I'm wondering about when the colimit and the homotopy colimit agree with diagrams of simplicial sets. I know that hocolim$(F)=$colim$(F_c)$ where $F_c$ is the cofibrant replacement of $F$. However, it is not always necessary for $F$ to be cofibrant for the colimit and homotopy colimit to be the same. For example, let $\mathcal{C}=b\leftarrow a\rightarrow c$. Then the cofibrant $\mathcal{C}$-diagrams of simplicial sets are ones $Y\leftarrow X\rightarrow Z$ with $X,Y,Z$ cofibrant and the maps $X\to Y$ and $X\to Z$ cofibrations. However, by the left-properness of simplicial sets, I believe we only need one of the maps to be a cofibration for $\text{colim}F$ to agree with $\text{hocolim}F$.

Are there other properties or tools that can say something about the colimit and homotopy colimit of a diagram of simplicial sets being the same?



3 Answers 3


I don't think we can expect to have one general answer to this question, only a collection of unrelated specialized results. Here are two more:

  • In the category of simplicial sets all filtered colimits are homotopy colimits. Added: The gist of the argument can be found in Proposition 1.3 in Quillen's Higher Algebraic K-theory I. I can't seem to find a more precise reference, but I can give a complete alternative argument. Let $J$ be a filtered category. First, observe that a levelwise acyclic Kan fibration of $J$-diagrams induces an acyclic Kan fibration on colimits (since acyclic Kan fibrations are detected by maps out of finite simplicial sets). By K. Brown's Lemma the colimit functor preserves levelwise weak equivalences between $J$-diagrams of Kan complexes. (That functor is not a right Quillen functor but the assumptions of K. Brown's Lemma are really much weaker, see Lemma 1.1.12 in Hovey's Model Categories.) It now follows that the colimit functor preserves weak equivalences between arbitrary $J$-diagrams since there are filtered colimit preserving fibrant replacement functors on the category of simplicial sets, e.g. $\mathrm{Ex}^\infty$. In particular, the colimit of any $J$-diagram is weakly equivalent to the colimit of its projectively cofibrant replacement.
  • Splittings of idempotents, which are colimits over the category freely generated by one idempotent, are always homotopy colimits. Added: This follows since splittings of idempotents are retracts and weak equivalences of simplicial sets are closed under retracts.

On the other hand, your example with homotopy pushouts does fit into the general framework of cofibrant diagrams. There may be more than one notion of cofibrant diagram suitable for deriving the colimit functor. In this case, if we denote the objects of the indexing category as $a_0 \leftarrow a_1 \to a_2$ and make it into a Reedy category by declaring the degree of $a_i$ to be $i$, then Reedy cofibrant diagrams are spans where one leg is a cofibration and the colimit functor is a Quillen functor with respect to the resulting Reedy model structure. (Added: See the proof of Lemma 5.2.6 in Hovey's Model Categories for the details.) Perhaps Gregory's example with cubes can also be described in a similar manner, but it is less obvious to me.

  • $\begingroup$ These are great results. Can you provide a reference? Thanks! $\endgroup$ Apr 7, 2016 at 11:43
  • $\begingroup$ @David I added some comments and references. $\endgroup$ Apr 7, 2016 at 12:14
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    $\begingroup$ @KarolSzumiło: There is a subtlety here: Reedy cofibrancy forces the objects a_i to be cofibrant. Pushouts along cofibrations in left proper model categories (which is what the OP mentioned) are always homotopy pushouts, even if a_i are not cofibrant. $\endgroup$ Apr 7, 2016 at 12:30
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    $\begingroup$ @Dmitri Thanks for the references. They are both very recent and are about much stronger results There should be some earlier references too, but I can't find any. The argument I gave above is also much more elementary. (And I find it quite amusing that we can prove homotopy invariance of filtered colimits by using fibrant replacements.) $\endgroup$ Apr 7, 2016 at 12:43
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    $\begingroup$ BTW, I have noticed that the argument I gave is the same as the one in Charles Rezk's answer to the question you linked. $\endgroup$ Apr 7, 2016 at 12:47

There is a rather obvious generalization of your example to cubical pushouts of any dimension. Let $I_n$ be the poset of proper subsets of $\{1, ..., n\}$, and $\chi\colon I \to $Sset a diagram indexed on $I$. Let $I_{n-1}$ be the subposet of sets not containing $n$, and $I^n$ the subposet of sets containing $n$ (you can think of these subposets as the back and front faces of a punctured cubical diagram). Suppose that the restrictions of $\chi$ to $I_{n-1}$ and $I^n$ are both cofibrant. Then the map $\mbox{ hocolim }\chi \to \mbox{colim }\chi$ is an equivalence. The reason is that $\mbox{hocolim } \chi$ is equivalent to the homotopy pushout $$\mbox{hocolim}_{I_{n-1}}\,\, \chi\leftarrow \mbox{hocolim}_{I_{n-1}^1} \,\,\chi \rightarrow \mbox{hocolim}_{I^n} \,\,\chi $$ Here $I_{n-1}^1$ is $I_{n-1}$ minus its final object, which is the set $\{1, \ldots, n-1\}$. There is a similar decomposition of colim $\chi$. It is isomorphic to the strict pushout $$\mbox{colim}_{I_{n-1}}\,\, \chi\leftarrow \mbox{colim}_{I_{n-1}^1} \,\,\chi \rightarrow \mbox{colim}_{I^n} \,\,\chi $$ Our assumptions guarantee that the left map in this pushout is a cofibration, so pushout=homotopy pushout. The assumptions also guarantee that the natural map from the first pushout diagram to the second is a pointwise equivalence, so it induces an equivalence of homotopy pushouts. So it induces an equivalence hocolim $\chi \to$ colim $\chi$.

Note that $I^n$ is isomorphic to $I_{n-1}$, so by induction you can replace the hypothesis that $\chi$ restricted to ${I^n}$ is cofibrant with a weaker one, so long as $n-1>1$.

The case $n=2$ is equivalent to your example.

Another example: Let $G$ be a finite group (probably can be a more general class of groups - how general?) acting on a pointed simplicial set $X$. You can think of this action as defining a diagram of pointed simplicial sets. The colimit of this diagram is the orbit space $X/G$. The homotopy colimit is the pointed homotopy orbit space $X\wedge_G EG_+$. $X$ is cofibrant if the action of $G$ is free except for the basepoint. However, the weaker assumption that $X^H$ is contractible for all non-trivial subgroups $H\subset G$ suffices to conclude that the map from the homotopy orbits space to the strict orbits space is an equivalence.

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    $\begingroup$ Hi Greg, perhaps you mean $X^H$ is contractible for all non-trivial subgroups $H\subset G$ ($G_+$ is cofibrant, but is not contractible for all proper subgroups unless $G$ is trivial). $\endgroup$ Apr 7, 2016 at 6:57
  • $\begingroup$ @JustinNoel Oops, thanks. Corrected. $\endgroup$ Apr 7, 2016 at 6:59

Another proof of the first fact of @KarolSzumiło's post (in the category of simplicial sets, all filtered colimits are homotopy colimits) can be found in section 14.2 of "Notes on homotopy colimits and homotopy limits (a work in progress)" by Hirschhorn.


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