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

    (Copied from MO)


    Recall:

    Let FU :CatCat ΔFU_\bullet:Cat\to Cat_\Delta be the bar construction assigned to the comonad FUFU determined by free-forgetful adjunction F:QuivCat:UF:Quiv\rightleftarrows Cat:U. The restriction of FU FU_\bullet to the full subcategory Δ\Delta (which is isomorphic to the category of finite nonempty ordinals) naturally determines a colimit-preserving functor :Set Δ=Set Δ opCat Δ\mathfrak{C}:Set_\Delta=Set^{\Delta^{op}}\to Cat_\Delta. The right adjoint of this functor is called 𝔑\mathfrak{N}, the homotopy-coherent nerve.

    Identify CatCat (not by FU FU_\bullet) with the full subcategory of Cat ΔCat_\Delta spanned by those simplicially enriched categories with discrete hom-spaces.

    Also, recall the definition of the right cone X X^\triangleright on a simplicial set XX is the join XΔ 0X\star \Delta^0. This determines an obvious natural map XX X\to X^\triangleright.

    Let XΔ 1=𝔑([1])X\to \Delta^1=\mathfrak{N}([1]) be an object of (Set ΔΔ 1)(Set_\Delta\downarrow \Delta^1), and let ε:(Δ 1)=(𝔑([1])[1]\epsilon:\mathfrak{C}(\Delta^1)=\mathfrak{C}(\mathfrak{N}([1])\to [1] be the counit (here [1][1] is the category determined by the ordinal number 22 (two objects, one nonidentity arrow). Form the pushout MM of the span (X )(X)(Δ 1)[1]\mathfrak{C}(X^\triangleright) \leftarrow \mathfrak{C}(X)\to \mathfrak{C}(\Delta^1)\to [1] (the two arrows in the same direction are replaced by their composite, so this is M=(X ) (X)[1]M=\mathfrak{C}(X^\triangleright)\coprod_{\mathfrak{C}(X)} [1]).

    This determines a functor St εX:[1]Set ΔSt_\epsilon X:[1]\to Set_\Delta defined as iM(i,p)i\mapsto M(i,p) where pp is the image of the cone point of (X ).\mathfrak{C}(X^\triangleright).

    Question:

    The book I’m reading asserts that St εX(0)St_\epsilon X(0) can be identified with St *(X× Δ 1Δ 0)St_*(X\times_{\Delta^1} \Delta^0) (where Δ 0Δ 1\Delta^0\to \Delta^1 is the map 𝔑(λ)\mathfrak{N}(\lambda) where λ:[0][1]\lambda:[0]\to [1] is the map choosing the object 00 of [1][1]) where St *SSt_*S is simply defined to be the analogous construction when ε\epsilon is replaced with the identity [0]=(Δ 0)[0][0]=\mathfrak{C}(\Delta^0)\to [0]. (Note that here we can identify functors [0]Set Δ[0]\to Set_\Delta with simplicial sets themselves, and suggestively, that under this identification, St εX(0)=λ *St εXSt_\epsilon X(0)=\lambda^*St_\epsilon X).

    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 sSetsSet (being a Grothendieck topos) we have pullback stability of colimits.

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

    More in detail:

    Since \mathfrak{C} is left adjoint we can essentially compute the pushout before applying \mathfrak{C}. Let me call the analog of MM obtained this way PP

    X X Δ[1] P \array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& P }

    We have a canonical map PΔ[1] P \to \Delta[1]^{\triangleright} induced from the commutativity of

    X X Δ[1] Δ[1] . \array{ X &\to& X^{\triangleright} \\ \downarrow && \downarrow \\ \Delta[1] &\to& \Delta[1]^{\triangleright} } \,.

    For evaluating P(0,p)P(0,p) we just need the fiber over {0} \{0\}^{\triangleright}, hence the pullback of the diagram

    P {0} Δ[1] . \array{ && P \\ && \downarrow \\ \{0\}^{\triangleright} &\hookrightarrow& \Delta[1]^{\triangleright} } \,.

    Now, since colimits commute with pullbacks in sSetsSet, this pullback is the pushout of the corresponding pullbacks of XX, and X X^{\triangleright}. But that pullback of XX is X× Δ[1]Δ[0]X \times_{\Delta[1]} \Delta[0]. Because you can compute it as this consecutive pullback:

    X× Δ[1]{0} X {0} Δ[1] {0} P \array{ X \times_{\Delta[1]} \{0\} &\to& X \\ \downarrow && \downarrow \\ \{0\} &\to& \Delta[1] \\ \downarrow && \downarrow \\ \{0\}^{\triangleright} &\to & 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 × (Δ 1) {0} =(X× Δ 1{0}) X^\triangleright\times_{(\Delta^1)^\triangleright} \{0\}^\triangleright=(X\times_{\Delta^1}\{0\})^\triangleright? 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 × (Δ 1) {0} =(X× Δ 1{0}) X^\triangleright\times_{(\Delta^1)^\triangleright} \{0\}^\triangleright=(X\times_{\Delta^1}\{0\})^\triangleright? 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]Δ aX i×Hom Δ a([k],[i][j]). X^{\triangleright} : [k] \mapsto \int^{[i],[j] \in \Delta_a} X_i \times Hom_{\Delta_a}([k],[i] \boxplus [j]) \,.

    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!