Commit 97f10611 authored by Leonard Guetta's avatar Leonard Guetta
Browse files

Beaucoup de typos corrigés. Il faudrait relire la preuve du Lemme 4.6.5

parent 02a5625a
This diff is collapsed.
No preview for this file type
...@@ -724,7 +724,7 @@ We can now prove the following proposition, which is the key result of this sect ...@@ -724,7 +724,7 @@ We can now prove the following proposition, which is the key result of this sect
Conversely, if $E$ is an $(n+1)$\nbd{}base of $C$, then we can define an $(n+1)$\nbd{}functor $C \to \E_E^*$ that sends $E$, seen as a subset of $C_{n+1}$, to $E$, seen as a subset of $(\E^*_E)_{n+1}$ (and which is obviously the identity on cells of dimension strictly lower than $n+1$). The fact that $C$ and $\E^*$ have $E$ as an $(n+1)$\nbd{}base implies that this $(n+1)$\nbd{}functor $C \to \E^*$ is the inverse of the canonical one $\E^* \to C$. Conversely, if $E$ is an $(n+1)$\nbd{}base of $C$, then we can define an $(n+1)$\nbd{}functor $C \to \E_E^*$ that sends $E$, seen as a subset of $C_{n+1}$, to $E$, seen as a subset of $(\E^*_E)_{n+1}$ (and which is obviously the identity on cells of dimension strictly lower than $n+1$). The fact that $C$ and $\E^*$ have $E$ as an $(n+1)$\nbd{}base implies that this $(n+1)$\nbd{}functor $C \to \E^*$ is the inverse of the canonical one $\E^* \to C$.
\end{proof} \end{proof}
\begin{paragr} \begin{paragr}\label{paragr:cextlowdimension}
We extend the definitions and the results from \ref{def:cellularextension} to We extend the definitions and the results from \ref{def:cellularextension} to
\ref{prop:criterionnbasis} to the case $n=-1$ by saying that a $(-1)$-cellular \ref{prop:criterionnbasis} to the case $n=-1$ by saying that a $(-1)$-cellular
extension is simply a set $\Sigma$ (which is the set of indeterminates) and $(-1)\Cat^+$ is the category of sets. Since a $0\Cat$ is also the category of sets, it makes sense to define the functors extension is simply a set $\Sigma$ (which is the set of indeterminates) and $(-1)\Cat^+$ is the category of sets. Since a $0\Cat$ is also the category of sets, it makes sense to define the functors
......
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment