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VOL. 53, 1965 : L. F. McAULEY 255

9 Groen, P., "Two fundamental theorems on gravity waves in inhomogeneous incompressible fluids," Physica, 14, 294-300 (1948); Groen, P., and B. Heyna, "On short period internal gravity waves," Physica, 24, 383-389 (1958). 10 Eckart, C., Hydrodynamics of Oceans and Atmospheres (New York: Pergamon Press, 1960). See chap. 8. 11 Leighton, R., R. W. Noyes, and G. W. Simon, "Velocity fields in the solar atmosphere," Astro- phys. J., 135, 474 (1962).

LIFTING DISKS AND CERTAIN LIGHT OPEN MAPPINGS* BY Louis F. MCAULEY

RUTGERS, THE STATE UNIVERSITY, NEW BRUNSWICK, NEW JERSEY Communicated by G. T. Whyburn, December 22, 1964 Introduction.-The concept of lifting mappings is of fundamental importance in the various fiber and theories. This concept is expressed in a number of ways, but usually as some type of covering property.' These properties give characterizations of very useful classes of spaces and - pings related to the following problem. General lifting problem: Suppose that f is a mapping of a space X onto a space Y. Furthermore, suppose that g is a mapping (more generally, g is replaced by a class of mappings, e.g., a homotopy) of a space Z into Y. Does there exist a mapping h of Z into X such that g = fh? For what spaces Z and for what map- pings g: Z -- Y do there exist mappings h: Z -> Y such that g = fh? Under what conditions on X, Y, and f do there exist such mappings h? Here, we shall say that when such an h exists, then h is a lifting of g to X. A special lifting problem: Suppose that Z is an n-cell and that g is a homeo- of Z into Y. Under what conditions on X and f does there exist h: Z X so that g = fh? In case that g is a homeomorphism, we shall say that h is a lifting of g(Z). For example, if Z is an n-cell and g is a homeomorphism, we shall speak of h as a lifting of either g or g(Z). Answers have been provided for special cases of these two problems. Success is obtained by restricting the mappings f:X -- Y to covering maps (X being a covering space of Y') and by requiring the space Y to satisfy conditions such as semilocal simple .2 A covering map is a which is a very special kind of light open mapping. A mapping (continuous) f:X -- Y is light iff f-lf(x) is totally disconnected for each x in X and f is open iff for each U in X, f(U) is open relative to f(X). At the present time, most definitions of fiber spaces require some kind of covering homotopy property.' Thus, some answers to these problems are built into the structure of the theory. Furthermore, fiber spaces with totally disconnected fibers have fiber mappings which are light and open. Definitions appear in the literatures of generalized covering spaces whose covering mappings are quite special cases of light open mappings. It therefore seems natural to develop some lifting properties for light open Downloaded by guest on September 30, 2021 256 MATHEMATICS: L. F. McAULEY PROC. N. A. S.

mappings in an attempt to provide answers to more general lifting problems-a step closer to the two problems stated above. In the case that each of X and Y is a Peano (locally connected com- pact continuum), G. T. Whyburn has shown in reference 3 that if f is a light open mapping of X onto Y and h is a homeomorphism of I into Y, then h can be lifted to X. That is, 1-cells may be lifted. In fact, if A is a simple arc (topological 1-cell) in Y, then f-'(A) is the union of a collection G of simple arcs such that ft g is a homeomorphism of g onto A for each g in G. This is remarkable since light open mappings are so much more general than either fiber mappings with 0-dimensional fibers or covering mappings. There is a characterization of open mappings due to Floyd4 which is as follows: Suppose that each of X and Y is a locally connected compact metric continuum and that f: X -- Y is a light mapping of X onto Y. Then a necessary and sufficient condi- tion that f be open is that for each map g:I -> Y where I = [0,1] and for each x in f-lg(O), there exists a map h:I -- X such that g = fh and h(O) = x. Thus, Floyd's characterization says that a light mapping f is open iff all maps g:1 -* Y can be lifted to X (with X and Y as above). In this paper, we shall give some conditions under which a (topological 2-cell) in Y can be lifted to X whenever f is a light open mapping of X onto Y. Unless stated otherwise, all mappings are continuous and all spaces are metric. One should observe that there is an at most two-to-one light open mapping of the projective X onto a disk Y such that if J = f-1(Bd Y), then fj J is a homeomorphism of J onto Bd Y. However, the disk Y cannot be lifted to X. A Condition Ensuring the Lifting of Disks.-Suppose thatfis a light open mapping of a space X onto a space Y and that D is a disk in Y. We say that (A,B) is a proper pair if and only if B is an arc spanning D (B c D and x E B n Bd D if and only if x is an end point of B) and ft A is a homeomorphism of A onto B. We say that f has Property P relative to D if and only if there is a simple closed J in X such that ft J is a homeomorphism of J onto Bd D and f(X- J) n f(J) = 0 and for each E > 0 and proper pair (A,B) there is a a > 0 such that if B' is an arc spanning D and B' c Na(B), then there is a proper pair (A',B') with A' c N,(A). Although Property P is somewhat more complicated than we desire, it is strong enough to yield the following theorem. THEOREM 1. Suppose that f is a light open mapping of X onto Y where each of X and Y is a Peano continuum. Furthermore, suppose that f has Property P relative to some disk D in Y. Then D may be lifted to X. Since no loss in generality results, we assume that Y = D and that D is the in the plane, i.e., D = (Xy) X2 + y2 < 1}. Hereafter the term "proper pair" will refer only to those proper pairs (A,B) such that B is a straight line . We use H(A,A') for the Hausdorff distance between subsets A and A' of a space, i.e., H(A,A') = inf {e > 0tA c N,(A') and A' c NE(A)}. We precede the proof of the theorem by four lemmas. In each we assume the hypothesis of the theorem. Lemmas 2-4 are strengthenings of Property P. LEMMA 1. If I (A,,B)} is a sequence of proper pairs {IB}B B, and {AX}A A, then (A,B) is a proper pair. Proof: Since A is compact, it suffices to show that ft A is 1-1. If this is not the Downloaded by guest on September 30, 2021 VOL. 53, 1965 MATHEMATICS: L. F. McAULEY 257

case, then there are sequences {pi} -* p and {qi} -. q with q,,pi E Ai, p $ q, and f(p) = f(q) = y e B. Now {f(p1)} -- y and {f(qi)} -- y. Let Ai' be the subarc piqi of Ai and Bi' be the subinterval f(pj)f(qi) of Be. Some subsequence { Ai,'} of { A,'} converges to C a subcontinuum of A which contains p and q. Furthermore, since {f(Aki')}-] y, we have f(C) = y. This contradicts the fact that f is light and completes the proof. LEMMA 2. If (A,B) is a proper pair, A pq, and e > 0, then there is an e' > 0 such that if (A',B') is a proper pair, A' = p'q', A' c Nf,(A), d(p,p') < e', and d(q,q') < e', then H(AA') < e. Proof: Suppose that the lemma is not true. Then there is a sequence (An',Bn')} of proper pairs such that (1) An' c Nl1n(A) and A ¢ N,(An') hold for each n, and (2) if An' = pnq", then d(pn',P) < 1/n and d(qn',q) < 1/n. Consider any subsequence { (AknBt) } such that { Ajtj - C and { Bknj -- E. Clearly C c A and I p,q} c C. Since C is connected, we see that E = B and, by Lemma 1, that C = A. From the compactness of X we conclude that { An A and thus reach the contradiction that A c NE(An') for sufficiently large n. LEMMA 3. For each e > 0 and proper pair (A,B) there is a 6 > 0 such that if B' is a straight line interval spanning D and H(B,B') < 6, then there is a proper pair (A',B') with H(A',A) < e. Proof: Let e > 0 and a proper pair (A,B) with A = pq be given. Consider E as guaranteed by Lemma 2. Let E"1 = min { evr and consider 61 =6 (A,B),e" as guaranteed by Property P. Choose 62 SO that if y e N~,(f(p)) n Bd D, then f-(y) e N.,(p) and if ye N,(f(q)) n Bd D, then f-1(y) E Ne,(q). Choose B3 SO that if B' is a straight line interval spanning D with H(B,B') < 63, then B' = uv with u E N2(f(p)) and v e Nh(f(q)). Let 6 = min {61,6} and suppose that B' is a straight line interval spanning D with H(BB') < 6. Since 6 < 61, there is a proper pair (A',B') such that A' c N,,(A). Furthermore, since 6 < 63, A' = p'q' with d(p,p') < e' and d(q,q') < E'. By Lenmma 2, A c NE(A') and we have H(A,A') < e. LEMMA 4. For each E > 0, there is a 6e > 0 such that if (A,B) is a proper pair and B' is a straight line interval spanning D with H(B,B') < be, then there is a proper pair (A',B') with H(A,A') < e. Proof: Suppose that the lemma is not true. Then there is an E > 0 and a sequence { (AnBn)} of proper pairs and a sequence {Bn'} of straight line intervals spanning D such that for each n, (1) H(Bn',Bn) < 1/n, (2) there is no proper pair (An'IBn') such that H(An',An) < e, and (3) there are proper pairs (An',Bn'). Choose a subsequence { (AAnB")} such that { Akj s A and { B1nQ - B. Now, we have either (Case 1) B is a straight line interval spanning D, or (Case 2) B is a point on Bd D. Downloaded by guest on September 30, 2021 258 MATHEMATICS: L. F. McAULEY PROC. N. A. S.

Inl Case 1, by Lemma 1, (A,B) is a proper pair. Applying the result of Lenima 3, there is a a > 0 such that if B' is a straight line interval spanning D with H(B',B) < 6, then there is a proper pair (A',B') with H(A,A') < e/2. Choose an integer kn so large that (1) H(Bk^,B) < a/2, (2) H(Ak,,A) < (/2, and 1/kn < 5/2. From (3) we get H(Bk,',BkD) < a/2 and with (1) this yields H(Bk,,',B) <5. Thus, there is a proper pair (A',Bk.') with H(A,A') < E/2. Using (2), we get H(A ',Akn') < -, a contradiction. In Case 2, {B} - z E Bd D. Hence, {Akj} -- x E J and, also, {B.'n} -- z. Choose a proper pair (An,Bkn) for each n. Clearly { Ak"'} -- x. For sufficiently large n, we have the contradiction that H(Ak,',Akn) < E. Proof of Theorem 1: Let 6s denote the set of all horizontal straight line intervals spanning D and for B e (B let ir(B) be the y-coordinate of points in B. Find positive integers k1,k2,k3,. . . such that kn is a divisor of kn+1 and 2/kn < 3i1/, where a,/,7 is as given by Lemma 4. For any positive integer n and j with 1 < j < kn- 1, let Bnj be that element kn-1 of 6B with 7r(Bnj) = -1 + j(2/kn). Let Fn = U Bnj and note that {Fn} is an 3=1 increasing sequence and U Fn is dense in D. n=1 Applying Lemma 4 for each n, we choose arcs Anl,An2,. .. Ankn-' such that for each j, (AnjBnj) is a proper pair and H(AnjAnj+l) < 1/n forj with 1 < j < kn- 2. kn-1 Let Cn = U An; and { Crnj be a convergent subsequence of { Cn} with { -> C. j=l Cr.1} Clearly, C is a compact continuum and f(C) = D. It remains to show that fI C is 1-1.

Suppose that f is not 1-1. Then there are points p and q in C with p ; q and f(p) = f(q) = z e Int D. Without loss of generality, we may assume that there are sequences {Arnsj and {Arntn} with {Arnsn-} AP, {Arntn} -> Aq, p E AP c C, and q e A" C C. Letting B be the horizontal straight line interval spanning D and containing z, we see that {BrnRn} -+ B and {Brnt,} B. By Lemma 1, (AP,B) and (A ,B) are proper pairs. Since f-'(z) is a totally disconnected compact set containing p and q, there are open sets U and V such that p e U, q e V, f-'(z) C U U V, and d(UV) > 0. Let K be the vertical straight line interval spanning D and containing z. We choose an integer N so large that for n > N, we have f-'(K) nfArn C U U V for all j for which either (1) j = s,,, (2) j = tn, or (3) j lies between Sn and tn. This can be done because otherwise, by compactness, we can find an x f U U V such that f(x) = z. Now, we choose M > N so that f-'(K) nf Arn,, C U and f-'(K) n Arntn C V hold for n > M. For each n > M, there are integers sn' and tn' such that (1) Sn' = t4' ± 1, (2) Sn = snt or Snt lies between sn and tn, (3) tn = tn' or tn lies between Sn and tn, (4) ft'(K) nfArlsn C U, and (5) f-'(K) n Arntn' C V. Again, without loss in generality, we assume that {Arnsn)} A. Since Sn' = tn' 4 1, we see that {Arntn'} -- A. Also tBrnn,'} -# B and {Brtn'} B. Thus (AB) is a proper pair. Setting Pn = f-l(K) n Arnsn, and qn = f-'(K) n Arntn' we have {Pn} P' E U ln Downloaded by guest on September 30, 2021 VOL. 53, 1965 MATHEMATICS: L. F. MCAULEY 259

A and {q.I -e V fn A, and f(p') = f(q') = z. Thus,f is not 1-1 on A contra- dicting the fact that (A,B) is a proper pair and completing the proof. Twist-Free Mappings and Lifting Disks.-Whyburn has developed a rather complete and remarkable theory of light open mappings of 2- onto 2- manifolds. 3, 5, 6 He has shown that such mappings are either local homeomorphisms or locally like the mapping w = Zn on {Iz zf = 1}, i.e., a twisting of a special type. His work motivates our definition below. Suppose that f:X -> Y is light and open. We shall say that f is twist-free if and only if for each simple closed curve J in Y and p e f-1(J), J can be lifted to a simple closed curve J1 in X with p in J1, i.e., there is a simple closed curve J1 ill X such that p e Ji and ft J1 is a homeomorphism of J1 onto J. Now consider the following theorem which shows that under certain simple restrictions on light open mappings which are twist-free, disks can be lifted. THEOREM 2. Suppose that each of X and Y is compact, X is locally connected, f is a light open mapping of X onto Y, and f is twist free. Furthermore, the set of points x in X such that f is not a local homeomorphism at x is the kernel Kf off. Then Y can be lifted to X. In fact, X is the union of a collection of sets such that f re- stricted to each of these is a homeomorphism of that set onto Y. First, we establish a lemma. LEMMA 5. Suppose that S is the set of all points x in X such that f is a local homeo- morphism at x and that S is nonempty. Then for each component C of S, ft C is a homeomorphism. Proof: It is clear that each of S and f(S) is open. Assume that there exists some component C of S such that fJ C is not 1-1. Let xy denote an arc in C such that fI xy - y is 1-1 and f(x) = f(y). It follows that f(xy) is a simple closed curve J. And, J can be lifted by hypothesis to J1 in X which contains x. Since f is locally 1-1 on C, C D xy, and x e J1, we have that J1 D xy - y and therefore xy. This contradicts the fact that fI J1 is 1-1. Hence, ft C is 1-1 and consequently a homeomorphism since f is open. Proof of Theorem 2: Since f is open, the kernel K, of f (the set of all x in X such that f-lf(x) = x) is closed. And, f(Kf) is closed in Y. There are at most a countable number of components K1,K2,..., of Y - f(Kf). For each K,, choose one component Ci of ft'(Ki) such that f(Ci) = Ki. Since f is twist-free, it follows that ft Ci is 1-1 on Ci. Now, let C = U Ci U Kf. Since each Ci is open and ft Kf is 1-1 on K,, it i=l follows that ft C is 1-1. Furthermore, C is closed and fj C is a homeomorphism of C onto Y. It should be clear that the conclusion of Theorem 2 follows by considering all choices for Ci and consequently for C, but f restricted to each such C is a homeo- morphism of C onto Y. Thus, X is a union of sets (not necessarily disjoint) such that each of these maps homeomorphically onto Y under f. Question: Does the conclusion of Theorem 2 remain true if the condition that the set of points x in X where f is not a local homeomorphism at x be the kernel of f is omitted from the hypothesis? Consider the following . Conjecture: Suppose that f: X -> Y is a finite-to-one open mapping of a compact locally connected continuum onto a Y. is {n- J disk Furthermore, f twist-free. Downloaded by guest on September 30, 2021 260 MATHEMATICS: R. E. GOMORY PROC. N. A. S.

Then X contains a disk D so that fj D is a homeomorphism of D onto Y. These results indicate a direction which may lead to a characterization of a certain class of light open mappings as generalized fiber spaces with totally disconnected fibers. The author wishes to express deepest appreciation to Professor G. T. Whyburn for stimulating conversations and encouragement. * Initial results were presented to the American Mathematical Society, January 1963, and re- search for these was accomplished at the University of Virginia while the author held an ONR Research Fellowship. 1 Hu, Sze-Tsen, Homotopy Theory (New York: Academic Press, 1959). 2 Browder, Felix E., "Covering spaces, fiber spaces, and local homeomorphisms," Duke Math. J., 21, 329-336 (1954). 3Whyburn, G. T., Analytic (Providence: American Mathematical Society, 1942). 4Floyd, E. E., "Some characterizations of interior maps," Ann. Math., 51, 571-575 (1950). 5 Whyburn, G. T., Topological Analysis (Princeton: Princeton University Press, 1958). 6Whyburn, G. T., "Open mappings on 2-dimensional manifolds," J. Math. Mech., 10, 181- 198 (1961). Related references: Anderson, R. D., "A characterization of the universal curve and a proof of its homogeneity," Ann. Math., 67, 313-324 (1958); Cernavskii, A. V., "Finitely multiple open mappings of manifolds," Soviet Math. Doklady, 4, 946-949 (1963); Dyer, E., and M.-E. Hamstrom, "Completely regular mappings," Fund. Math., 45, 103-118 (1957); Whyburm, G. T., "On sequences and limiting sets," Fund. Math., 25, 408-426 (1935).

ON THE RELATION BETWEEN INTEGER AND NONINTEGER SOLUTIONS TO LINEAR PROGRAMS* BY R. E. GOMORY THOMAS J. WATSON RESEARCH CENTER, YORKTOWN HEIGHTS, NEW YORK Communicated by R. Courant, December 22, 1964 We will refer to the ordinary linear programming problem maximize zi = cx (1) Ax = b, x > 0 as problem P1. In (1) b is an integer m-vector, c is an m + n vector, and A is an m X (m + n) integer matrix. x is an m + n vector, all of whose components are required to be nonnegative. We assume that A is of the form (A', I) with I an m X m identity matrix, so that in (1) Ax = b is equivalent to the m inequalities in n variables A 'x' < b. We will say that x is feasible if it satisfies the equality and non- negativity conditions of (1) and optimal if it also maximizes. A problem closely related to P1 is the integer programming problem P2 which is P1 with the added condition that the components of x be integers. Because of the comparative ease with which P1 is solved' and the comparative difficulty of P2,2, it is natural to consider getting from the solution of P1 to the solution to P2 by some sort of a "rounding" process through which the noninteger components of the x solving P1 are rounded either up or down to produce a solution to P2. This Downloaded by guest on September 30, 2021