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    • CommentRowNumber1.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 3rd 2010
    • (edited Nov 3rd 2010)

    (Copied from MO)


    Recall:

    Let FU:CatCatΔ be the bar construction assigned to the comonad FU determined by free-forgetful adjunction F:QuivCat: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 :SetΔ=SetΔopCatΔ. 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 XΔ0. This determines an obvious natural map XX.

    Let XΔ1=𝔑([1]) be an object of (SetΔΔ1), and let ε:(Δ1)=(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 (X)(X)(Δ1)[1] (the two arrows in the same direction are replaced by their composite, so this is M=(X)(X)[1]).

    This determines a functor StεX:[1]SetΔ defined as iM(i,p) where p is the image of the cone point of (X).

    Question:

    The book I’m reading asserts that StεX(0) can be identified with St*(X×Δ1Δ0) (where Δ0Δ1 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]=(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).

    • CommentRowNumber2.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 3rd 2010

    =(. 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.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeNov 4th 2010

    This should be the statement that in sSet (being a Grothendieck topos) we have pullback stability of colimits.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeNov 4th 2010
    • (edited Nov 4th 2010)

    More in detail:

    Since is left adjoint we can essentially compute the pushout before applying . Let me call the analog of M obtained this way P

    XXΔ[1]P

    We have a canonical map PΔ[1] induced from the commutativity of

    XXΔ[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}P
    • CommentRowNumber5.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 4th 2010

    Thanks!

    • CommentRowNumber6.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 4th 2010
    • (edited Nov 4th 2010)

    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!

    • CommentRowNumber7.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 4th 2010
    • (edited Nov 4th 2010)

    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?

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeNov 4th 2010
    • (edited Nov 4th 2010)

    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 kk) over a diagram whose vertices are sets of the form X i×Hom Δ a([k],[i][j])X_i \times Hom_{\Delta_a}([k],[i'] \boxplus [j]).

    I think therefore the argument that colimits are stable under pullback applies to this case, too. Yes.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeNov 4th 2010
    • (edited Nov 4th 2010)

    Maybe again more in detail: with the above argument we find first that

    X × Δ[1] {0} =(X× Δ[1] {0} ) X^{\triangleright} \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} = \left( X \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} \right)^{\triangleright}

    And that remaining pullback is easily seen to be

    X× Δ[1] {0} =X× Δ[1]{0}. X \times_{\Delta[1]^{\triangleright}} \{0\}^{\triangleright} = X \times_{\Delta[1]} \{0\} \,.

    (All equality signs denote isomorphisms.)

    • CommentRowNumber10.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 4th 2010

    Thanks again!