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Suppose that I have a diagram of simplicial sets $X_\bullet:\mathscr{C} \to Set^{\Delta^{op}},$ with $\mathscr{C}$ a small category such that for each $C \in \mathscr{C},$ $X_\bullet(C)$ is a Kan complex. Let $X_0$ denote the diagram of $0$-simplices, but regarded again as a simplicial diagram on $\mathscr{C}$ (which is simplicially constant). There is a canonical natural transformation $X_0 \to X_\bullet.$

Question: Under what conditions will the induced map $$\mathbf{hocolim} X_0 \to \mathbf{hocolim} X_\bullet$$ between the homotopy colimits (computed in Quillen model structure on simplicial sets) be a weak equivalence?

To be clear, I'm not interested in pathalogical conditions like $X_\bullet$ is itself simplicially constant. I have a very specific example in mind which is not of this trivial form.

Here is a non-trivial example where it does work:

Example:

Take $\mathbf{M}$ to be a (almost) simplicial model category in which every object is cofibrant, and suppose that there is a cosimplicial object $I^\bullet$ such that $I^0$ is terminal and the simplicial enrichment is given by $$Map(C,D)_n=Hom(C \times I^n,D).$$ (For example, take the opposite category of the projective model structure on commutative dg-algebras). Let $D$ and $E$ be fibrant. Consider the category $\mathscr{C}$ to be the (opposite of the) category of trivial fibrations over $D$ with the arrows being commutative triangles. Let $X_\bullet$ be defined by

$$X_\bullet(\varphi:D' \to D)=Map(D',E)_\bullet.$$ Then both homotopy colimits are equivalent to the mapping space $Map(D,E).$

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  • $\begingroup$ Unless I misinterpreted something, the question essentially asks when the inclusion C→Δ^op×C is homotopy cofinal. Is the standard criterion for homotopy cofinality not sufficient here? $\endgroup$ – Dmitri Pavlov Nov 30 '14 at 15:48
  • $\begingroup$ @DmitriPavlov: I'm not asking for what $\mathscr{C}$ is this true for all $X_\bullet$. I'm asking, given $\mathscr{C}$, for what $X_\bullet$ is this true. $\endgroup$ – David Carchedi Nov 30 '14 at 21:46
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    $\begingroup$ Could whoever downvoted please tell me why? If there's an error in one of my answers, I definitely want to know! (I couldn't care less about the points) $\endgroup$ – David Carchedi Feb 26 '15 at 21:40
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I welcome further input, but since I have figured out a reformulation of this condition, I have posted it:

I claim the necessary and sufficient condition for this to be true can be formulated as follows:

For each $n$ let $\tilde X_n$ be the simplicial set $N\left(\int_{\mathscr{C}} X_n\right)$- the nerve of the Grothendieck construction. This assembles into a bisimplicial set $\tilde X_\bullet$ (written here as a simplicial simplicial set).

Then $X_0 \to X_\bullet$ induces a homotopy equivalence $$\mathbf{hocolim} X_0 \to \mathbf{hocolim} X_\bullet$$ if and only if the canonical map of simplicial sets

$$\tilde X_0 \to diag\left(\tilde X_\bullet\right)$$ is a weak equivalence.

Proof:

Let $$\hat{X}:\mathscr{C} \times \Delta^{op} \to Set_\Delta$$ be the functor $$(C,n) \mapsto X_n(C),$$ where we regard $X_n(C)$ as a constant simplicial set. Then $$\mathbf{hocolim}_\mathscr{C} \mathbf{hocolim}_{\Delta^{op}} X_n(C) \simeq \mathbf{hocolim}_{\Delta^{op}}\mathbf{hocolim}_\mathscr{C}X_n(C).$$

The LHS is canonically equivalent to $\mathbf{hocolim}_{\mathscr{C}} X_\bullet$, and each $\mathbf{hocolim}_\mathscr{C}X_n(C)$ can be modelled by $N\left(\int_{\mathscr{C}} X_n\right),$ so the RHS can be modelled by $diag\left(\tilde X_\bullet\right)$. Since in particular $\tilde X_0=N\left(\int_{\mathscr{C}} X_0\right)\simeq \mathbf{hocolim} X_0,$ we are done.

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(Too long for a comment)

I think I may have figured out what makes the example I listed above work:

For each $n$ let $\tilde X_n = \mathbf{hocolim} X_n$ (to this functorially, take the Nerve of the Grothendieck construction of $X_n$). Then $$\mathbf{hocolim} \tilde X_\bullet \simeq \mathbf{hocolim} X_\bullet$$ where the left homotopy colimit is over $\Delta^{op}$ and the right one is over $\mathcal{C}.$ So, a sufficient condition would be that $\tilde X_n$ is homotopically simplicially constant, which it is in the example above, since $\tilde X_n \simeq Map(D \times I^n,E)\simeq Map(D,E).$

I still want to hold out to make sure no one else has any input (or comments about this observation).

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