Constructing the classifying space for a groupoid G I am working with the classifying space BG of the groupoid G. One definition is as follows:
$$
BG = \bigsqcup _n (G_n \times \Delta ^n) /(d_i(g),x) \sim (g, \delta _i (x)).
$$
Where the $d_i$ are face operators and the $\delta _i$ linear embeddings (I hope that's specific enough).
I am looking at another paper which defines it as
$$
BG = \bigsqcup _n (G_n \times \Delta ^n) /(\alpha ^*(g),x) \sim (g, \alpha _* (x)).
$$
for $g\in G_n, x \in \Delta ^m$ and injective monotone functions  $\alpha : [m] \rightarrow[n]$.
First I'm not entirely sure what theses functions really are, is it literally $\alpha (m) = n$ and that is all there is to it? And I'm struggling to see what the induced maps $\alpha ^*$ and $\alpha _*$ are. I imagine this is essentially the same thing as above but wanted to be sure I understood it.
 A: The difference between these definitions boils down to two different descriptions of the category of simplices, and both yield the same $BG$. The first description of $BG$ "thinks about" simplices as topological simplices, especially with language like "linear embedding".
For the notation $\alpha:[m]\to [n]$ in the second description, see simplex category. The objects are the sets $[n]:=\{0,1,\ldots,n\}$ (corresponding to the standard $n$-simplex) and morphisms are monotone functions. Any injective morphism $[m]\to[n]$ can be written as a composition of the face maps $\delta_i:[n-1]\to [n]$ (in this context, the monotone map whose image is everything except $i$).
In the first description, you are only explicitly writing the equivalences due to the face maps, which imply the equivalences for all injective morphisms (i.e. linear embeddings from any subsimplex into the standard simplex, not just a codimension $1$ face) by transitivity. In the second description, you are explicitly listing equivalences for all injective morphisms.
With this, the meanings of $\alpha^\ast$ and $\alpha_\ast$ are the same as $d_i$ and $\delta_i$, respectively, when you are using face maps, and the analogous definitions when you are using more general maps. So if $\alpha:[1]\to[3]$ maps $0 \to 0$ and $1 \to 3$, then $\alpha^\ast(g_0,g_1,g_2,g_3)=(g_0,g_3)$.
