Recall from Lemma \ref{lemma:adjlambdasusp} that $\lambda_n \circ\iota_n : n\Cat\to\Ab$ (which we abusively wrote as $\lambda_n$) is left adjoint to the functor $B^n : \Ab\to n\Cat$. In particular, for every $\oo$\nbd-category $C$ and every abelian group $G$, we have

Then, it follows from an argument similar to the proof of Lemma \ref{lemma:nfunctortomonoid} (see also the proof of Lemma \ref{lemma:adjlambdasusp}) that this last set is naturally isomorphic to the set of functions $f_n : C_n \to G$ such that:

\begin{itemize}[label=-]

\item for every $0\leq k <n $ and every pair $(x,y)$ of $k$-composables $n$\nbd-cells of $C$, we have

\[

f(x \comp_k y)= f(x)+ f(y),

\]

\item for every $\alpha : x \to y$, $(n+1)$\nbd-cell of $C$, we have

\[

f(x)=f(y).

\]

\end{itemize}

By definition of $\lambda_n(C)$ and of $\partial : \lambda_{n+1}(C)\to\lambda_n(C)$, we thus have

Hence, we have $\lambda_{\leq n}(\tau^{i}_{\leq n}(C))_n \simeq\tau^{i}_{\leq n}(\lambda(C))_n$ and a thorough analysis of naturality shows that this isomorphism is nothing but $\beta_n$.

%% Let $C$ be an $\oo$\nbd-category. Since for every $n \geq 0$, the abelian group $\lambda_n(C)$ only depends on $C_n$, we obviously have that

%% \[

%% \tau^{i}_{\leq n}(

%% \]

\end{proof}

With this lemma at hand we can prove the important following proposition which basically says that if an $\oo$\nbd-category $C$ is free up to dimension $n-1$, then for any $k$ such that $0\leq k \leq n$ there is no need to find a cofibrant replacement in order to compute $H^{\pol}_k(C)$.