I am reading Prof. McCarthy's paper proving additivity (1992). In this paper he proves a version of Quillen's theorem A for simplicial sets (this is the unique result labeled as proposition of the paper for those interested). There is a step which I am failing to make more explicit. Let $D$ be a category with a $0$ object, a specified class of morphisms called cofibrations (which are the morphisms of interests, which we denote by $\rightarrowtail$) such that every cofibration admits a cokernel (which we write as a quotient). We consider a bisimplicial $X$ set whose $(m,n)$ simplicies are diagrams in $D$ of the form
$$0=D_0\rightarrowtail D_1\rightarrowtail \cdots\rightarrowtail D_m\rightarrowtail E_0\rightarrowtail \cdots \rightarrowtail E_n$$
and we map this to to
$$0=E_0/E_0\rightarrowtail E_1/E_0 \rightarrowtail\cdots \rightarrowtail E_n/E_0.$$
We view this as a map $X\to N_\bullet(D)R$ where the codomain is the bisimplicial set which is constant in the first coordinate equal to the nerve of the category $D$. I want to show this map is a homotopy equivalence.
Prof. McCarthy states that it suffices to fix a simplicial direction, and then observe it is a homotopy equivalence by noticing this yields a simplicial map between nerves of contractible categories and using the realization lemma, I imagine we fix the $m$ direction. I believe that doing this yields a map between the nerves of $\bigsqcup_{0=D_0\rightarrowtail D_1\rightarrowtail\cdots \rightarrowtail D_m} D_m/D$ and $D$. Although $D$ is contractible as there is a $0$ object, I really don't see what trick shows that the domain category is contractible, rather than a disjoint union of contractible categories.
Any help is appreciated.
McCarthy's argument is elaborated in a later paper of his, The cyclic homology of an exact category (1994) - see lemma 3.4.4. Briefly, to show that the bisimplicial map $\rho: S_\bullet \mathrm{id}_\mathscr{D} \mid \mathscr{D} \to S_\bullet \mathscr{D} R$ is an equivalence, it suffices by the realization theorem to show that $\rho(-, [n]): (S_\bullet \mathrm{id}_\mathscr{D} \mid \mathscr{D})(-, [n]) \to S_\bullet \mathscr{D}R(-, [n])$ is an equivalence for each $n$.
The codomain of this map is a constant simplicial set $S_n \mathscr{D}$. Consider the map $\nu: S_n \mathscr{D} \to (S_\bullet \mathrm{id}_\mathscr{D})(-, [n])$ that sends an element $(0 = F_0 \rightarrowtail F_1 \rightarrowtail \cdots \rightarrowtail F_n)$ to $(\underbrace{0 \rightarrowtail 0 \rightarrowtail \cdots \rightarrowtail 0}_{m \text{ zeroes}} \rightarrowtail F_0 \rightarrowtail F_1 \rightarrowtail \cdots \rightarrowtail F_n)$. We claim that $\nu$ is a homotopy inverse for $\rho(-, [n])$. One direction is easy: we see that $\rho(-, [n]) \circ \nu = \mathrm{id}_{S_n \mathscr{D}}$.
The other composition $\nu \circ \rho(-, [n])$ sends $(0 = D_0 \rightarrowtail D_1 \rightarrowtail \cdots \rightarrowtail D_m \rightarrowtail E_0 \rightarrowtail E_1 \rightarrowtail \cdots \rightarrowtail E_n)$ to $(\underbrace{0 \rightarrowtail 0 \rightarrowtail \cdots \rightarrowtail 0}_{m \text{ zeroes}} \rightarrowtail E_0/E_0 \rightarrowtail E_1/E_0 \rightarrowtail \cdots \rightarrowtail E_n/E_0)$. We want to show that this composition is homotopic to the identity. For $0 \leq i \leq m$, define a map $h_i: (S_\bullet \mathrm{id}_\mathscr{D} \mid \mathscr{D})([m], [n]) \to (S_\bullet \mathrm{id}_\mathscr{D} \mid \mathscr{D})([m + 1], [n])$ that sends $(0 = D_0 \rightarrowtail \cdots \rightarrowtail D_m \rightarrowtail E_0 \rightarrowtail \cdots \rightarrowtail E_n)$ to $(0 = D_0 \rightarrowtail \cdots \rightarrowtail D_i \rightarrowtail \underbrace{E_0 \rightarrowtail E_0 \rightarrowtail \cdots \rightarrowtail E_0}_{m - i + 2 \text{ $E_0$'s}} \rightarrowtail E_1 \rightarrowtail \cdots \rightarrowtail E_n)$. Then $h$ assembles to a simplicial homotopy (check!) between $d_0 h_0 = \nu \circ \rho(-, [n])$ and $d_{m+1} h_m = \mathrm{id}_{(S_\bullet \mathrm{id}_\mathscr{D} \mid \mathscr{D})(-, [n])}$.
This proves that $\rho(-, [n])$ is a homotopy equivalence, and hence so is $\rho$.