicial sets and since the functor is contravariant we get the ’globular structure equations’ for S:=F(s), T:=F(t) and I:=F(i)
1.) SS=ST
2.) TT=TS
3.) SI=Id
4.) TI=Id
and hence get a globular set. BUT the thing is, that for general globular sets, we obviously can choose S≠T in most cases, while in the category G as given above, this is not true!
In particular from is=Id and it=Id, we see that the costructure maps s and t have left inverses and hence are injective. But then follows s=t from ss=ts and tt=st at least for [n]≤[1].
So the contradiction is, while there are globular sets with S≠T for n≤1, there are no corresponding maps s and t in G! Hence such a globular set, can’t be a functor F;Gop→Set.
Remark: (Here I wrote just s, t as well as S and T for what is really sn, tn, Sn and Tn. Just to not discourage the reader by indices.)
Maybe an example is a good thing here:
Consider a two cell α:f⇒g:A→B. (Unfortunately I can’t write such a diagram in my browser). Then the globular structure is clear:
For n=2 we have S(α)=f, T(α)=g
For n=1 we have S(f)=A, S(g)=A as well as T(f)=B, T(g)=B and I(f)=id2f, I(g)=id2g.
For n=0 we have I(A)=idA and I(B)=idB
Obviously the globular structure equations are satisfied.
========================
Now consider A≠B and try to express this 2-cell as a functor F:Gop→Set, where you explicitly give the maps s1:[1]→[2] and t1:[1]→[2]. I would say that it is not possible!
Mirco, I’m sorry, but you seem to be pretty confused here.
But then follows s=t from ss=ts and tt=st at least for [n]≤[1].
You could conclude that if s were an epimorphism. But not from the fact (that you correctly observed) that s is a monomorphism.
To understand better the globular category, it might help to picture n as representing an n-disk, s:n→n+1 as the inclusion of an n-disk onto the “northern hemisphere” of the boundary of an (n+1)-disk, and t:n→n+1 as the inclusion onto the southern hemisphere. The retraction n+1→n could be pictured as the projection of an (n+1)-disk or ball onto an equatorial cross-section. (And indeed, there is a covariant functor G→Top which carries exactly this interpretation.)
Todd, thanks for your comment. Intuitively the hemisphere POV is what I have in mind about the globular category. However just to make sure, that my imagination really works,I tried, to explicitly find the appropriate maps and now I’m not entirely sure if it works anymore.
To see why, here is the brute force method, giving first all possible injective maps for n≤1:
f00:[0]→[1];0→0,
f01:[0]→[1];0→1,
f10:[1]→[2];(0,1)→(0,1),
f11:[1]→[2];(0,1)→(0,2),
f12:[1]→[2];(0,1)→(1,0),
f13:[1]→[2];(0,1)→(1,2),
f14:[1]→[2];(0,1)→(2,0),
f15:[1]→[2];(0,1)→(2,1).
There are no other, right?
Now I ’brute force’ the combinatorics, checking all compositions, if there are some satisfying fj∘fi=fk∘fi as well as fk∘fl=fj∘fl, since then t1:=fk, t0:=fl, s1:=fj and s0:=fi.
That is
f0∘f1=1=f5∘f1 but f5∘f0=2≠0=f0∘f0
f1∘f1=2=f3∘f1but f3∘f0=1≠0=f1∘f0
f2∘f1=0=f4∘f1 but f4∘f0=2≠1=f2∘f0
f3∘f1=2=f1∘f1 but f1∘f0=0≠1=f3∘f0
f4∘f1=0=f2∘f1 but f2∘f0=1≠2=f4∘f0
f5∘f1=1=f0∘f1 but f0∘f0=0≠2=f5∘f0
============
f0∘f0=0=f1∘f0 but f1∘f1=2≠1=f0∘f1
f1∘f0=0=f0∘f0but f0∘f1=1≠2=f1∘f1
f2∘f0=1=f3∘f0 but f3∘f1=2≠0=f2∘f1
f3∘f0=1=f2∘f0 but f2∘f1=0≠2=f3∘f1
f4∘f0=2=f5∘f0 but f5∘f1=1≠0=f4∘f1
f5∘f0=2=f4∘f0 but f4∘f1=0≠1=f5∘f1
=====================
This means that the pair of equations fj∘fi=fk∘fi and fk∘fl=fj∘fl has no solution for fi≠fl here.
Please tell me you know what I’m up to! (Such that this tedious Tex-exercise was not for nothing :-) ) There are no injective maps t1;[1]→[2], t0:[0]→[1], s1:[1]→[2] and s0:[0]→[1], satisfying the structure equations. With t0≠s0…
So to proof me wrong, it is enough to just give maps s0:[0]→[1], t0:[0]→[1] and s1:[1]→[2] as well as t1:[1]→[2] such that they are injective with s1s0=t1s0 and t1t0=s1t0. But you can’t, at least under the assumption s0≠t0 and that’s an implication of the INJECTIVITY, call me confused or not.
(At least you should take the time to proof that your insult is satisfied, so I have to take it as a correct observation, not as blinded arrogance)
Re #6, I don’t think s, t and i are maps of sets, rather that G is just the free category generated by this data. Then s and t are monomorphisms, but not injective maps.
P.S. You have proved that G is not concrete, as Δ is.
Yes of course you can just define a category to have objects n and those morphisms. But at least to my eyes, people mostly think of the objects n as ’the ordinals’ [n]={0,…,n}. (At least that is the POV of the nLab entry)
However,.. thanks David to clear that point! These tiny bits on the lowest level, commonly taking for granted, can really screw me, if they don’t work out as expected.
….
But I think we can still use the ordinals [n] as we know them from the simplex category, if we shift the grading. Maybe this will require some negative thinking. I’ll post it, if I know what I mean…
Yeah. Show me where the objects n are regarded as ordinals in that article. Show me any place in that article that said we are supposed to regard the n as finite sets and the morphisms as functions.
If you find such a concrete representation (which might be possible; I’m not denying the possibility), then go ahead and add your finding to the entry.
I would like to add, Mirco, that I never insulted you. On the other hand, if you claim that many scores of people who have examined this material before you are wrong about something, then obviously the burden of proof is on you. You made repeated claims that the injectivity of s and t forced s=t, to which I responded (and where I thought you were confused).
From http://ncatlab.org/nlab/show/globular+set
The globe category 𝔾 is the category whose objects are the natural numbers, denoted here [n]∈ℕ and whose morphisms are generated from
σn:[n]→[n+1]
τn:[n]→[n+1]
for all n∈ℕ subject to the relations. I bet that most mathematicians would read this as [n]={0,…,n} and σ/τ:[n]→[n+1] as a map (of sets). Especially regarding the article http://ncatlab.org/nlab/show/ordinal+number, where its said, that any natural number is a finite ordinal and ordinals can be seen as ordered sets.
I don’t say, that there are no other possibilities, but at least it should be mentioned, as David pointed out, that the globe category can’t be seen as a concrete category that way and that σn as well as τn are not set maps.
@Mirco
it should be mentioned,
go ahead :-)
SIGH.
I bet most mathematicians who read category theory know about defining a category using generators and relations, which is what the article does. It says so right there. They are furthermore used to the idea that not all categories are described as concrete categories in their definitions (and also the fact that there might not be any concretizations, or there may be many).
I doubt that most mathematicians would make your mistake, and even doubt many of them would. But if you INSIST they would, then one remedy might be to change all those [n]’s to something else like n’s that don’t carry the same associations.
Ok maybe its better to let these speculations about mathematicians aside. I see nothing to gain here anymore and I shouldn’t have started into that direction.
If this is not a common pitfall but an act of isolated confusion, there is no need to change the entry.
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