The A2 Davis complex is a hexagon whose boundary is the Coxeter complex circle

By [F9], the relations are $s^2=t^2=1$ and $(st{)}^3=1$. Put $r=st$; then $srs=r^{-1}$, so every word reduces to $r^k$ or $s{r}^k$ with $0\le k<3$. The map $s\mapsto (12)$, $t\mapsto (23)$ is onto $S_3$, so the six normal forms are distinct and $\mathrm{\mid }W\mathrm{\mid }=6$, in agreement with [F10]. Thus $W_{\{s\}}=\{1,s\}$ and $W_{\{t\}}=\{1,t\}$ each have index $3$, while $W_S=W$. The cosets are $W_{\{s\}},t{W}_{\{s\}},st{W}_{\{s\}};W_{\{t\}},s{W}_{\{t\}},ts{W}_{\{t\}};W,$ and the six singleton cosets $w{W}_{\mathrm{\varnothing }}$. The equality criterion [F2] shows that these are distinct within each type, and different types give different cells. Since every subset of $S$ is spherical, these are all cells; their dimensions are $0,1,2$ by [F3],[F5]. This gives $6+3+3+1=13$ cells, agreeing with the coarser counts of [F13].

In the rank-two reflection plane, the reflecting lines bound a sector of angle $\pi \mathrm{/}3$: by [F7] their unit normals have inner product $-\cos (\pi \mathrm{/}3)=-1\mathrm{/}2$, so the normals meet at angle $2\pi \mathrm{/}3$ and their perpendicular lines meet at angle $\pi \mathrm{/}3$. By [F8], the action defining $C_S$ sends $s,t$ to these two reflections. The distances of $x_S$ from the two walls are $B(x_S,e_s)=d_s$ and $B(x_S,e_t)=d_t$ by [F3]. If $d_s=d_t$, then $x_S$ lies on the sector's internal angle bisector. Choose angular coordinates with the walls at $0$ and $\pi \mathrm{/}3$; then $x_S$ has angle $\theta =\pi \mathrm{/}6$. The reflections preserve $B$, and their product has determinant $1$ and trace $2\cos (2\pi \mathrm{/}3)=-1$ by [F7], so in this Euclidean plane it is a rotation by $2\pi \mathrm{/}3$ or $-2\pi \mathrm{/}3$. The orbit angles are therefore $\theta +2k\pi \mathrm{/}3$ and $-\theta +2k\pi \mathrm{/}3$ for $k=0,1,2$. These are the six equally spaced angles $\theta +j\pi \mathrm{/}3$, $0\le j<6$. Their convex hull is a regular hexagon, as also recorded in [F12].

The proper nonempty faces of $C_S$ are its six vertices and six edge cosets by [F3] and the lists in step 1.1. The boundary walk alternates the two labels and has vertices $1,s,st,sts,ts,t,1$: they are distinct up to the repeated endpoint by the six normal forms in step 1.1. Hence the proper cells form a $6$-cycle, which is the rank-two Coxeter complex described by [F11]. Since $S$ itself is spherical, [F1] makes $L$ the single $1$-simplex on $s,t$. At each hexagon vertex there are exactly two incident edges, one of each label, and the unique $2$-cell joins their directions; its local link is therefore one closed interval, the same $1$-simplex $L$.

The order complex of the four-element poset $\mathbb{S}$ has the two triangles $\mathrm{\varnothing }<\{s\}<S$ and $\mathrm{\varnothing }<\{t\}<S$, glued along $\mathrm{\varnothing }<S$, so its realization is the stated square chamber $K$. By [F4], the subposet below the maximal coset $W$ (which is all of $WS$) realizes the barycentric subdivision of $C_S$; consequently $\mathrm{\Sigma }$ is homeomorphic to the closed disk $C_S$. Its coarse cellulation has the six vertices, six edges, and one face from step 1.1, giving Euler characteristic $1$, in agreement with [F13]. Since $C_S$ is convex and contains $0$ by [F3], the homotopy $H(u,z)=(1-u)z$ contracts it to $0$. The quotient statement follows from [F6]. All group and coset lists are finite and explicit, so no Choice is used.