# Three questions on $\operatorname{hocolim}$

I posted this on math.stack.exchange but didn't get a helpful response, so please let me try it here.

Let $D$ be a small category and $F:D\to sSets$ a functor.

There is a bisimplicial set indicated by $$...\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0)$$ which i like to call $sres(F)_{\bullet\bullet}$. By considering either the horizontal or the vertical index first, I get two functors $H:\Delta^{op}\to sSets$ and $V:\Delta^{op}\to sSets$ (the ''vertical'' direction is the one of the simplicial sets $F(d)$, i.e. each object in the diagram above should be imagined as a vertical column).

The homotopy colimit $\operatorname{hocolim}F$ of $F$ can be defined in different ways (up to weak equivalence) and one way is $(\operatorname{hocolim}F)_n=sres(F)_{n n}$, the diagonal of $sres(F)_{\bullet\bullet}$.

Is there a weak equivalence between between $\operatorname{hocolim}F$ and $\operatorname{hocolim}H$? (I don't think there is a weak equivalence between $\operatorname{hocolim}F$ and $\operatorname{hocolim}V$, or is it?)

Let me call a simplicial set $A$ $k$-dimensional, if it is equal to its $k$-skeleton $\mathbf{sk}_k$ (If I am not mistaken, $k$ should be the smallest integer such that all elements of $A_{k+1}$ are degenerated simplices).

Suppose that the nerve $N(D)$ of $D$ is weakly equivalent to a simplicial set $A$ of dimension $k$. Is it true that $\operatorname{hocolim}F$ is weak equivalent to the homotopy colimit of the diagram $$\coprod_{d_{k+1}\to ...\to d_0}F(d_{k+1})\begin{array}{c}\to \\ \vdots\\\to\end{array}...\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0),$$ (the truncated $H$) or in other words, can I ''stop'' at the $k+1$th stage of the diagram in the horizontal direction to calculate the homotopy colimit? If yes, why?

In particular, if $D$ is linear (for example $\mathbb{N}$ or $\cdot\to\cdot\to\cdot$), this would mean, that $\operatorname{hocolim}F$ is the homotopy colimit of the simple diagram $$\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0).$$

My third question is a little vague and like the second one but not ''for the source'' of $F$ but ''for the target''. Please don't hesitate to post an answer only to the first two questions, if this third one is not well formulated. I wondered, why one takes only two-limits of stacks and not $k$-limits. The nerve of a category is a 2-coskeletal simplicial set and this is where the reason comes from, I guess.

Suppose the functor $F$ factorize through the nerve $F:D\to CAT\xrightarrow{N} sSets$. Is it true that $\operatorname{hocolim}F$ is weakly equivalent to the homotopy colimit of the diagram $$\coprod_{d_2\to d_1\to d_0}F(d_{2})\begin{array}{c}\to \\ \to\\\to\end{array}\coprod_{d_1\to d_0}F(d_1)\begin{array}{c}\to\\\to\end{array}\coprod_{d_0}F(d_0)$$ and what is the reason? If not, what did I misunderstand here?

• Now that MSE and MO are on the same networks you can flag for a moderator attention on MSE and request the question to be migrated here. – Asaf Karagila Jun 28 '13 at 1:07
• 2-limits have to do with 2-category theory and are related to homotopy limits in certain cases, but they are not the same thing. For instance, 2-limits can be defined uniquely up to isomorphism. – Zhen Lin Jun 28 '13 at 4:01

First question$\newcommand{\op}{{#1}^{\mathrm{op}}}$$\newcommand{\sSet}{\mathrm{sSet}}$$\newcommand{\Grpd}{\mathrm{Grpd}}$$\newcommand{\Cat}{\mathrm{Cat}}$$\newcommand{\NN}{\mathbb{N}}$$\newcommand{\sres}{\mathrm{sres}}$$\newcommand{\hocolim}{\operatorname{hocolim}}$$\newcommand{\diag}{\operatorname{diag}}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\real}{\lvert #1 \rvert} The homotopy colimit of a functor X:\op{\Delta}\to\sSet is equivalent to its geometric realization \real{X}. This holds because every functor \op{\Delta}\to\sSet is cofibrant in the Reedy model structure. In particular, geometric realization of simplicial simplicial sets preserves objectwise weak equivalences. You may consult corollary 15.8.8 and theorem 18.7.4 in Hirschhorn's book "Model categories and their localizations" for these results. The equivalence between the homotopy colimit and the geometric realization is given explicitly by the Bousfield–Kan map (in section 18.7 of Hirschhorn's book). On the other hand, it is a very interesting well-known fact that the geometric realization of a simplicial simplicial set X:\op{\Delta}\to\sSet is canonically isomorphic to the diagonal \diag X of the associated bisimplicial set. You will find this as theorem 15.11.6 of Hirschhorn's book, or as exercise IV.1.4 in the book "Simplicial homotopy theory*" by Goerss and Jardine. [Further, if you are very categorically inclined, you may see this quite formally as a consequence of geometric realization being a coend whose weight is the Yoneda embedding.] From the preceding paragraphs, we conclude that there is a natural weak equivalence \hocolim X\overset{\sim}{\To}\diag X for every simplicial simplicial set X. This is stated explicitly as corollary 18.7.7 of Hirschhorn's book. What I said so far is in part briefly described in section XII.3.4 of the book "Homotopy limits, completions and localizations" by Bousfield and Kan. Answer to first question: We conclude from the previous discussion that the homotopy colimits of both H and V are naturally equivalent to their diagonals. Each of these diagonals is by definition also the diagonal of \sres(F), i.e. your proposed definition for the homotopy colimit of F. [As a consistency check, the definition of homotopy colimit of a diagram F:D\to\sSet given in both Bousfield–Kan and Hirschhorn is isomorphic to your definition as the diagonal of \sres(F). You can see a brief explanation of this fact in section 4.13 of Dan Dugger's notes "A primer on homotopy colimits". These are very readable notes, and I highly recommend them. You can also read about the Bousfield–Kan map in section 17.2 of those notes.] Second question The answer to your second question is no. Take the very simple category D described schematically as 0\overset{a}{\to} 1\overset{b}{\to} 2. Then the category D has a terminal object so its nerve is contractible, and we may take k=0 in your second question. Consider the terminal functor F=\ast : D\to \sSet whose value is always the terminal simplicial set with a single vertex, and no non-degenerate higher dimensional simplices. The homotopy colimit of this diagram is equivalent to the nerve of D, and thus contractible. On the other hand, the 1-truncated simplicial object you write in your question is levelwise discrete, so we can think of it as a simplicial set with: • three vertices 0, 1, and 2 • three non-degenerate edges: a from 0 to 1, b from 1 to 2, and ab from 0 to 2. Non-coincidentally, this is just the 1-skeleton of the nerve of D. It is quite easy to see that its realization is a circle. Thus the homotopy colimit of your 1-truncated simplicial object is not contractible. The problem is of course that we "erased" the non-degenerate 2-simplex from the nerve of D, thus changing its homotopy type. On a positive note, I believe a formula like the one you state does hold. Under the stronger assumption that the nerve of D is k-skeletal, then the homotopy colimit of F:D\to\sSet is equivalent to the homotopy colimit of a k-truncated simplicial object like the one you write (except you write a (k+1)-truncated simplicial object instead). [By a k-truncated simplicial object, I mean a functor indexed by the full subcategory of \op{\Delta} spanned by the n-simplices with n\leq k.] Sketch of proof: If the nerve of D is k-skeletal, \sres(F) is also (horizontally) k-skeletal. Consequently, the diagonal of \sres(F) coincides with the geometric realization of the horizontal k-truncation of \sres(F) (in your question you actually write down the (k+1)-truncation). This is equivalent to the homotopy colimit of the k-truncation for the same reasons as before: every k-truncated simplicial object in simplicial sets is Reedy cofibrant (which actually follows from the analogous result for simplicial simplicial sets). Intermezzo I will make a couple of simple comments regarding your examples in between the second and third questions. The homotopy colimit of any functor F indexed by a category D with a terminal object d is weakly equivalent to F(d). So the homotopy colimit of any functor F indexed by 0\to 1\to 2 is weakly equivalent to F(2). Moreover, any filtered colimit in simplicial sets is actually a homotopy colimit, essentially because filtered colimits commute with homotopy groups in \sSet. In other words, homotopy colimits in \sSet along filtered categories (such as \NN) are no harder than ordinary colimits. Third question For this question, the relevant part about stacks is not that they have values in categories. It is that they have values in groupoids. Moreover, one needs to take the homotopy colimit in groupoids and not in simplicial sets. I will elaborate below. First, here is a counter-example to your third question as it stands. Consider any category C whose nerve is equivalent to S^3, for example. [Any simplicial set is weakly equivalent to the nerve of some category, for example its category of simplices. Here, however, is an explicit example for S^3: simply take the join of categories C = 2\ast 2 \ast 2 \ast 2, where 2 denotes a discrete category with two objects.] Then the homotopy colimit along C of the nerve of the constant terminal functor \ast:C\to\Cat is simply$$ \hocolim N(\ast) = \hocolim \ast \simeq BC \simeq S^3$$However, the homotopy colimit of the corresponding$2$-truncated simplicial set you write will be equivalent to its geometric realization, which is$2$-dimensional. In particular, it cannot have non-trivial homology in dimension$3$, and cannot be equivalent to$S^3$. What is true is that the homotopy colimit in groupoids (with respect to the natural model structure) of a functor$F:D\to\Grpd$(from$D$to the category of groupoids) is equivalent to the homotopy colimit in groupoids of the$2$-truncated simplicial object you write down. If you want to further compose$F:D\to\Grpd$with the nerve functor, then you will have to apply the fundamental groupoid functor to the homotopy colimit of$N\circ F$in simplicial sets to recover the homotopy colimit of$F$in groupoids. This holds because the fundamental groupoid functor is part of a Quillen adjunction between simplicial sets and groupoids. The same holds for the homotopy colimit of the$2$-truncated simplicial object you write down. • Dear @Ricardo Andrade, thank you for the very helpful answer! In the proof of the ''positive note'', what means$k$-skeletal for a bisimplicial set like$sref(F)$? Does it mean ''horizontally$k$-skeletal''? – Ronald Bernard Jun 28 '13 at 10:32 • Good catch! I was looking at the way you wrote$sres(F)$, which looks like a simplicial object in simplicial sets. I meant$k$-skeletal in the first direction, which I guess is horizontally. I will edit the answer. Finally, you are welcome. I am glad to help! – Ricardo Andrade Jun 28 '13 at 10:47 • Dear @Ricardo Andrade, concerning the ''lemma'' in the second answer and third answer: Is the nerve of a groupoid$3$-skeletal? – Ronald Bernard Jun 28 '13 at 19:00 • @Ronald: The nerve of any category with non-identity isomorphisms has non-degenerate simplices in every dimension. For example, the nerve of a groupoid with two objects and an isomorphism between them is infinite dimensional, and thus not$k$-skeletal for any$k$. When I have the time, I may try to add some details explaining why homotopy colimits in groupoids can be computed by a 2-truncated simplicial object (but I am afraid it may take me a while). Finally, what do you mean by the lemma in the second answer? – Ricardo Andrade Jun 28 '13 at 19:43 • @Ronald: The reason is indeed different. Briefly, it happens because the inclusion into$\Delta$of its full subcategory spanned by the$k$-simplices for$k\leq 2\$ is "homotopy 1-final", in the sense that the relevant over-categories have simply connected nerve (homotopy final would mean instead that the over-categories have contractible nerve). I am afraid I will not be able to add details concerning this stuff to my answer over the next few days, but I will do so when I have some time. – Ricardo Andrade Jun 28 '13 at 23:37