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:


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 $Map(D,E).$

  • $\begingroup$ 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? $\endgroup$ Nov 30, 2014 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. $\endgroup$ Nov 30, 2014 at 21:46
  • 1
    $\begingroup$ 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) $\endgroup$ Feb 26, 2015 at 21:40

2 Answers 2


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.


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.


(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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.