ACYCLIC CONTINUA IN THE PLANE AND QUASI-COMPLEXES James E. Keisler "Qualified requesters may obtain copies of this report from the ASTIA Document Service Center, Arlington Hall Station, Arlington 12, Virginia. Department of,.Defense contractors must be established for ASTIA services,. or hav e' their':nieed-to-know' certified by the cognizant, milit'ary- agendc'of their project or contract." TEE UNIVERSITY OF MICHIGAN ANN ARBOR, MICHIGAN July 1958

ACYCLIC CONTINUA IN THE PLANE AND QUASI-COMPLEXES James E. Keisler Quasi-complexes, defined by Lefschetz [2, p. 323], form an extensive class of spaces to which the Lefschetz fixed point formula applies. This class includes all finite Euclidean complexes, all parallelotopes, and all snake-like continua [1, p. 667]. One might hope to.prove that all acyclic plane continua possess the fixed point property by showing that they are quasi-complexes. It is the purpose of this paper to show that there exist acyclic plane continua which are not quasi-complexes. Indeed, a plane continuum may be T-like and yet fail to be a quasi-complex. In this paper covering is used to mean finite open covering. An E-covering of a metric space is a covering such that the diameter of each element of the covering is less than c. For coverings A1, A2 such that Al refines A2 (Al > A2), N(Ai) denotes the complex which is the nerve of the covering Ai and Tc(1,2) denotes one of the projection chain maps 1t(l,2): N(Al) + N(A2) induced by the inclusion map (see [2, pp. 244, 245]). For a an element of the covering A, Sta is the union of all those elements of A which intersect a. Al star refines A2 (A >> A2) if the covering C, whose elements are the sets Sta for all a which are elements of Al, refines A2. A simplex a belongs to a chain C if the coefficient of a in C is not zero. K(a) denotes the point set which is the intersection of the vertices of a and K(C) is the union of the sets K(a) for all the simplices which belong to C. A finite collection of open sets which can be ordered, al, a2,... an, so that ai intersects aj if and only if li-j|l | is called a snake. If the collection is so ordered, we refer to it as an ordered snake. A metric continuum is snake-like if for each positive e it has an e-covering which is a snake; that is, such that the nerve of the covering is an arc. Similarly, a metric continuum is T-like if for each positive c it has an c-covering whose nerve is a "T. " A quasi-complex, to be denoted by QC, is a compact Hausdorff space., X, for which there exists a cofinal family of coverings, say kUcd, a c A, partially ordered by refinement, such that in addition to the projection chain maps, 1

t()a), P > a, there exists a collection of antiprojection maps w)(ac,): N(Ua) + N(Up), > a with the properties: 1) the antiprojections are chain maps, 2) given a, there is an f(a) > a such that an antiprojection w(af) exists, 3) if g > f > a and there exist antiprojections wD(a,f) and. (f,g)> then the composition map o(fg) c(a,f) is an antiprojection, 4) given two antiprojections w(a,B) and. c(ac,), then c(a,P) is homologous to c((c,), 5) w(a,) t((,a) is homologous to the identity chain map, 6) given 0, there is a g=g(a) > a and given b, there is an h=h(a,g,b)> b,g such that antiprojections c(g,h) exist and the collection tK(a) U K[w(g,h)a]}, for all simplices a belonging to N(Ug), refines Up. (All indices belong to the set A.) Lemma: Let {Oi} denote a cofinal sequence of coverings of the QC X, such that 0i+l refines Oi and the maps TE H[N(Oj)] + H[N(Oi)] induced by the projection maps, are onto. (H[N(U)] denotes the homology groups of N(U).) If such a sequence {Oi} exists, then there exists a subsequence, say {Vi, such that antiprojections may be defined among the N(Vi), which with their projections show that X is a QC, Proof: For a properly chosen subsequence of the ~oi} we construct the required antiprojections and show that this system satisfies properties 1)-6). Since X is a QC, there exists a cofinal family of coverings {1UTJ and associated projection and antiprojection maps which satisfy properties l)-6). We construct a cofinal sequence {CiJ of coverings which relate the coverings Ua with the coverings Oio In this construction, the choices are made so that Ci+i refines Cio Ci is the element OC(i) of the sequence tOij if i=4j for some integer jo Otherwise, Ci is an element of the collection {Ua }. The g and h functions are those functions given in property 6) of the definition of a quasi-complex. Select any Uc; call it C_5. Choose C-1 such that -1 = g(-5). Let Co be any one of the coverings 0i which refines C-1. Choose C1 so that it is a star refinement of Co. Let C2 be chosen so that 2 = g(-l). Our conventions require that C2 refine C1o This is possible, even though only one index g(-l) is assumed in property 6). For if P > a then g(P) > 5 > c and 2

the collection {K(a) U K[ii(gh)a], for all simplices a belonging to N(Ug), refines Ug which refines Uoa That is, for P > aC any g(3) is also a g(a)o Let C3 = C20 Choose C4 such that C4 refines C3 and 0O, i < 4, and C(4) > C(O)o For k = 4j, j _ 1, given Ck, select Ck+l such that Ck+l is a star refinement of Cko Choose Ck+2 such that k+2 = h(k-9, k-5, k+l). Choose Ck+3 such that k+3 = g(k-l). Choose Ck+4 such that Ck+4 refines Ck+3 and Oi, i _ k+4, and C(k+4) C(k). f By induction, this selects a sequence CCi3. Let Vi = C4io Then fVi3 is a cofinal subsequence of {Oi3. We shall denote projection maps among the Ci: t (kj)o N(Ci) + N(Cj) i > j o Thus a projection map from N(Vi) to N(Vj) will be written t((4i,4j). Given Vi, Vi+l we define an antiprojection map s(i,i+l) from N(Vi) to N(Vi+l) by s(i,i+l) = t(4i+6, 4i+4) o(4i-l, 4i+6) t(4i, 4i-1) The antiprojections are all finite products of the maps s(i,i+l) s(ii,j) = s(j-l,j).. s(ii+l) In particular, note that w(4i-l, 4i+6) exists, since 4i+6 = h(4i-5, 4i-1, 4i+5) o These antiprojections are chain maps, since they are defined as a product of chain maps, so that 1) is satisfied. Requirement 2) is satisfied by letting f(i) = i+l. Requirement 3) follows by the definition of our antiprojections. To prove 4), suppose that we are given s(i,i+l) = 1t(4i+6, 4i+4) c(4i-l, 4i+6) 7t(4i, 4i-1) and s(ii+l) = T(4i+6, 4i+4) i(4i-1, 4i+6) T(4i, 4i-l). 5

That Tr(4i, 4i-1) is homologous to T(4i, 4i-1) and r(4i+6, 4i+4) is homologous to 7(4i+6, 4i+4) is a well-known property of the projection maps [2, po 245], and that w(4i-1, 4i+6) is homogolous to J(4i-l, 4i+6) is assured since 4) holds for the fU[n I and their antiprojectionso Since product maps of homologous maps are homologous, s(i,i+l) is homologous to s(il,i+l), and therefore finite products of these antiprojections are homologouso Hence, 4) is satisfied. To prove 5), we use the requirement that the homology maps, induced by the projections, are onto. This is the only use that is made of this requiremento We designate the identity chain map by Io That two maps, or two chains, are homologous is stated by A - B. We first show that s(ii+l) tc(4i+4, 4i) I- i Let Z be any cycle of N(Vi+l)o Since the projection map jt(4i+8, 4i+4) induces a homology map which is onto, there exists a cycle, say Z1, of N(Vi+2) such that t Z1 = t(4i+6, 4i+4) A(4i+8, 4i+6) Zl = z - Let Z2 denote t(4i+8, 4i+6) Z1o Since Z ~ Z: s(i,i+l) r(4i+4, 4i) Z - s(i,i+l) t(4i+4, 4i) Z But~ s(i,i+l) t(4i+4, 4i) Z = t(4i+6, 4i+4) w(4i-l, 4i+6) i(4i+6, 4i-l) Z2 and i(4i-l, 4i+6) t(4i+6, 4i-l) I by 5) for {U}o Thus s(i,i+l) t(4i+4, 4i) Z - t(4i+6, 4i+4) Z2 and t((4i+6, 4i+4) Z2 Z o Therefore s(i,i+l) t(4i+4, 4i) Z - Z z That is, s(i,i+l) t(4i+4, 4i) -I In general: s(j,k) c(4k, 4j)- [s(k-l,k) ooo s(j,j+l) A(4j+4, 4j) 0o, k(4k, 4k-4)] [s(k-l.,k) ooo s(j+2, j+l) 3(4j+8, 4j+4)... A((4k, 4k-4)] - oo - [s(k-l,k) t(4k, 4k-4)] - I Thus, the s(j,k) satisfy 5).

To prove 6) we define g(i) = i+2, and given Vb let h(i,g,b) be the maximum of g+l, bo Then s(gh) exists since s(i,j) exists for i < jo Let a be an arbitrary, fixed simplex which belongs to N(Vg)o The proof that 6) is satisfied will be complete if we can show that there exists an open set Vi,o which is an element of Vi such that K(a) U K[s(g,n)a] C Vi, for all no For generic ca such that ax is an element of N(Vn) and ac is also an element of s(g,n)a, let ac denote rj(4n, 4n-l)aao In addition, aa is the image under i(4n+2, 4n) of some simplex which is an element both of N(C4n+2) and of c(4n-5, 4n+2) i(4n-4, 4n-5) s(g,n-l)oa Let a& denote such a simplex. Denote a(4g, 4g-l)a by To Denote elements of Ci by Ci,ko Property 6) will be established by induction. First, {K(a) U K[ao(4g-l, 4g+6)a]] refines C4g.5 since 4g-1 = g(4g-5) and 4g+6 = h(4g-5, 4g-1, 4g+5). Thus K(a) U K[w(4g-l, 4g+6)Y] C C4g.-5 Since C4g+6 > C4g+4 = Vg+l > C4g+3 > C4g-5 >> C4g_8 = Vi, K(a) U K[((4g+4, 4g+3) jt(4g+6, 4g+4) (4g-l, 4g+6)i] C St C4g-5,2 C Vi,o Thus K(a) U K[s(g,g+l)a] C Vi,o Let ai' be an element of c(4g-1, 4g+6).o Then K( j') C C4g_5, Also K(oj) U K[w(4g+3, 4g+l0) j] C C4g_, a(j) since 4g+3 = g(4g-l) and 4g+10 = h(4g-1, 4g+35 4g+9) o Since C4g+lO > C4g+8 > C4g+7 > C4g- >> C4g-5 5

K(aj) K[t(4g+8,4g+7) T(4g+10,4g+8) )(4g+3,4g+l0)oj] C St C4gl,a(j) C C4g_5,b(j). But K(aj') is contained in the intersection of C4g-5,b(j) and C4g-5,Q. Therefore C4g-5,b(j) C St C4g_51, C Vi,o Thus K(a) U Up {K(oj) jcGP J K[s(g+l, g+2)aj]) C Vi,o where P is the set of all those simplices aj which belong to s(g,g+l)a. That is, K(a) U K[s(g,g+2)a] C Vi,. We are now ready for the general inductive step. Assume that K(a) U K[s(g,n)a] C Vi o where if a2 is an element of s(g,n)a, there is a simplex al, which belongs to s(g,n-l)a, for which a2 belongs to s(n-l,n)ac, and is such that K(ca) U K(a2) C C4n-9,a(ai) and K(1) U K(a2) C St C4n9,a(1i) C C4n-13,b(al) C Vi,o (n 2 g + 2. We have seen that these assumptions are satisfied for n = g+2.) K(a2) U K[w(4n-l, 4n+6)a] C C4n-5,a(a2) since 4n+6 = h(4n-5, 4n-l, 4n+5) Since C4n+6 > C4n+4 > C4n+3 > C4n-5 >> C4n9 for s3' an element of co(4n-l, 4n+6)a2, we have: K(a2) U K(U3) C St C4n.5,a(a2) C C4n-9,b(a2) But K(a2') C C4n_9,b(a2) C4n-9,a(ja) or K(i2) U K(s3) C St C4n_9,a(o) C Vi,o 6

Thus K(c) U K[s(n,n+l) s(g,n)o] = K(a) U K[s(gn+l)] C Vi,o By replacing n by n+l, a0 by c2, and a2 by C3 in the induction assumptions, we see that all conditions are fulfilled, and the induction is complete. This completes the proof of the lemma. Corollary 1: The maps T: H[:N(Vj)] + H[N(Vi)], which are required to be onto maps, are actually isomorphisms. Proof: We have st I. The isomorphism can fail only if there exists a nonzero homology class Z such that isZ ~ 0. Suppose so. Then stZ = s(tZ) = s(O) = 0, and srZ - IZ = Z. But Z L O. Therefore, cZ ~ 0 only if Z 0, and TT is an isomorphism. Let T1 = {(x,y) x = 0, 0 ~ y _ 2}, T2 = ((x,y)| 0o x 2 2, y = 2}, T3 = ((x,y) -2 x x O, y = 2], T = T U T2 T3 C E2, the planeo Let R denote a ray which spirals down on T, R C E2. R = ((2-1 < x x 2, y = -2-1) U n {(x = 2- n -2n y 2 -2n) n=l U (2-n < x 2 + 2-n y 2 -2-n) U (x = 2+2-n, 2-2-n < y 2+2-n) U (-2-2-n x < 2+2-n, y = 2+2-n) U (x = -2-2-n, 2-2-n y 2+2n) U -2-2 -n2 U (-2-2'n x < -2-n, y = 2-2n) U (x= 2-n, 2-(n+l) y 22-n) (-2-n " x 2-(n+l), y -2-(n+l))l Obviously R = R U T. Denote R by X. Corollary 2: There exist a T-like plane continuum which is not a QC. 7

Proof~ X is obviously a T-like plane continuum. Under the assumption that X is a QC we will arrive at a contradiction. Let PO = (2, -2-1), PI = (1, -2-1) P2 = (0, ), P3 = (0, 1), P4 = (0, 2), P5 = (1l 2) P6 = (2, 2), P7 = (-1, 2), and P8 (-2, 2). It is clear that for each positive e, there is an c-covering of X, say Uc = Ue 1 *'. U,n(e) such that: i) Pi e UP(e) P(i,e) i,) < P(i+l,e), P(O,e) = 1, P(8,e) = n(); ii) UE 1..l UE P(2 e) is an ordered snake; iii) UCp(2, ). * Ucp(4,e) is an ordered snake which covers T1; iv) UC p(4,e) e. U P(G6 ) is an ordered snake which covers T2; v) UeP(6,c)+l... UcE,n() is an ordered snake which covers Ts; and vi) the nerve of Ug is a "T." Obviously, ~U2.nI is a cofinal sequence of coverings of X for which the homology maps induced by the projections are onto maps, since each nerve is acyclic. Thus, by the lemma, some subsequence of {U2-n} carries the QC structure of X. Denote this subsequence by Vi, which is assumed chosen, as in the proof of the lemma, such that i > j implies that Vi refines Vj. We adopt the notation Vi = Vi1 ~.. Vi,n(i) Pj e Vi,P(j,i) satisfying i)-vi) above. For b > a and k _ P(3,b) let: 8

Vb,P(n;l,;b,a) be the first Vb,k VbP(5;j,2;ba) be the first Vb,k with k > P(6;j,l;b,a); Vb,p(4;j,2;b,a) be the first Vb,k with k > P(5;j,2;b,a); VbP(7;j,2;b,a) be the first Vb,k with k > P(8;,l;b,a); VbP(4;j,3;b,a) be the first Vbk with k> P(7;j,2;b,a); vb,P(3;j,2;b,a) be the first Vbk with k > P(4;j,5;ba); VbP(nj,l;b,a) be the first Vb,k with k > P(3;j-l,2;b,a); such such that that Vb, k Vb,k Vbgk intersects intersects Va,P(n,a); Va,P(5,a) such that Vb,k intersects Va,P(4,a) such that Vb,k intersects Va,p(7,a) such that Vb,k intersects Vap(4,a) such that Vb,k intersects Va,p(3,a) such that Vb,k intersects Va,P(n,a) and N(b,a) be the maximum j such that P(2;jl;ba) is defined by the above procedure. It is obvious that for each pair (ba), N(ba) exists and that given an integer M and an index a, there exists an index b such that N(b,a) > M. Let Va be an element of 7Vi} such that Va refines U2-3 g = g(a); and h = h(ag,b) where b is chosen so that N(ba) > N(ga) + 4. We assume the existence of an antiprojection Wo N(Vg) + N(Vh) and show that w cannot satisfy all of the properties 1)-6)o For any zero chain, n ai Vn,i i=l define KI(Zai Vn,i) = Zait For any one chain A, KI(OA) = 0 where a denotes the usual boundary operator. By property 5) KI [w Vgi] =1. Let Vni + Vn denote the one chain Vn i Vn,i+l + Vn,i+l Vn,i+2 +..+ Vn Vn,j where Vn, i ntersects V k if and only if |I-k| _ 1, i _., k s j. By property 6) K(w Vgp(i,g)) C st Va,p(i,a); K(uo Vg,P(i;j,k;g,a)) C St Va;P(i,a); and P(2,a) K[(Vg,1 + Vg,p(2;ll;g,a))] C U StVPa 9 ~i=l

Given the complex K with n-dimensional simplices aol.. ar and the nchains r C1 = E ai ci 9 i=l r C2 = Z bi ai, C1 * C2 i=l denotes the n-chain r C3 = Z c i ci i=l where ci = ai if bi 0, ci = bi tersect if there exists an n-simplex A 1-chain between Vn,i and Vn,j a2 Vn jo if bi = 0. Two n-chains are said to inwhose coefficient in each chain is nonzero. is a 1-chain C such that aC = a, Vni + Since the distances I|PoP1, IP1P2I and IPoP21 are each greater than 1/2. CVg,p(2; 1,1; g,a) = D(Vg, + Vg,P(2;I,l;g,a))* Vg~v g, n(h) P(,h) i=P(1,h) Vh,i Also W(Vg,1 + Vg,(2;ll;g,a)) does not intersect N(h,a)-l n=l (Vhp(3;n,l;ha) + Vh,P(3;n,2;h,a)) + n(h) P(,h) i=P(3, h) Vh,i Vh,k Therefore: KI [Vg,P(2; 1,1; g, a)* P(3;n+l,1;h,a) z Vh,i] i=P(3;n,2;h,a) =0 n=l...N(h,a)-2 and KI [COVg,p(2; 1,1; g,a)* P(3,h) Z=( 7 N )Vh,i] i=P(3;N(h,a)-1,2;h,a) = 0 Hence: KI [WCVg,p(2; 1,1; g,a)* P(3;l,l;h,a) i=P(l,h) Vh,i] = 1 10

Assume that P(3; j,l;h,a) KI [OVg, p( 2; n., 1; ga)* i=P(3; j-l, 2; h, a) Vh,i] = 5 j for n < N(g,a) = N, where 5n j = O, n B j; 5n, = 1, n = J3 We have seen that this assumption is valid for n=l, if we let P(l,h) = P(i;O,j;h,a). Now consider the 1-chain C = Vg,P(2;n,l;g,a) Vgp(4, n, 1;g,a) By property 6): P(4,a) K[WCC] C U St Va,i i=P(2,a) Thus:'(C*[h, P(5; j,1;h,a) + Vh,p(5; j,2;h,a) + Vh, P(7; j,;h,a) Vh,P(7; j,2;h,a)] = 0 This implies that P(5; j,l;h,a) KI[oC- * E Vh,i] = 0 i=P(7, j-l,2;h,a), and P(7; j,l;h,a) KI [cC * Z i=P(5;j,2;h,a) Vh,i] = 0 Since also K[~V g,P(2;n,1;g,a)] C St Va, P(2,a) we have: and KI [VgI, p( 4; n,!; g, a)* i: KI [cVg, p( 4;n,l; g,a)* KI^vgp(4nnlgva P(5; j,l;h,a) z Vhi] =P(7; j-l,2;h,a) P(3;j-l,2;h,a) E Vh, i=P(7; J-1,2;h,a) = n,j i] -KI [wVg,p( 4; n,1 g,a)* P( 5; j. 1; h, a) i=P(3;j,lh,a) 1.1 Vh, iI for j ~ n

Assume that P(3 j,2;h,a) KI [Cvg, p( 4;n,l; g,a)* Vhi( 7, i=P(7; 2;h, a) In this case W(Vg,P(4;n,l;g,a) + Vg,P(6;n,l;g,a)) must contain a 1-chain between an element in P(3; j,2;h,a) z Vh,i i=P(7; j,2;h,a) and an element either in P(7; j,l;h,a) E Vh,i i=l or in n(h) C Vh,i i=P(3; j:+l, l;h, a) But: P(6,a) K[C(Vg,P(4;n,1;g,a) + Vg,P(6;n,l;g,a))]C U st Va, i=P( 4, a) and no such 1-chain s exists with the property that P(6,a) K(s) C U St V, i=P(4,a) Therefore: P(3; j,2;h,a) KI[coVg,p(4;n,l;g,a)* i Vh,i] = i=P(7; j,2;h,a) P(7; j,l;h,a) KI [CVg, p(4;n,l; ga)* Vh,i] = 0 i=P(5;j,2;h,a) and 12

P(5;j,l;h,a) KIT[OVg,p(4;n,1;g,a)* Z Vh,i] = n,j i=P(3;j,l;h,a) In this same manner, successively considering: (Vg,p(4;n,1;g,a) + Vg,P(6;n,1;g,a)) which does not intersect Vh,,P(7; j, l;h,,a) -"Vh, P(7; j, 2;h,a) nor Vh,P(3;j, 2;h, a) + Vh,P(3;j+1,1;h,a); c(Vg,P(6;n,l;g,a) which does not intersect Vh,P(3; j,2;h,a) + ((Vg,p(8;n,l;g,a) which does not intersect + Vg,P(8;n, l;g,a)) Vh,P(3; j+l,l;h,a); + Vg, P(4;n,3;g,a)) Vh, P(3; j, 2; h, a) + Vh, P(3; j+l,1;h, a) nor Vh,P(5;j,l;h,a) + Vh,P(5;j,2;h,a); and (D(Vg,P(4;n, 3;g,a) + Vg,P(2;n+l,l;g,a)) which does not intersect Vh,P(7; j,l;h,a) + Vh,P(7;j,2;h,a) nor Vh,P(5; j,l;h,a) + Vh,P(5;j,2;h,a); we finally conclude that KI [Vg,p(2;n+l, 1; g,a)* P(; j+l,l;h,a) E i=P(53; j,2;h,a) 13 Vh,i] = n, j

In particular, this holds for n+l = N, by inductiono Since K[cD(Vg, p(2;N,1;g,a) + Vgp(4,g))] C P(4, a) U St Vi i=P(2,a) we have: KI[6co(Vg,p(2;N,1; g,a) + Vgp(4,g))* P(5;N,l;h,a) i=P(7;N-1,2;h,a) Vh,i] -O Therefore, either: (A) KI[ Vg,p(4,g)* P(3;N-1,2;h,a) Z i=P(7;N-1,2;h, a) Vh,i] # 0 or (B) K [WVgp(4,g)* P(5;N,l;h,a) P(;N,l;h,a) i=P(3; N, 1; h, a) Vh, i Suppose that (A) holds. Since P(6,a) K[w(Vgp(4,g) VgP(6,g)) ]C P i=P( 4, a).St Vai g((Vg,p(4, g) + Vg,P(6,g)) must contain a 1-chain s between some element in and some element either in P(5;N-1,2;h,a) E Vh,i i=P(7; N-1, 2; h, a) P(7;N-l,l;h, a) Z Vh,i i =1 n(h) Z Vhgi' i=P(5;N,l;h, a) or in 14

But no such s exists with the property that P(6,a) K(s) C U St Vai i=P(4,a) Hence (A) does not hold. By analogously considering g(Vg,P(4,g) Vg, P(6,g)+l + Vg,p(6,g)+l +Vg,P(8,g)) which, by property 6) has the property. K[D( Vgp(4,g) Vg,p(6,g)+1 + VgP(6,g)+l + VgP(,g)) P(8,a) C U St Vai USt Va, p(4, a), i=P(6,a)+1 we see that (B) does not hold. That is, the required antiprojections do not exist. Therefore, X is not a quasi-complex. Using polar coordinates in the plane, let Ri = {(rQ) I = l/r-l, 1 < r c 2 C = {(rQ)I |r = 1} D = {(rQ)I Irl l}. The above methods can be used to show that neither RiU C nor RlU D is a QC. REFERENCES 1. E. Dyer,"A Fixed Point Theorem,"Proc. Amer. Math. Soco, 7 666-672 (1956). 2. S. Lefschetz, "Algebraic Topology," Amero Math. Soc. Colloquium Publications, 27 (1942). 15

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