The D4 groupoid and the spherical cell complex

Use the dual sphere of [F2] for $D$, with disk attachments interpreted as in [F3]. Its one-skeleton is exactly the chamber adjacency graph, and every dual two-cell is an atomic square or hexagon. Conversely every atomic polygon is one of these cyclic faces. Thus filling all faces, including whichever face becomes outermost after removing a point for a planar drawing, gives this entire sphere. Omitting the outermost disk would remove a disk from this model, so it must be included in the stated spherical construction.

For genuinely distinct opposite directed edges, use the directed bigon construction of [F2], with the quotient topology of [F3]. Its polygon attaching loops use a coherent sphere orientation, with the two adjacent polygons using opposite directed copies of each shared edge. The bigon imposes that the two crossing arrows are inverses. A polygon disk identifies its two opposite geodesic arcs after these inverse identifications, exactly the codimension-two relation in [F1]. The construction therefore has the specified directed attaching maps and is homeomorphic to $S^2$.

For either $X=D$ or $X=K$, steps 1.1 and 2.1 give a homeomorphism $h:X\to S^2$. By [F6], both $h$ and $h^{-1}$ are continuous. For any $a,b\in X$, compose a path from $h(a)$ to $h(b)$ with $h^{-1}$; such a path exists by [F4]. Every loop in $X$ at $a$ maps under $h$ to a loop at $h(a)$, which is homotopic to the constant loop with basepoint fixed by [F4]; composing that homotopy with $h^{-1}$ gives the corresponding homotopy in $X$. If $p,q:a\to b$ are path classes, then $q^{-1}\circ p$ is therefore the identity at $a$. The groupoid inverse laws in [F2] imply $p=q$. Thus the fundamental groupoid has exactly one arrow between every ordered pair of points, for both spaces; in particular this agrees with the supplied vertex conclusion for $K$.

For each $X\in \{D,K\}$ send each chamber object to its corresponding vertex, and its unique universal wall-groupoid arrow to the unique path class between those endpoints in $X$. By [F1] and step 3.1 this is bijective on objects and on each hom-set. It preserves identities and composition because the relevant hom-sets are singletons; its inverse on objects and arrows is a functor for the same reason. The two functors are inverse on every object and arrow, so their composites are the identity functors. Hence it is an isomorphism in the sense of [F7] to the full vertex subgroupoid of [F5]. Each generating crossing maps to the actual crossing-edge path class, since that class has the required endpoints.

Fix one chamber vertex $v_0\in V$. For each $X\in \{D,K\}$ let $I:{\mathrm{\Pi }}_1(X;V)\to {\mathrm{\Pi }}_1(X)$ be the inclusion and let $R:{\mathrm{\Pi }}_1(X)\to {\mathrm{\Pi }}_1(X;V)$ send every object to $v_0$ and every arrow to its identity. These preserve endpoints, identities and composition, so are functors by [F7]. Define the component at $v\in V$ of $1\Rightarrow RI$ to be the unique arrow $v\to v_0$, and the component at $x\in X$ of $IR\Rightarrow 1$ to be the unique arrow $v_0\to x$. These arrows exist by step 3.1 and are invertible by [F2]. Every naturality square commutes because the two composites have the same endpoints and there is only one arrow between them. For each of these component families, the inverse component arrows also form a natural transformation: their naturality squares commute by the same uniqueness of arrows with specified endpoints. Both vertical composites have identity components by the inverse laws, so the families are natural isomorphisms by [F8]. Thus $I$ and $R$ give the claimed equivalence by [F5]. Together with step 4.1 this proves all groupoid claims, with the asserted isomorphisms using precisely the 32 chamber objects.