# Are the unwound thin realization and fat realization homotopy equivalent?

This is a question about a theorem (proposition 2) in the article--On the homotopy type of classifying spaces

Recall some definitions first:

Given a category $\mathcal{C}$ internal in $\operatorname{Top}$, the unwound thin realization is defined by

$$\vert N_{\cdot}\mathcal{C}_{\mathbb{N}}\vert,$$

where $\mathcal{C}_{\mathbb{N}}$ is the subcategory of the product category $\mathcal{C}\times \mathbb{N}$ obtained by deleting all the morphisms $(f,i\leq i)$ with $f$ not an identity (this is defined, I think, first by Segal) and $N_{\cdot}$ stands for the nerve construction.

This is equivalent to the fat realization $\vert\vert-\vert\vert$ of the following product simplicial space $$N_{\cdot}\mathcal{C}\times S_{\cdot}$$ where $S_{\cdot}$ is the semi-simplicial set
$$\triangle_{in}\rightarrow \mathbb{N}$$
sending $[k]$ to $Hom([k],\mathbb{N})$, the set of all injective increasing maps.

Now it is proved in On the homotopy type of classifying spaces (proposition 2) by tom Dieck that the canonical projection

$$\pi:\vert\vert N_{\cdot}\mathcal{C}\times S_{\cdot}\vert\vert\rightarrow \vert\vert N_{\cdot}\mathcal{C}\vert\vert$$

is a homotopy equivalence. Actually his proposition is more general--for any simplicial space $X_{\cdot}$.

In his proof, he constructed a map $$\rho:\vert\vert X_{\cdot} \vert\vert\rightarrow\vert\vert X_{\cdot} \times S_{\cdot}\vert\vert$$ and then showed it is the homotopy inverse of $\pi$.

There $\rho$ is defined by
$$\rho(x)=[y,i_{n},s_{0}(x),...,s_{n}(x)]$$

where we assume $x$ is represented by $$(y,t_{0},...,t_{n})\in X_{n}\times \triangle^{n},$$ $i_{n}(k)=k+1$, and $s_{i}$ is the partition of unity constructed by considering the composition $$X_{n}\times \triangle^{n}\rightarrow \triangle^{n}\xrightarrow{\sigma_{j,n}} [0,1],$$ where $$\sigma_{j,n}\equiv \sum_{E\subset [n],card(E)=j+1} \operatorname{Max}(0,\operatorname*{Min}_{j\in E}t_{j}-\operatorname*{Max}_{j\in [n]\setminus E}t_{j}),$$ for all $j\leq n$. Since $\sigma_{j,n}$ are compatible for all $n$, it defines a map $$\sigma_{j}:\vert\vert X_{\cdot}\vert\vert\rightarrow [0,1]$$ and $s_{j}\equiv (j+1)\sigma_{j}$

Now my questions are:

1. Is the map $$\rho:\vert\vert X_{\cdot}\times S_{\cdot}\vert\vert\rightarrow \vert\vert X_{\cdot}\vert\vert$$ well-defined?

For example, if $t_{k}=0$, then $x$ might also be represented by $(d_{k}y,t_{0},...,t_{n-1})$, whereas $(d_{k}y,i_{n-1},s_{0}(x),...,s_{n}(x))$ doesn't always represent the same element as $(y,i_{n},s_{0}(x),...,s_{n}(x))$.

Or do I actually misunderstand what he means there?

1. What actually concerns me is the case when $X_{\cdot}$ is $N_{\cdot}\mathcal{C}$.

Is there an easier way to see $$\vert\vert N_{\cdot}\mathcal{C}\times S_{\cdot}\vert\vert$$ and $$\vert \vert N_{\cdot}\mathcal{C}\vert\vert$$ are actually homotopy equivalent?

1. Or if not, is there any counterexample?

Any suggestion or comment concerning this question will be very much appreciated.

Thank you so much.

• I think you are right and $\rho$ is not well-defined. However, I think that one can construct a well-defined map by mapping one simplex of the first barycentric subdivision of each $n$-simplex $y$ of $\|X\|$ (say, the one with $t_0\ge\cdots\ge t_n$) as specified, and the remainder onto pairs of a degenerated face of $y$ and $\Delta[n]$. For example, the first half of a one-simplex $y$ is mapped as specified, then the midpoint ends up at $d_0y\times\{1\}$. The second half has to be mapped "backwards" to $s_0d_0y\times i[1]$, so $p$ maps it constantly to $d_0y$. The formula will be complicated. – Sebastian Goette May 26 '16 at 13:45
• @Sebastian: Thank you very much. I think that is the map I want to have, it is indeed a bit hard to write down in detail through. After working out the detail, I will post it! Thanks again. – yisheng May 28 '16 at 10:36