A small category $\mathcal C$ having $O$ as its set of objects is called free if there exists a set $S$ of non-identity maps in $\mathcal C$ such that every non-identity map in $\mathcal C$ can uniquely be written as a finite composition of maps in $S$.
If such a set $S$ exists, it is clearly unique; its elements are called the generators of $\mathcal C$.
Let $N\mathcal C$ be the nerve of a free category. For every integer $k\geqslant 1$, the $k-$dimensionnal sub-complex $N^k\mathcal C\subset N\mathcal C$ generated by the $k-$simplices $$X_k\rightarrow X_{k-1}\rightarrow \cdots\rightarrow X_0$$ of $N\mathcal C$, for which each of the maps $X_i\rightarrow X_{i-1}$ is either a generator of $\mathcal C$ or an identity map.
Claim: for every integer $k\geqslant 1$, the geometric realization $|N^k\mathcal C|$ is a strong deformation retract of $|N^{k+1}\mathcal C|$.
I have trouble proving this claim. Is there a general argument proving this fact?
The inclusion N^k(C)→N^{k+1}(C) can be represented as a pushout of coproduct of acyclic cofibrations of maps A_{k+1}→Δ^{k+1}, (one map for each nondegenerate n-simplex, in which all maps are generators, none are identities), where A_{k+1} is a subset of Δ^{k+1} obtained by removing those simplices that contain both 0 and k+1 as a vertex (these have more than k free generators inside them, so cannot belong to N^k(C)). It remains to observe that each inclusion A^{k+1}→Δ^{k+1} is an acyclic cofibration (e.g., A^{k+1} is contractible to any inner vertex), which completes the proof.