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(Copied from MO)
Recall:
Let FU•:Cat→CatΔ be the bar construction assigned to the comonad FU determined by free-forgetful adjunction F:Quiv⇄Cat:U. The restriction of FU• to the full subcategory Δ (which is isomorphic to the category of finite nonempty ordinals) naturally determines a colimit-preserving functor ℭ:SetΔ=SetΔop→CatΔ. The right adjoint of this functor is called 𝔑, the homotopy-coherent nerve.
Identify Cat (not by FU•) with the full subcategory of CatΔ spanned by those simplicially enriched categories with discrete hom-spaces.
Also, recall the definition of the right cone X▹ on a simplicial set X is the join X⋆Δ0. This determines an obvious natural map X→X▹.
Let X→Δ1=𝔑([1]) be an object of (SetΔ↓Δ1), and let ε:ℭ(Δ1)=ℭ(𝔑([1])→[1] be the counit (here [1] is the category determined by the ordinal number 2 (two objects, one nonidentity arrow). Form the pushout M of the span ℭ(X▹)←ℭ(X)→ℭ(Δ1)→[1] (the two arrows in the same direction are replaced by their composite, so this is M=ℭ(X▹)∐ℭ(X)[1]).
This determines a functor StεX:[1]→SetΔ defined as i↦M(i,p) where p is the image of the cone point of ℭ(X▹).
Question:
The book I’m reading asserts that StεX(0) can be identified with St*(X×Δ1Δ0) (where Δ0→Δ1 is the map 𝔑(λ) where λ:[0]→[1] is the map choosing the object 0 of [1]) where St*S is simply defined to be the analogous construction when ε is replaced with the identity [0]=ℭ(Δ0)→[0]. (Note that here we can identify functors [0]→SetΔ with simplicial sets themselves, and suggestively, that under this identification, StεX(0)=λ*StεX).
Why is this true?
Edit: (Blah, modulo the inevitable sign error here).
=(. If someone could pry himself/herself away from the argument over at universe and explain this even for a moment, I’d really appreciate it.
This should be the statement that in sSet (being a Grothendieck topos) we have pullback stability of colimits.
More in detail:
Since ℭ is left adjoint we can essentially compute the pushout before applying ℭ. Let me call the analog of M obtained this way P
X→X▹↓↓Δ[1]→PWe have a canonical map P→Δ[1]▹ induced from the commutativity of
X→X▹↓↓Δ[1]→Δ[1]▹.For evaluating P(0,p) we just need the fiber over {0}▹, hence the pullback of the diagram
P↓{0}▹↪Δ[1]▹.Now, since colimits commute with pullbacks in sSet, this pullback is the pushout of the corresponding pullbacks of X, and X▹. But that pullback of X is X×Δ[1]Δ[0]. Because you can compute it as this consecutive pullback:
X×Δ[1]{0}→X↓↓{0}→Δ[1]↓↓{0}▹→PThanks!
Oh, by the way, I posted your answer on MO (with attribution and a link, as well as making it community wiki). If you’d prefer to answer it there yourself, I will delete the copy. Thanks again!
Meanwhile: Is it obvious for any formal reason that the pullback X▹×(Δ1)▹{0}▹=(X×Δ1{0})▹? I mean, I think I can just show it by a computation, but can you derive it again from the fact that colimits are universal?
Is it obvious for any formal reason that the pullback X▹×(Δ1)▹{0}▹=(X×Δ1{0})▹? I mean, I think I can just show it by a computation, but can you derive it again from the fact that colimits are universal?
If we invoke the description of the join in terms of Day convolution we have the coend expression
X▹:[k]↦∫[i],[j]∈ΔaXi×HomΔa([k],[i]⊞A coend is just a certain kind of colimit, so on the right this is some colimit of sets (for each ) over a diagram whose vertices are sets of the form .
I think therefore the argument that colimits are stable under pullback applies to this case, too. Yes.
Maybe again more in detail: with the above argument we find first that
And that remaining pullback is easily seen to be
(All equality signs denote isomorphisms.)
Thanks again!
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