Consider the collection of symmetric groups $\{\Sigma\_n\}\_{n\geq1}$ as a semi-simplicial set (i.e. a simplicial set without degeneracies) as follows. Consider $i\in\{1,\dots,n+1\}$ and $\pi\in\Sigma\_{n+1}$ represented as a sequence $(\pi(1),\dots,\pi(n+1))$, then
$$d\_{i-1}(\pi)=(\pi(1)-\epsilon\_1,\dots, \widehat{\pi(i)},\dots,\pi(n+1)-\epsilon\_{n+1})$$ 
where for every $j\in\{1,\dots,n+1\}$
$$\epsilon\_j=\begin{cases} 0 & if\ \ \pi(j)<\pi(i)\\ 1 & if\ \ \pi(j)>\pi(i).
\end{cases}$$ 



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> I am wondering what it's known about the homotopy type of the geometric realization of this semi-simplicial set?
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Homotopy type of the semi-simplicial set of symmetric groups
It is contractible. To see this first observe that it is simply connected. It has 2 arcs $a = (1,2)$ and $b = (2,1)$, however the tirangle $(3,1,2)$ gives us the relation that $ab = b$ and symmetrically $ba = a$ so $\pi\_1 = \{1\}$ for this space. To see that the homology vanishes you can use the homotopy operator on the homology complex given by $H((i\_1,...,i\_{n+1})) = (i\_1,...,i\_{n+1},n+2)$. 
It is pretty clear that $dH \pm Hd = Id$ in this case, just note that all the first face maps just turn formally $n+2$ into $n+1$ (giving precisely the Hd term) while the last face map eliminate the $n+2$-st term, giving back the original chain. 


So, this simplicial complex is a simply connected acyclic space, hence contractible. 


