Explicitly calculating the homotopy fiber of a 3-cocycle in the category of simplicial sets

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?

• 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 $\pi_2(BG)=A$. Nov 25, 2021 at 3:36
• 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 $\tau_0G$. Nov 25, 2021 at 3:39
• 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. Nov 25, 2021 at 4:17