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    • CommentRowNumber1.
    • CommentAuthorHarry Gindi
    • CommentTimeFeb 9th 2011
    • (edited Feb 9th 2011)

    Reposted from MO:


    The mapping simplex: We define a functor M n:Fun([n],sSet)sSetM^n:Fun([n],sSet)\to sSet sending the functor ϕ:[n]sSet\phi:[n]\to sSet classifying the sequence

    A 0A 1A nA^0\leftarrow A^1\leftarrow\dots\leftarrow A^n

    (notice that A k=ϕ(nk)A^k=\phi(n-k).) to the simplicial set M n(ϕ)M^n(\phi) defined as a representable functor as follows:

    Hom sSet(Δ k,M n(ϕ))= {g:[k][n]}Hom sSet(Δ k,A g(k))Hom_{sSet}(\Delta^k,M^n(\phi))=\coprod_{\{g:[k]\to[n]\}}Hom_{sSet}(\Delta^k,A^{g(k)})

    The identity map M n(ϕ)M n(ϕ)M^n(\phi)\to M^n(\phi) then determines a unique map M n(ϕ)Δ nM^n(\phi)\to \Delta^n by the fact that simplicial sets are colimits of their simplices.

    Also, given a map Δ mΔ n\Delta^m\to \Delta^n corresponding to a functor F:[m][n]F:[m]\to [n] (by full and faithfulness of the Yoneda embedding), it’s not hard to see that

    M n(ϕ)× Δ nΔ mM m(F *ϕ)M^n(\phi)\times_{\Delta^n} \Delta^m\cong M^m(F^*\phi)

    Let ϕ:[n]sSet\phi:[n]\to sSet be a functor classifying a sequence of composable maps

    A 0A 1A nA^0\leftarrow A^1\leftarrow\dots\leftarrow A^n

    Then let c A nc_{A^n} be the constant functor [n]sSet[n]\to sSet at A nA^n. Then by the functoriality of M nM^n, we have a map M n(c A n)M n(ϕ)M^n(c_{A^n})\to M^n(\phi) induced by the obvious natural transformation α:c A nϕ\alpha:c_{A^n}\to \phi, which is defined componentwise as the composite map α i:A nA i\alpha_i:A^n\to A^i for each ii (the naturality of this map is immediate).

    Notice that M n(c A)M^n(c_A) for any simplicial set AA is canonically isomorphic to the product A×Δ nA\times \Delta^n since

    Hom(Δ m,A×Δ n)Hom(Δ m,A)×Hom(Δ m,Δ n)= {[m][n]}Hom(Δ m,A).Hom(\Delta^m,A\times \Delta^n)\cong Hom(\Delta^m,A)\times Hom(\Delta^m,\Delta^n)=\coprod_{\{[m]\to [n]\}}Hom(\Delta^m,A).


    Let ι:[n1][n]\iota:[n-1]\to [n] be the obvious inclusion on the last n1n-1 objects of [n][n], and let ϕ:[n]sSet\phi:[n]\to sSet be a functor parameterizing a sequence of composable maps

    A 0A 1A n.A^0\leftarrow A^1\leftarrow\dots\leftarrow A^n.

    Let ϕ=ι *ϕ\phi'=\iota^*\phi, which parameterizes the sequence:

    A 0A 1A n1A^0\leftarrow A^1\leftarrow\dots\leftarrow A^{n-1}

    (yes, the indexing is annoying, since A k=ϕ(nk)A^k=\phi(n-k)).

    Also, let c A n:[n]sSetc_{A^n}:[n]\to sSet be the constant functor at A nA^n, and let c A n=ι *c A nc'_{A^n}=\iota^*c_{A^n}. We also define a third sequence d A n:[n]sSetd_{A^n}:[n]\to sSet to be the sequence

    A nA nA nA^n\leftarrow A^n\leftarrow\dots\leftarrow A^n\leftarrow \emptyset

    which extends c A nc'_{A^n} by the empty object on the right. It is immediate from the definition that we have a canonical isomorphism M n1(c A n)M n(d A n)M^{n-1}(c'_{A^n})\cong M^n(d_{A^n}). This gives us a natural inclusion M n1(c A n)M n(d A n)M n(c A n)M^{n-1}(c'_{A^n})\cong M^n(d_{A^n})\hookrightarrow M^n(c_{A^n})

    Then we apparently have the following pushout square:

    M n1(c A n) M n(c A n) M n1(ϕ) P\begin{matrix}M^{n-1}(c'_{A^n})&\hookrightarrow &M^n(c_{A^n})\\ \downarrow&&\downarrow\\ M^{n-1}(\phi')&\hookrightarrow& P\end{matrix}

    Claim: PM n(ϕ)P\cong M^n(\phi)

    Then why does the claim hold?

    • CommentRowNumber2.
    • CommentAuthorHarry Gindi
    • CommentTimeFeb 10th 2011

    Does anyone know how to show this? It’s geometrically obvious, but I’m having a rough time showing it to be true. The mapping properties are in exactly the wrong directions =(.

    • CommentRowNumber3.
    • CommentAuthorHarry Gindi
    • CommentTimeFeb 11th 2011

    Alright, I think I answered my own question on MO. Any chance anyone can tell me if the proof works?