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Lemma: Any left anodyne map in SSet/S is a covariant equivalence.
Proof: We can consider the case of a left horn inclusion, since these generate all left anodyne maps.
Then we must show that any map
i:LeftCone(Λnj)∐ΛnjS→LeftCone(Δn)∐ΔnSis a categorical equivalence. However, i is a pushout of of the map Λn+1j+1→Δn+1, which is inner anodyne, so we’re done.
Questions:
How do we show that i is a pushout as described in the bolded sentence?
A pushout in the category of arrows? Can we replace n+1 and j+1 by n and j in the bold statement?
The LeftCone(X) is the join Δ0⋆X. Also, that definitely won’t work, since if we replace n+1 and j+1 by n and j, we don’t get an inner anodyne map when i=0.
Also, I’m not sure if the pushout is in the category of arrows or not, or if it means that it’s a pushout by some morphism (although the second one sounds more plausible).
(HTT, lemma 2.1.4.6)
The input data is a morphism σ:Δn→S.
The pushout diagram in question is
Λn+1i+1→(Λni)◃∐ΛniS↓↓Δn+1→(Δn)◃∐ΔnS.Writing out the cells here is, as usual, obvious but tedious. Think about it in low dimensions, where you can visualize the simplices:
Set n=2. Start with a 2-simplex σ in S. Then (Δ2)◃∐Δ2S is the original simplicial set S together with a tetrahedron Δ3 built over σ. One face of the tetrahedron is the original 2-simplex σ in S, the three others “stick out” of S:
The simplicial set (Λ21)◃∐Λ21S is accordingly the simplicial set S with only two of the three faces of this tetrahedron over σ erected.
The map (Λ32)→(Δ2)◃∐Δ2S identifies the horn of this tetrahedron given by these two new faces and the original face σ.
The pushout therefore glues in the remaining face of the tetrahedron and fills it with a 3-cell.
Thanks! The trick I was missing was that since we’re gluing the cone over the horn back to S, we can take the original simplex as a face of the higher dimensional horn. That is, I was having a hard time seeing what the map
Λn+1i+1→(Λni)◃∐ΛniSwas.
I was about to ask how to construct all of this formally, but a quick second with a pencil and paper made it clear.
Thanks so much, I really appreciate it!
The key point here is that
Λn+1i+1≅(Λni)◃∐ΛniΔnin a compatible way.
I added this, with a tiny bit of further details, to model structure for left fibrations– Properties – Weak equivalences.
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