I am stuck at a critical step in my master's thesis. If someome can help me out here, I will give appropriate credit.

We know that the data of a 2-group $G$ can be given by a group $\tau_0(G)$ and a 3-cocycle $\alpha:\tau_0(G)^3\rightarrow A$ into some abelian group $A$, which serves as the associator of $G$. I would like to link this result to the more general internal theory of the topos of $\infty$-groupoids $Grpd_\infty$ by showing that the homotopy fiber of $\alpha$ as a morphism $\tau_0(G)\rightarrow B^3 A$ in $Grpd_\infty$ is homotopy-equivalent to $B G$. For this I wanted to explicitly calculate the pullback of $\alpha$ in $Set^{\Delta^{op}}$ along the decalage morphism of the image of the chain complex $A[2]$ under the Dold-Kan-correspondence, $w:W(DK(A[2]))\rightarrow \bar{W}(DK(A[2]))$. As can be read on the nlab page for simplicial classifying spaces, the pullback of $\alpha$ along $w$ should represent the homotopy fiber of $\alpha$. However, I seem to be getting a wrong result, so please read critically so you might find hidden mistakes.

I proceeded to calculate this pullback step by step, starting with the Dold-Kan image of $A[2]$, using this explicit formula. $A[2]$ is concentrated in all degrees except $A[2]_2=A$, so the formula can be specialized to $DK(A[2])_{n}=\bigoplus_{[n]\rightarrow [2]_{surj}} A$. So $DK(A[2])$ is concentrated in degrees $0$ and $1$, consists of one copy of $A$ in degree $[2]$, and and three copies $A_0$, $A_1$, $A_2$ in degree $3$, corresponding to the three degeneracy maps $\sigma_{i}, i=0,1,2$ between degree 3 and 2 mapping $i$ and $i+1$ to the same node $i$. Following the formula further, we see that the face maps of $DK(A[2])$ simplify on each factor $A_j$ of the product $DK(A[2])_3$

$$\delta_i|_{A_j}\rightarrow A=\begin{cases} id & \text{if } i=j, j+1 \\ * & \text{otherwise} \end{cases}$$

and can be extended to $DK(A[2])_3$ by $\delta_i(a,b,c)=\delta_i|_{A_0}(a)+\delta_i|_{A_1}(b)+\delta_i|_{A_2}(c)$.

Next I calculated $W(DK(A[2]))$ using again the procedure given on the nlab page of simplicial classifying spaces. Since degrees 0 and 1 of $DK(A[2])$ are trivial, the general formula simplifies to $W(DK(A[2]))_{2}\cong DK(A[2])_2=A$ and $W(DK(A[2]))_3\cong DK(A[2])_3\times DK(A[2])_2\cong A_0\times A_1\times A_2 \times A_3$, where $A_0,A_1,A_2$ come from $DK(A[2])_3$ and $A_3$ comes from $DK(A[2])_2$, for objects of degree 2 and 3. Since the face maps from degree 2 to degree 1 of $DK(A[2])$ are all trivial, the face maps between $\bar{W}(DK(A[2])_3$ and $\bar{W}(DK(A[2]))_2$ are given by $\delta_0(a,b,c,d)=\delta_0(a,b,c)d=ad$, $\delta_1(a,b,c,d)=\delta_1(a,b,c)=ab$, $\delta_2(a,b,c,d)=\delta_2(a,b,c)=bc$ and $\delta_3(a,b,c,d)=\delta_3(a,b,c)=c$.

$\bar{W}(DK(A[2]))$ finally is given in degree 2 by the quotient of $A$ with itself, thus trivial, and in degree 3 by $DK(A[2])_3/ DK(A[2])_3\times DK(A[2])_2$, thus isomorphic to $A$, and the decalage morphism $w$ is (equivalent to) projection onto the fourth factor.

Thus, for a 3-cocycle $\alpha: W \tau_0(G)\rightarrow \bar{W}(DK(A[2]))$, the pullback of $\alpha$ along $w$ consists of one 0-cell, the objects of $\tau_0(G)$ as 1-cells, for each $f,g\in \tau_0(G),a\in A$, 2-cells given by triangles with two sides labeled by $f, g$ and the third by the composition $fg$, and, for $\alpha(f,g,h)=d$ and $a,b,c\in A$, 3-cells with edges labeled by $f, g, h, fg, gh, fgh$ and faces labeled by $\delta_0=ad$, $\delta_1=ab$, $\delta_2=bc$ and $\delta_3=c$.

So this is as far as I got and I'm not really sure how to interpret this result. Did I make a mistake somewhere? Were my presumptions invalid in some way? If not, how do I proceed?

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    $\begingroup$ The thing you're getting in the end sounds more or less like a 2-group, doesn't it? You haven't said anywhere what notion of 2-group you're trying to compare it to, but for your 2-group $G$ the space $BG$ should precisely have $\pi_1(BG)=\tau_0(G)$ and $\pi_2(BG)=A$. $\endgroup$ Nov 25, 2021 at 3:36
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    $\begingroup$ In the post, there are some notational issues, maybe that's the confusion? Your cocycle is a map $B\tau_0G\to B^3 A$, and the fiber should be $BG$. Also, $G$ in the end of the post sounds like it refers to $\tau_0G$. $\endgroup$ Nov 25, 2021 at 3:39
  • $\begingroup$ Ah, sorry, I should have been more clear. The model for 2-groups is the globular. Any 3-cocycle, given as a special function $G^3\rightarrow A$, describes a skeletal 2-group that has it as an associator (as a bicategory), and this gives all 2-groups up to equivalence. What I'm trying is get from the globular picture into the simplicial. So I'm starting with $\alpha$ given as a special function on $G^3$ and would like to show that the $\alpha:B G\rightarrow B^3 A$ represented by it has as fiber the same 2-group. In particular, I have to derive the associator from the simplicial pullback. $\endgroup$ Nov 25, 2021 at 4:17

1 Answer 1


For anyone interested, I've found a solution. It is the central construction in https://arxiv.org/abs/2203.11392.


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