In Segal's paper "Classifying Spaces and Spectral Sequences" he claims that Milnor's join construction for the classifying space of a topological group is homeomorphic to taking the geometric realization of the nerve of the category $G_\mathbb{N}$. $G_\mathbb{N}$ is the category obtained by taking the product category of $\mathbb{N}$ (as an ordered set) with $G$ (treated as a category with one object and a morphism for every element) and removing all morphisms $(n,*)\rightarrow (n,*)$ except the identity morphism. I'm trying to construct a map between Milnor's model $\mathbb{B}G$ and $BG_\mathbb{N}$.
I know elements in $\mathbb{B}G$ are of the form $[t_1g_1\oplus t_2g_2\oplus \cdots]$ where each $t_i\in [0,1]$, $\sum_i t_i=1$, all but finitely many of the $t_i$ are $0$, and the defining equivalence relation says $[t_1g_1\oplus t_2g_2\oplus \cdots]$=$[s_1h_1\oplus s_2h_2\oplus \cdots]$ if $t_i=s_i$ for all $i$ and there is some $k\in G$ so that $h_i=g_i k$. Elements of $BG_\mathbb{N}$ are of the form $((n_1,*)\xrightarrow{g_1}(n_2,*)\xrightarrow{g_2}(n_3,*)\xrightarrow{g_3}\cdots\xrightarrow{g_{k-1}}(n_k,*),t\in\Delta^k)$ where $\Delta^k$ is the standard k-simplex, under the geometric realization equivalence relations given in Milnor's "Geometric Realization of a Semi-Simplicial Complex". My intuition is to map $[t_1g_1\oplus t_2g_2\oplus \cdots]$ to $((n_1,*)\xrightarrow{g_{n_1}}(n_2,*)\xrightarrow{g_{n_2}}(n_3,*)\xrightarrow{g_{n_3}}\cdots\xrightarrow{g_{n_{k-1}}}(n_k,*),(t_{n_1},\dots,t_{n_k})\in\Delta^k)$ where the $t_{n_i}$ are the non-zero values. But then it seems that this map would not be well defined on the equivalence relation for $\mathbb{B}G$.
Any ideas on how to fix this? A similar map seemed to work when showing Milnor's $EG$ is isomorphic to $B\bar{G}_{\mathbb{N}}$ as described in Segal.