Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
1 to 21 of 21
Let X→Δn be a right fibration (has the right lifting property with respect to right horn inclusions), and let
Δ{n−i}↪Δ{n−i,…,n}↪Δn(for a fixed i:0≤i≤n) be the obvious inclusion maps.
Then why is the induced map:
X×ΔnΔ{n−i}↪X×ΔnΔ{n−i,…,n}a deformation retract?
(This is from the end of the proof of proposition 2.2.3.1 of HTT. The statement should be true out of the context in the book with the hypotheses I’ve given, but if not, there’s the source.)
And if you’d like to answer it on Math Overflow, go ahead.
Here’s hoping someone will give it a shot.
Bump
In all seriousness, Tom Goodwillie gave an answer on MO, but I can’t see how it works, since it doesn’t seem like the inclusion of the point into the n-simplex is actually a deformation retract (although it is in fact anodyne). The trouble is that the n-simplex is not a kan complex.
It’s fine: it is a deformation retract. Let’s take the inclusion of the zeroth vertex (as Tom Goodwillie did), and look at it this way: starting off in Cat, we have a natural transformation from the composite [n]→[1]→[n] to the identity ([n] is the n-element total order), where [1]→[n] names the initial object. This natural transformation can be re-expressed as a functor [2]×[n]→[n]. Take the nerve of this functor; it gives
hom(−,[2])×hom(−,[n])→hom(−,[n])which is a simplicial homotopy from ir to id (i the inclusion of the point, r the retraction).
Thanks!
And it is the fact that we are dealing with right fibrations that allows to lift this to a retract of the objects in question: the composition cells that are guaranteed to exist allow to lift the simplicial homotopy that Todd indicated.
Hey Todd, is your indexing one different from the usual indexing (where [0] is usually taken to be the terminal category and [1] is taken to be the interval category)?
@Urs: I still don’t really follow what’s lifting what, but I’m going to give it a shot.
Harry: yes (which is why I bothered to defined what [n] meant (-: ).
Could someone help me set up the lifting problem? I don’t see what exactly we’re trying to lift.
We can easily reduce to the case where it’s the inclusion of the initial vertex (right fibrations are stable under pullback), so we’ve got a pullback diagram
X→Δn←{0}, where the first map is a right fibration and the second map is a deformation retract.
Call the pullback of this diagram P. Then we want to show that the projection P→X is a deformation retraction.
Now, I don’t see where we should set up the lifting problem.
We look at the diagram
X×ΔnΔ{n−i}↪X×ΔnΔ{n−i,…,n}↓↓Δ{n−i}↪Δ{n−i,…,n}.We want to show that the top morphism is a deformation retract. We know that the bottom morphism is. We want to see that this lifts from the bottom to the top.
What we know is that the two vertical morphisms are right fibrations, because they are both pullbacks of right fibrations and these are stable under pullback.
I think one can – but I haven’t taken pen and paper to actually do – see that using the right horn fillers that the right fibratio property guarantees, we ca build the simplicial map
Δ[2]×X×ΔnΔ{n−i,…,n}→X×ΔnΔ{n−i,…,n}that covers the one that Todd describes ad exhibits the desired deformation retract.
I don’t feel I have the leisure to work this out further, I have other exercises that I badly need to be looking into myself, but this is what I am envisioning the strategy here would be.
I think I’ve got it, but it’s pretty complicated.
@Urs: I don’t think that will work.
Alright, the trouble is, I can’t show that my new homotopy sends the 0 projection to the idempotent composite.
What I did was notice that in the diagram where we’re trying to show a lifting (X×Δ1←X×∂Δ1(fg,idx)→X), we can factor the (implied) terminal maps as X→Δn→* and X×Δ1→Δn×Δ1→* because X lives over Δn. It’s easy to check that the factorization is compatible with the the map (fg,id_X), which gives us a lift X×Δ1→Δn, making a commutative square.
Now it looks like we’re in business. Using the thing Tom Goodwillie mentioned, we can look at the union and induce maps on the righthand inclusion, which we choose to be the identity and the projection composed with the map f, which makes the diagram commute. We now have by the lifting property a homotopy H such that H|X×{1}=idX and also that H behaves as a projection composed with the inclusion f:P→X on P×I where P=X×Δn{0} from the beginning.
Well it looks like we’re finally on to something, but now we must verify that H|X×{0} is the composite fg, where g is the map splitting f. It’s not clear to me how we can show this. The composite of the restriction with the map X→Δn is equal to the composite of fg with the map X→Δn, but I can’t seem to show that the restriction itself is actually equal to fg.
Do we need to do some kind of double homotopy trick?
Maybe it has to do with the uniqueness of the maps f and g, since they are induced by the pullback?
Let X be a simplicial set, and let ℓ:X→Δn be a right fibration.
Consider the inclusion of the 0th vertex {0}↪Δn. This is a deformation retract, since we have a homotopy h:Δn×Δ1→Δn, where h|Δn×{0}=Δn→{0}↪Δn is the composite of the terminal map and the inclusion. and h|Δn×{1}=idΔn.
Taking the pullback X×Δn(−), we would like to show that the induced map f:X×Δn{0}↪X×ΔnΔn=X is also a deformation retract. It’s clear that f admits a retraction, g, so we’d like to give a homotopy fg→idX.
We have the obvious setup to construct the homotopy as the span X×Δ1↩X×∂Δ1(fg,idX)→X. We notice that the implied terminal maps of the diagram can be factored into X→Δn→* and X×Δ1→Δn×Δ1→*. We notice further that we can factor the the composite X×∂Δ1↪X×Δ1→Δn×Δ1 instead as
X×∂Δ1→Δn×∂Δ1→Δn×Δ1,and the composite X×∂Δ1→X→Δn as
X×∂Δ1→Δn×∂Δ1→Δn.These factorizations let us fit the homotopy h into the diagram, since it is easy to see that these maps are exactly the ones that h lifts. This gives a commutative square for our span with Δn in the bottom right corner, where the maps into it are the obvious ones induced by the homotopy.
We’d now like to give the homotopy we’ve been looking for. Since ℓ is a right fibration, we can lift maps of the form K×Δ1∪L×{1}→L×Δ1 for any inclusion K↪L.
Let A:=X×Δn{0}. We have the inclusion f:A↪X, so applying the pushout above, we have the map A×Δ1∪X×{1}→X×Δ1. We give a map from the pushout to X by specifying it on the components. The map A×Δ1→X is given by the composite of the projection onto A with the map f. The map X×{1}→X is just the identity on X. Then we have a commutative square with the pushout at the top left, X×Δ1 at the bottom left, X at the top right, and Δn in the bottom right. The by the lifting property, we have a lift H:X×Δ1→X making the whole diagram commute.
We’d like to show that H is the required homotopy. By construction, H|X×{1} gives the identity and H∘(f×Δ1) is equal to the first projection composed with f.
Then we need only show that H|X×{0}=fg. It’s easy to show that the composite ℓH|X×{0}=ℓfg by the commutativity of the diagrams, and also that H|X×{0}f=f by the last sentence of the previous paragraph. Can we prove that H is the required homotopy, or do we need to do another construction and another homotopy?
If this is the wrong way to prove this, let me know.
Also, I’m really stuck, so if you have time, I entreat you, please help. I’ve been stuck for 28 hours, and I’m now obsessing over it. =(.
It’s clear that f admits a retraction, g
You’re much deeper into this problem than I am, but even this much isn’t obvious to me. In general, it will not be true that given a diagram
X↓hY→fZthe pullback h*Y→X of f along h will admit a retraction if f does. For an example in Top, take f=incl:{0}→[0,1] and h to be the projection to the coordinate z of the northern hemisphere X={(x,y,z):x2+y2+z2=1,z≥0}. Then the fiber h−1(0)↪X is not a retract of X, since X is contractible but the fiber is not. (It is easy to mimic this example in simplicial sets.)
Is there some extra hypothesis you’re using to get the retraction g of f?
No, I thought that was clear. Hmm…
(Although it was bothering me. For some reason, I thought we got this for free but couldn’t seem to verify it.)
So does being incorrect count as an extra hypothesis? =D!
I’m sure Mike (because he’s an algebraic topologist) knows the actual answer/could sketch out a proof (for the problem of pulling back the inclusion of the 0-vertex into an n-simplex by a right fibration and showing it’s a deformation retract), so Mike, I beseech you, please help.
What gave you the idea I was an algebraic topologist? (-: I’m a category theorist who knows (and occasionally does) some algebraic topology. My advisor was an algebraic topologist, but a topological one, not a simplicial one, so my simplicial-fu is not that great.
Didn’t May write a book on simplicial stuff? Also, isn’t he famous for inventing some simplicial stuff as well?
The book was a long time ago, methods are a lot different now, and he’s working on different things.
I still think the idea that I sketched works.
I’m not sure how you could build up the map, could you elaborate?
I can’t guarantee that I’ll find time to think about this, but for my education, could someone explain the notation Δ{n−i}, Δ{n−i,…,n}, and also explain what a right horn is?
Alright, first off, you should submit your answer (if you come up with one) on MO, since I’m offering a bounty like last time (by 2AM EST if you’d like to claim it)).
The notation Δ{n−i,...,n} is used only in the context of morphisms. That is, the map Δ{n−i,...,n}↪Δn is the inclusion of the n-i simplex spanned by the vertices {n−i,...,n} (note that the order here does matter).
The ith horn for 0≤i≤n of an n-simplex is given by the subfunctor of Λni⊂Δn≔HomΔ(−,[n]) where (Λni)([m])⊆Hom([m],[n]) is subset of all maps p:[m]→[n] such that p[m]∪{i}≠[n] (viewed as sets). Geometrically this corresponds to the boundary of the n-simplex without the ith face.
A right horn inclusion is just the inclusion Λni→Δn for a horn 0<i≤n.
Edit: My original proof was almost right. Tom Goodwillie has answered it on MO. In particular, we don’t need to assume that the new retraction exists. The result follows immediately.
1 to 21 of 21