### qlkjfdlqkj

parent 3e739f45
 ... ... @@ -907,23 +907,6 @@ The previous proposition admits the following corollary, which will be of great \] This functor is fully faithful and $\Ch^{\leq n}$ may be identified with the full subcategory of $\Ch$ spanned by chain complexes $K$ such that $K_k = 0$ for every $k >n$. Notice now that for an $n$\nbd-category $C$, the chain complex $\lambda(\iota_n(C))$ is such that $\lambda(\iota_n(C))_k=0$ for every $k > n$ and hence $\lambda(\iota_n(C))$ can be seen as an object of $\Ch^{\leq n}$. Thus, we can define a functor $\lambda_{\leq n } : n\Cat \to \Ch^{\leq n}$ as \begin{align*} \lambda_{\leq n} : n\Cat &\to \Ch^{\leq n}\\ C&\mapsto \lambda(\iota_n(C)). \end{align*} %% this means that there exists a unique functor $\lambda_{\leq n} : n\Cat \to \Ch^{\leq n}$ such that the following square is commutative %% $%% \begin{tikzcd} %% n\Cat \ar[d] \ar[r,"\lambda_{\leq n}"] & \Ch^{\leq n} \ar[d] \\ %% \oo\Cat \ar[r,"\lambda"] & \Ch. %% \end{tikzcd} %%$ %% This functor being Recall from \ref{paragr:defncat} that for every $n \geq 0$, the canonical inclusion $\iota_n : n\Cat \to \oo\Cat$ has a left adjoint $\tau^{i}_{\leq n} : \oo\Cat \to n\Cat$, where for an $\oo$\nbd-category $C$, $\tau_{\leq n }^{i}(C)$ is the $n$\nbd-category whose set of $k$\nbd-cells is $C_k$ for $k n$ and hence $\lambda(\iota_n(C))$ can be seen as an object of $\Ch^{\leq n}$. Thus, we can define a functor $\lambda_{\leq n } : n\Cat \to \Ch^{\leq n}$ as \begin{align*} \lambda_{\leq n} : n\Cat &\to \Ch^{\leq n}\\ C&\mapsto \lambda(\iota_n(C)). \end{align*} We tautologically have that the square $\begin{tikzcd} n\Cat \ar[d,"\iota_n"] \ar[r,"\lambda_{\leq n}"] & \Ch^{\leq n} \ar[d,"\iota_n"] \\ \oo\Cat \ar[r,"\lambda"] & \Ch \end{tikzcd}$ is commutative. \end{paragr} \begin{lemma} The square \[ ... ... @@ -965,7 +975,7 @@ The previous proposition admits the following corollary, which will be of great is commutative (up to a canonical isomorphism). \end{lemma} \begin{proof} \todo{À écrire} Let $C$ be an $\oo$\nbd-category. For \$k
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