, Since they are finite and gated, by the Helly property for gated sets, X is nonempty. If X = H, by Claim 10, there exists an edge of H whose carrier does not include all maximal cells, a contradiction. Hence H = X, i.e., H is a finite cell. Consequently, Z ? (G) is a finite cell of G
Every automorphism ? of G maps maximal halfspaces to maximal halfspaces, thus ? ,
By Lemma 21, Z ? (G) is a finite cell, thus every automorphism of G fixes the cell Z ? (G) ,
, A non-expansive map from a graph G to a graph H is a map f : V (G) ? V (H) such that for any x, y ? V (G) it holds d H (f (x), f (y)) ? d G (x, y)
i k ), and w = (j 1 , j 2 , . . . , j k ), where i m , j m ? i m , n m ? j m < n m /2 for all 1 ? m ? k. Since f (u ) = u , f (v ) = v , f (w ) = w and any vertex of X lies on a shortest path between one of the pairs of u , v , w , we conclude that f (X) ? X. It remains to prove that f (X) = X. Without loss of generality assume that among all j m ? i m with 1 ? m ? k, the difference j 1 ? i 1 is minimal. The vertex y = (i 1 , 0, 0, Lemma 22. Let G be a hypercellular graph and f be a non-expansive map from G to itself. Let u, v, w be any three vertices of G and let X = u , v , w be their median-cell. If f (u) = u, f (v) = v, f (w) = w, then f fixes each of the apices u = (uvw), v = (vwu), w = (wuv) and f (X) ,
Both must be at distance at most two from each other, at distance at most n 1 /2?1 from y, and a must be adjacent or equal to ,
In each case y, a, b spans a cycle, since the length of C 1 is at least six and f is a non-expansive map. Thus f (C 1 , x 2 + 1, x 3 , . . . , x k ) is a cycle of the form (f 1 (C) + s, y 2 , y 3 , . . . , y k ) or (f 1 (C), y 2 ,
, An endomorphism r of G with r(G) = H and r(v) = v for all vertices v in H is called a retraction of G and H is called a retract of G. Moreover, if r is just a non-expansive map from G to itself
, = H, where X is the median cell of u, v, w. This proves that H satisfies the median-cell property and by Theorem C, H is hypercellular. We continue with the proof of assertion (ii) of Theorem F. S is finite, also conv(S) is finite, therefore H is finite and nonempty. Notice that f (conv(S)) ? conv(S), thus H ? f (H) ? f (f (H)) ? . . ., but since for every v ? V (H) there exists n v such that f nv (v) = v, the inclusions cannot be strict. Thus f (H) = H and f acts as an automorphism on H, Proof. Let r be a weak retraction of G to H. For arbitrary vertices u, v, w of H it holds r(u) = u, r(v) = v, r(w) = w, thus by Lemma 22 X = r(X) ? r(G)
, If G is a finite regular hypercellular graph, then G is a single cell, i.e., G is isomorphic to a Cartesian product of edges and even cycles
, We will prove that also W (b, a) is minimal. The carrier N (E ab ) is the union of maximal cells of G crossed by E ab . For each such maximal cell X of N (E ab ) there exists a unique automorphism of X that fixes edges of E ab ? X and maps X ? W (a, b) to X ? W (b, a) and vice versa, G such that W (a, b) is an inclusion minimal halfspace
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