Let $(G,B,N,S)$ be a Tits system, and let $\varphi: G \rightarrow \hat{G}$ be a $B$-adapted homomorphism. This means the kernel of $\varphi$ is contained in $G$, and for every $g \in \hat{G}$, there exists an $h \in G$ such that
$$g \varphi(B)g^{-1} = \varphi(hBh^{-1})$$
Since $B$ is its own normalizer, if $h_1, h_2$ are in $G$ and satisfy the above, then $h_1B = h_2B$. It is claimed in (1.2.16) of the first paper of Bruhat-Tits on buildings that this induces an action of $\hat{G}$ on $W = N/B \cap N$. I am trying to figure out why this is the case, but I don't see it.
Okay I see. Let $n \in N$ be a representative of $W$. We can write
$$g \varphi(BwB)g^{-1} = g \varphi(B)g^{-1}(g \varphi(n)g^{-1})g \varphi(B)g^{-1} = \varphi(hBh^{-1}) g \varphi(n)g^{-1} \varphi(hBh^{-1})$$
and so
$$\varphi(h)^{-1}g \varphi(BwB)g^{-1} \varphi(h) = \varphi(B) \varphi(h)^{-1}g \varphi(n)g^{-1} \varphi(h) \varphi(B)$$
Since the image of $\varphi$ is a normal subgroup of $G$, the right hand side is the image of a double coset $Bh_0B$ of in $G$. Thus the right hand side is equal to $\varphi(Bw'B)$ for a unique $w' \in W$. This is the map $w \mapsto w'$ induced by $g \in \hat{G}$.
To see that this only depends on $g \in \hat{G}$, not on the choice of $h$, suppose that $h_1$ is another choice. Then $Bh_1^{-1} = Bh^{-1}$ and $h_1B = hB$, and so
$$\varphi(h)^{-1}g \varphi(BwB)g^{-1} \varphi(h) = \varphi(B) \varphi(h)^{-1}g \varphi(n)g^{-1} \varphi(h) \varphi(B) = \varphi(B) \varphi(h_1)^{-1}g \varphi(n)g^{-1} \varphi(h_1) \varphi(B)$$
$$ = \varphi(h_1)^{-1}g \varphi(BwB)g^{-1} \varphi(h_1)$$
To show that $\xi$ actually defines a group action, and therefore a bijection, comes down to the following observation. Let $g_1, g_2 \in \hat{G}$, and choose corresponding $h_1, h_2 \in G$. Let $h_0$ be an element of $G$ corresponding to $g_1g_2$. Then $\varphi(h_0) = g_1\varphi(h_2)g_1^{-1}\varphi(h_1)$.