Commit 927a3c8f authored by Leonard Guetta's avatar Leonard Guetta
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parent 4b99831d
......@@ -946,7 +946,7 @@ The previous proposition admits the following corollary, which will be of great
The isomorphism being obviously induced by the unit map $K \to \iota_n\tau^{i}_{\leq n}(K)$.
Notice now that for an $n$\nbd-category $C$, the chain complex $\lambda(\iota_n(C))$ is such that
Let $C$ be $n$\nbd-category $C$. An straigtforward computation shows that the chain complex $\lambda(\iota_n(C))$ is such that
......@@ -975,7 +975,61 @@ is commutative.
is commutative (up to a canonical isomorphism).
Let $C$ be an $\oo$\nbd-category. For $k<n$, we obviously have that
Notice first that we have a natural transformation
\beta : \tau^{i}_{\leq n}\circ \lambda \Rightarrow \lambda_{\leq n} \circ \tau^{i}_{\leq n}
defined as
\oo\Cat \ar[r,"\tau_{\leq n}^{i}"] \ar[rd,"\mathrm{id}"',""{name=A,right}] & n\Cat \ar[d,"\iota_n"] \ar[r,"\lambda_{\leq n}"] & \Ch^{\leq n} \ar[d,"\iota_n"'] \ar[dr,"\mathrm{id}",""{name=B,left}] & \\
&\oo\Cat \ar[r,"\lambda"] & \Ch \ar[r,"\tau^{i}_{\leq n}"'] & \Ch^{\leq n}
\ar[from=A, to=1-2,Rightarrow,"\eta"]
Since for every $\oo$\nbd-category $C$ and every $k <n$, we have
C_k=\iota_n(\tau^{i}_{\leq n}(C))_k
and for every chain complex $K$ and every $k<n$, we have
\iota_n(\tau^{i}_{\leq n}(K))_k=K_k,
it follows that for every $\oo$\nbd-category $C$ and every $k<n$, the morphism $\beta_k$ is nothing but the equality
\tau^{i}_{\leq n}(\lambda(C))_k=\lambda_{\leq n}(\tau^{i}_{\leq n}(C))_k.
Hence, all we have to prove is that
\beta_n : \tau^{i}_{\leq n}(\lambda(C))_n \to \lambda_{\leq n}(\tau^{i}_{\leq n}(C))_n
for every $\oo$\nbd-category $C$.
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
\Hom_{\Ab}(\lambda_n\iota_n\tau^{i}_{\leq n}(C),G) \simeq \Hom_{n\Cat}(\tau^{i}_{\leq n}(C),B^nG) \simeq \Hom_{\oo\Cat}(C,\iota_n(B^nG)).
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:
\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
By definition of $\lambda_n(C)$ and of $\partial : \lambda_{n+1}(C) \to \lambda_n(C)$, we thus have
\Hom_{\Ab}(\lambda_n\iota_n\tau^{i}_{\leq n}(C),G)\simeq \Hom_{\Ab}(\lambda_n(C)/{\partial(\lambda_{n+1}(C))},G).
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
%% \[
%% \tau^{i}_{\leq n}(
%% \]
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 $H^{\pol}_k(C)$.
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