I’m reading this book called group theory I by Michio Suzuki and part of a proof I just cannot figure out. It’s on page 322 proposition 3.18. I will include it here:
(3.18) Let $\Delta$ be a building, and let $\Sigma$ be an apartment of $\Delta$. Assume that the two chambers $C$ and $C'$ of $\Sigma$ are adjacent. Then, there are foldings $\varphi$ and $\varphi'$ of $\Sigma$ such that
$$\varphi(C')=C\quad\text{and}\quad \varphi'(C)=C'.$$
We have $\Sigma=\varphi(\Sigma)\cup\varphi'(\Sigma)$. Furthermore, $\varphi(\Sigma)$ and $\varphi'(\Sigma)$ have no chamber in common. The mapping $\sigma$ defined by
$$\sigma(a)= \begin{cases} \varphi'(A) & \text{if }A\in\varphi(\Sigma)\\ \varphi(A) & \text{if }A\in\varphi'(\Sigma) \end{cases}$$
is an automorphism of order $2$ of $\Sigma$.
Proof: Let $B$ be a common wall of $C$ and $C'$. By Axiom (B1) of a building, there is a third chamber $C''$ which contains $B$. By Axiom (B3) there is an apartment of $\Sigma'$ which contains $C$ and $C''$. Let $\varphi$ be the restriction of the product
$${\rm retr}(\Sigma', C){\rm retr}(\Sigma, C')$$
on the apartment $\Sigma$. Clearly, $\varphi(\Sigma)\subset \Sigma$ and $\varphi(C)=C$. But, we have $\varphi(C')\neq C'$. This may be proved as follows. [. . .]
I just can’t see why we have $\phi (\Sigma ) \subset \Sigma $. It makes sense that $\text{retr}(\Sigma , C’)(\Sigma ) = \Sigma $ as $\text{retr}(\Sigma , C’) $ fixed every element of $\Sigma $ but can’t then see why this containment is the case.