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Sep 21, 2023 at 7:46 comment added Tom De Medts @Anonmath101 Well, to define $\rho(\Sigma, C)(D)$ for some $D \in \Sigma'$, you can take any apartment containing $D$ and an isomorphism $\varphi'$ from that apartment to $\Sigma$. This is independent of the choice of that apartment, as explained on p. 320. Now you simply make the same choice $\Sigma'$ itself, for any $D$, and so also the same isomorphism $\varphi'$. In other words, $\rho(\Sigma, C)$ coincides with $\varphi'$ on $\Sigma'$.
Sep 20, 2023 at 10:59 comment added Anonmath101 I am wondering why the restriction of $\rho(\Sigma, C)$ to $\Sigma'$ is an isomorphism?
Sep 18, 2023 at 15:30 comment added Tom De Medts @Anonmath101 Are you asking why the composition of two isomorphisms is again an isomorphism? (Notice that by construction, the restriction of $\rho(\Sigma, C)$ to $\Sigma'$ is an isomorphism from $\Sigma'$ to $\Sigma$.)
Sep 18, 2023 at 15:17 comment added Anonmath101 Why is composition actually an isomorphism?
Sep 18, 2023 at 15:10 comment added Tom De Medts Because its restriction to $\Sigma'$ is an isomorphism from $\Sigma'$ to itself fixing $C$ and all subsets of $C$; it then follows from Suzuki's (3.15) that it is the identity on $\Sigma'$.
Sep 18, 2023 at 15:05 comment added Anonmath101 Why does $\rho(\Sigma', C ) \circ \rho(\Sigma, C)$ act as the identity on $\Sigma'$?
Sep 18, 2023 at 13:14 comment added Tom De Medts @Anonmath101 Indeed, you are right. I have now also added a proof that does not rely on types.
Sep 18, 2023 at 13:13 history edited Tom De Medts CC BY-SA 4.0
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Sep 18, 2023 at 11:41 comment added Anonmath101 Thank you for your reply. The only problem is that type isn't defined yet and is only defined later, so I'm not sure if this is the argument Suzuki uses here??
Sep 18, 2023 at 10:47 history answered Tom De Medts CC BY-SA 4.0