The homotopies of admissible multivalued mappings

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1 Cent. Eur. J. Math. 10(6) DOI: /s Central European Journal of Mathematics The homotopies of admissible multivalued mappings Research Article Mirosław Ślosarski 1 1 Institute of Electronics, Department of Electronics and Information Technology, Technical University of Koszalin, Śniadeckich 2, Koszalin, Poland Received 11 March 2012; accepted 28 May 2012 Abstract: Certain properties of homotopies of admissible multivalued mappings shall be presented, along with their applications as the tool for examining the acyclicity of a space. MSC: 54C60, 55P57, 54C20 Keywords: Absolute neighborhood multi-retracts Countable dimension Trivial shape TSA-extension TSA-homotopy TSA-map Versita Sp. z o.o. 1. Introduction A very useful notion of homotopy has been functioning in mathematical literature for many years. It is a very good topological tool for examining the acyclicity of a space. We know that every set that is contractible to a point is acyclic. There exist compact sets of trivial shape (contractible to a point in each of its own open neighborhood), acyclic in the sense of Čech homology, that are not contractible to a point. In 1980, A. Suszycki in the paper [6] introduced the notion of multi-homotopy. He proved that a compact, multi-contractible to a point set is acyclic in the sense of Čech homology. He gave an example of a compact set that is multi-contractible to a point but does not have a trivial shape, in particular, it is not contractible to a point. In 1976, L. Górniewicz introduced the notion of homotopy of admissible mappings. In this paper a homotopy on some class of admissible mappings shall be defined and it will be proven that it is a more general and more flexible topological tool than the Suszycki multi-homotopy. slosmiro@gmail.com 2187

2 The homotopies of admissible multivalued mappings 2. Preliminaries Throughout this paper all topological spaces are assumed to be metric. Let H be the Čech homology functor with compact carriers and coefficients in the field of rational numbers Q, from the category of Hausdorff topological spaces and continuous maps to the category of graded vector spaces and linear maps of degree zero. Thus H (X) = H q (X)} is a graded vector space, H q (X) being the q-dimensional Čech homology group with compact carriers of X. For a continuous map f : X Y, H (f) is the induced linear map f = f q }, where f q : H q (X) H q (Y ) [3]. A space X is acyclic if (i) (ii) X is non-empty, H q (X) = 0 for every q 1, and (iii) H 0 (X) Q. A continuous mapping f : X Y is called proper if for every compact set K Y the set f 1 (K) is non-empty and compact. A proper map p: X Y is called Vietoris provided for every y Y the set p 1 (y) is acyclic. A Vietoris map p: X Y is called a cell-like map provided for every y Y the set p 1 (y) is of trivial shape. The notion of shape is understood in the sense of Borsuk, see [2]. It can be proven that a compact set of trivial shape is acyclic. Let X and Y be two spaces and assume that for every x X a non-empty subset φ(x) of Y is given. In such case we say that φ : X Y is a multivalued mapping. For a multivalued mapping φ : X Y and a subset U Y, we let φ 1 (U) = x X : φ(x) U}. If for every open U Y the set φ 1 (U) is open, then φ is called an upper semi-continuous mapping; we shall write that φ is u.s.c. Let φ : X Y be a multivalued map. A pair (p, q) of single-valued, continuous maps is called a selected pair of φ (written (p, q) φ) if there exists a metric space Z such that the following two conditions are satisfied: (i) p: Z X is a Vietoris map, (ii) q(p 1 (x)) φ(x) for any x X, where q: Z Y. Definition 2.1. Let φ : X Y be a multivalued map. The map φ is called admissible provided there exists a selected pair (p, q) of φ. The map φ is called strongly admissible (s-admissible) provided there exists a selected pair (p, q) of φ such that for each x X, q(p 1 (x)) = φ(x) (written (p, q) = φ). Proposition 2.2 ([3]). If φ : X Y and ψ : Y T are admissible, then the composition ψ φ : X T is admissible and for every (p 1, q 1 ) φ and (p 2, q 2 ) ψ there exists a pair (p, q) ψ φ such that q 2 p 1 2 q 1 p 1 1 = q p 1. A space X is of countable dimension if where X n is compact and dim X n < for all n. X = X n, (1) n=1 Definition 2.3. We say that an admissible map φ : X Y is of TSA type (φ TSA) if there exists a selected pair (p, q) φ and a metric space Z such that p: Z X is a cell-like map. We say that an s-admissible map φ : X Y is of STSA type (φ STSA) if there exists a selected pair (p, q) = φ and a metric space Z such that p: Z X is a cell-like map. 2188

3 M. Ślosarski Proposition 2.4. Let φ : X Y be of TSA (STSA) type, f : P X, g: Y T be continuous maps and A X a non-empty set. Then the following conditions are satisfied: g φ TSA(STSA), φ f TSA(STSA), φ A TSA(STSA), where φ A : A Y, φ A (x) = φ(x) for each x A : We have the following diagram: X r Z s Y g T, where r : Z X is a cell-like map and s: Z Y is a continuous map such that (r, s) φ ((r, s) = φ). Let p = r and q = g s. Then (p, q) g φ ((p, q) = g φ) : We have the following diagram: P f X r Z 1 s Y, where r : Z 1 X is a cell-like map and s: Z 1 Y is a continuous map such that (r, s) φ ((r, s) = φ). Let Z = (y, z) P Z 1 : f(y) = r(z)}, p: Z P, p(y, z) = y for each (y, z) Z, and q: Z Y, q(y, z) = s(z) for each (y, z) Z. Then p: Z P is a cell-like map, since for each y P, p 1 (y) = y} r 1 (f(y)) = r 1 (f(y)) and (p, q) φ f ((p, q) = φ f) : It is sufficient to apply for f = i, where i: A X is an inclusion. Theorem 2.5. Let Y ANR (Y AR) and let X be a continuum. Assume that the set A X is non-empty, closed and of finite or countable dimension. Then a TSA-map φ : A Y has a TSA-extension onto U (respectively, X) such that φ U\A : U \ A Y (resp. φ X\A : X \ A Y ), φ U\A (x) = φ(x) (resp. φ X\A (x) = φ(x)) for each x U \ A (resp. x X \ A) is single-valued, where U A is some open set in X. Assume that Y ANR. Choose (p, q) φ such that p: Z A is a cell-like map. We observe that, from the assumption, the space Z is a continuum. Let h: Z Q be an embedding of the metric space Z in the Hilbert cube Q. Obviously Q AR. Let ψ : A Q be a map given by the formula ψ(x) = h(p 1 (x)) for each x A. It is clear that for each x A, ψ(x) is compact and of trivial shape, so there exists u.s.c. extension ψ : X Q such that the map ψ X\A : X \ A Q, ψ X\A (x) = ψ(x) for each x X \ A, (2) is single-valued, see [5]. 2189

4 The homotopies of admissible multivalued mappings Let q: V Y be a continuous extension of q h 1, where V Q is an open set such that h(z) V. We define a TSA-extension φ : U Y of φ as φ(x) for x A, φ(x) = q( ψ(x)) for x U \ A, where U = ψ 1 (V ). If Y AR then there exists an extension q: Q Y of q h 1. Hence and from (2) the proof is complete. Definition 2.6 ([3]). Two admissible maps φ, ψ : X Y are called homotopic (written φ ψ) provided there exists an admissible map χ : X [0, 1] Y such that χ(x, 0) φ(x) and χ(x, 1) ψ(x) for each x X. Theorem 2.7 ([3]). Let φ, ψ : X Y be two admissible maps. Then φ ψ implies that there exist selected pairs (p, q) φ and (p, q) ψ such that q p 1 = q p 1. Definition 2.8. Let X Q be a closed subset, where Q is the Hilbert cube. We say that X is movable in Q provided every neighborhood U of X admits a neighborhood U of X, U U, such that for every neighborhood U of X, U U, there exists a homotopy H : U [0, 1] U with H(x, 0) = x and H(x, 1) U for any x U. Proposition 2.9 ([3]). Let X be a metric space. If Id = g, where Id: X X is the identity map and g: X X is a constant map, then X is acyclic. Proposition Let X = A B, where A, B X are closed sets in X and A B. Let φ A : A Y and φ B : B Y be multivalued u.s.c. maps such that for each x A and for each x B, φ A (x) and φ B (x) are non-empty and compact sets. If φ A (x) φ B (x) for each x A B, then the map φ : X Y, φ A (x) for x A \ B, φ(x) = φ B (x) for x B, is u.s.c. Let U Y be an open set. We show that φ 1 (U) X is open. From the assumption we have A B φ 1 B (U) A B φ 1 A (U). (3) Let x A B φ 1 (U) = A B φ 1 B (U), then there exists an open ball K(x, r 1) X, r 1 > 0, such that and there exists an open ball K(x, r 2 ) X, r 2 > 0, such that Let r = minr 1, r 2 }. From (3) (5) we get K(x, r 1 ) A φ 1 A (U), (4) K(x, r 2 ) B φ 1 B (U). (5) K(x, r) = (K(x, r) (A \ B)) (K(x, r) B) φ 1 (U). If x (A \ B) φ 1 (U) or x (B\A) φ 1 (U), then the proof is obvious. 2190

5 M. Ślosarski 3. Main result Let φ, ψ : X Y. If ψ(x) φ(x) for every x X, then we shall write ψ φ. Definition 3.1. Let φ, ψ : X Y, φ, ψ TSA, see Definition 2.3. We say that an admissible map χ : X [0, 1] Y is a TSA-homotopy connecting φ with ψ if the following conditions are satisfied: χ TSA, χ(x, 0) = φ(x) and χ(x, 1) = ψ(x) for each x X. In this case the maps φ and ψ are said to be TSA-homotopic (φ TSA ψ). It is clear that [3, proof of Theorem 40.11, p. 203] Theorem 3.2. Let φ, ψ : X Y be two TSA-maps. Then φ TSA ψ implies that there exist selected pairs (p, q) φ and (p, q) ψ such that where p, p are cell-like maps. q p 1 = q p 1, Proposition 3.3. If in the condition the identities are replaced by inclusions, that is, χ(x, 0) φ(x) and χ(x, 1) ψ(x) for each x X. Then it will result in an equivalent definition, see Definition 2.6. We define a homotopy χ : X [0, 1] Y as φ(x) for 0 t 1 3, 1 χ(x, t) = χ(x, 3t 1) for 3 < t < 2 3, 2 ψ(x) for 3 t 1, where χ : X [0, 1] Y is a TSA-homotopy from Definition 3.1. Then we have χ(x, 0) = φ(x) and χ(x, 1) = ψ(x) for each x X. We show that χ TSA. Let (p, q) χ, where p: Z X [0, 1] is a cell-like map. We define η: X [0, 1] Z as p 1 (x, 0) for 0 t 1 3, η(x, t) = p 1 1 (x, 3t 1) for 3 < t < 2 3, p 1 2 (x, 1) for 3 t 1. The map η is u.s.c., see Proposition 2.10, and for each (x, t) X [0, 1] the set η(x, t) is compact and of trivial shape. We observe that for each (x, t) X [0, 1], q(η(x, t)) χ(x, t). Hence χ TSA and the proof is complete. 2191

6 The homotopies of admissible multivalued mappings The TSA-homotopy is, obviously, a particular case of the homotopy defined in Definition 2.6, but, in our opinion, with regard to the application of trivial shape in the definition, see Definition 2.3, it is of a topological, and not only homological, character. It is easy to verify that it is reversible and symmetric, but it does not have to be transitive. Example 3.4. Let R n+1 denote the (n + 1)-dimensional Euclidean space and let S n = x R n+1 : x = 1}. Let φ, ψ, η: S n S n, φ(x) = x, ψ(x) = S n, η(x) = x 0 for each x S n, where x 0 S n is a stationary point. Notice that φ TSA ψ and ψ TSA η, where χ 1, χ 2 : S n [0, 1] S n, are respectively homotopies connecting the mappings φ and ψ as well as ψ and η by the following formulas: x for t < 1 2, S n for t 1 2, χ 1 (x, t) = S n for t 1 χ 2 (x, t) = 2, x 0 for t > 1 2. We show that φ and η are not homotopic. Let us assume, to the contrary, that φ TSA η. Then from Theorem 2.7, Id = η where η is a constant map, but it is not possible, since S n is not acyclic, see Proposition 2.9. Definition 3.5. Let X = A B and A B. Let φ A : A Y, φ B : B Y be admissible mappings. The mappings φ A and φ B will be called associated (we write φ A φ B ) if there exist selected pairs (p 1, q 1 ) φ A (p 1 : Z A A, q 1 : Z A Y ), (p 2, q 2 ) φ B (p 2 : Z B B, q 2 : Z B Y ) such that the following conditions are satisfied: p 1 1 (x) p 1 2 (x) for each x A B, or p 1 2 (x) p 1 1 (x) for each x A B, q 1 (z) = q 2 (z) for each z Z A Z B, Z = Z A Z B is a metric space. If, additionally, in the above selective pairs the mappings p 1 : Z A A, p 2 : Z B B are of cell-like type, it will be said that the association of mappings φ A and φ B is of TSA type and it will be written (φ A φ B ) TSA. We say that an u.s.c. map ψ : X Y is of trivial shape if for each x X, ψ(x) is compact and of trivial shape. (6) Proposition 3.6. Let X = A B and A B. Let φ A : A Y, φ B : B Y be mappings of trivial shape. If φ A (x) φ B (x) for every x A B or φ B (x) φ A (x) for every x A B, then (φ A φ B ) TSA. Assume that for each x A B, φ A (x) φ B (x). Let Γ A be the graph of φ A and let Γ B be the graph of φ B. Let Z = Γ A Γ B. The set Z is a metric space, since Z X Y. We define selective pairs (p 1, q 1 ) φ A (p 1 : Γ A A, q 1 : Γ A Y ) and (p 2, q 2 ) φ B (p 2 : Γ B B, q 2 : Γ B Y ) given by formulas p 1 (x, y) = x, q 1 (x, y) = y, (x, y) Γ A, p 2 (s, t) = s, q 2 (s, t) = t, (s, t) Γ B. Let x A B. Then we have p 1 1 (x) = x} φ A (x) x} φ B (x) = p 1 2 (x). It is clear that for each (z 1, z 2 ) Γ A Γ B, q 1 (z 1, z 2 ) = q 2 (z 1, z 2 ). In the case of φ B (x) φ A (x), x A B, the proof is analogous. The same fact can be proven when the mappings are acyclic. 2192

7 M. Ślosarski Proposition 3.7. Let X = A B, A B and A, B be closed in X. Assume that φ A, φ B TSA and φ A (x) = φ B (x) for each x A B. Let φ : X Y be a map given by φ A (x) for x A, φ(x) = φ B (x) for x B. The map φ is of TSA type if and only if (φ A φ B ) TSA. If φ TSA then it is obvious that (φ A φ B ) TSA. Assume now that (φ A φ B ) TSA, then there exist (p 1, q 1 ) φ A (p 1 : Z A A, q 1 : Z A Y ) and (p 2, q 2 ) φ B (p 2 : Z B B, q 2 : Z B Y ) such that p 1, p 2 are cell-like maps and the conditions are satisfied. We define a map η: X Z as η(x) = p 1 1 (x) for x A \ B, p 1 2 (x) for x B, if p 1 1 (x) p 1(x) for each x A B, and 2 η(x) = p 1 1 (x) for x A, p 1 2 (x) for x B \ A, if p 1 2 (x) p 1 1 (x) for each x A B. The map η is of trivial shape, see (6), Proposition Let q: Z A Z B Y be a continuous map given by q1 (z) for z Z A, q(z) = q 2 (z) for z Z B. We observe that for each x X, q(η(x)) φ(x). Hence φ TSA. Let X be a continuum. If a map φ : X Y is of TSA type, then there exists (p, q) φ (p: Z X, q: Z Y ) such that p is a cell-like map. It is clear that Z is a compact metric space and it can be assumed that it is a subset of the Hilbert cube Q. Proposition 3.8. Let X be a continuum of finite or countable dimension and Y be a metric space. Let X = A B, where A B and A, B are closed in X, and let φ A : A Y be of TSA type. Assume that there exists (p, q) φ A (p: Z A, q: Z Y, Z Q) such that p is a cell-like map and there exists an extension q: Q Y of q. Then there exists φ B : B Y such that (φ A φ B ) TSA. From assumption there exists (p, q) φ A (p: Z A, q: Z Y, Z Q) such that p is a cell-like map and there exists an extension q: Q Y of q. Let ψ : A B Z Q be a map given by the formula ψ(x) = p 1 (x) for each x A B. From Theorem 2.5 there exists an extension ψ : B Q such that the map ψ B\(A B) : B \ (A B) Q is single-valued. Hence the map ψ is of trivial shape. Let φ B : B Y be a map given by φ B (x) = φa (x) for x A B, q( ψ(x)) for x B \ A. The map φ, see Proposition 3.7, is of TSA type, so the proof is complete. 2193

8 The homotopies of admissible multivalued mappings Example 3.9. Let R be the set of reals and let A = (, b], B = [a, ), where a, b R and a < b. Let f : A R, g: B R be continuous maps. Assume that f(x) g(x) for each x A B. We define two admissible maps φ A : A R, φ B : B R as φ A (x) = f(x) for x A \ B, f(x), g(x)} for x A B, φ B (x) = f(x), g(x)} for x A B, g(x) for x B \ A. We observe that for each x A B, φ A (x) = φ B (x). The mappings φ A and φ B are not associated because the mapping φ is not admissible, see Proposition 3.7. Let φ : X Y be a TSA-map and C Y a non-empty and compact set. The notation φ TSA C (7) will stand for a TSA-homotopy joining the mapping φ with the mapping C : X Y given by C(x) = C for each x X. The following fact, in its single-valued version, is very well known in mathematical literature. Proposition Let K n+1 = x R n+1 : x 1} and let Y be a metric space. A TSA-map φ : S n Y has a TSA-extension φ : K n+1 Y if and only if there exists a non-empty and compact set C Y such that φ TSA C, see (7), and (χ C) TSA, where χ is a homotopy connecting φ with C and C : K n+1 1} Y is a map C(x) = C for each (x, 1) K n+1 1}. Assume that we have a TSA-extension φ : K n+1 Y. We define a TSA-homotopy χ : S n [0, 1] Y as χ(x, t) = φ((1 t)x + tx 0 ) for each (x, t) S n [0, 1], see 2.4.2, where x 0 S n is a stationary point. It is clear that χ(x, 0) = φ(x) = φ(x), χ(x, 1) = φ(x 0 ) = φ(x 0 ) for each x S n. We show that (χ C) TSA, where C : K n+1 1} Y is a map, C(x, 1) = φ(x 0 ) for each x K n+1. Let (p, q) φ, where p: Z K n+1 is a cell-like map, q: Z Y is a continuous map and let Z S n [0,1] = (x, t, z) S n [0, 1] Z : (1 t)x + tx 0 = p(z) } K n+1 [0, 1] Z, Z K n+1 1} = K n+1 1} p 1 (x 0 ) K n+1 [0, 1] Z, p 1 : Z S n [0,1] S n [0, 1], q 1 : Z S n [0,1] Y, p 1 (x, t, z) = (x, t), q 1 (x, t, z) = q(z), (x, t, z) Z S n [0,1], p 2 : Z K n+1 1} K n+1 1}, q 2 : Z K n+1 1} Y, p 2 (x, 1, z) = (x, 1), q 2 (x, 1, z) = q(z), (x, 1, z) Z K n+1 1}. We observe that (p 1, q 1 ) χ, (p 2, q 2 ) C and the conditions are satisfied. For each (x, t) S n [0, 1], p 1 1 (x, t) = (x, t)} p 1 ((1 t)x + tx 0 ) and for each (x, 1) K n+1 1}, p 1 2 (x, 1) = (x, 1)} p 1 (x 0 ), 2194

9 M. Ślosarski so p 1 and p 2 are cell-like maps. Assume now that there exists a TSA-homotopy χ : S n [0, 1] Y such that χ(x, 0) = φ(x), χ(x, 1) = C Y for each x S n and (χ C) TSA. We define an extension map φ : K n+1 Y as From Proposition 3.7, φ TSA and the proof is complete. C for 0 x 1 2, φ(x) = ( ) x χ x, 2 2 x 1 for x 1. 2 A theorem on the extension of homotopies, known from mathematical literature in its single-valued version, is another fact that will be formulated here. Theorem Let H : A [0, 1] Y be a TSA-map, see Definition 2.3, where Y ANR, X is a continuum of finite or countable dimension, and A is a closed set in X. Let φ : X 0} Y be a TSA-extension of H A 0} such that (H φ) TSA, see Proposition 3.8, where H A 0} : A 0} Y is a map, H A 0} (x, 0) = H(x, 0) for each x A. Then there exists a TSA-extension H : X [0, 1] Y of H such that H(x, t) = H(x, t) for each (x, t) A [0, 1], H(x, 0) = φ(x) for each x X. (8) We define a map χ : (A [0, 1]) (X 0}) Y as χ(x, t) = H(x, t) for (x, t) A [0, 1], φ(x) for (x, 0) X 0}. From Proposition 3.7, χ TSA. From Theorem 2.5 there exists a TSA-extension χ : U Y of χ, where U X [0, 1] is an open set such that (A [0, 1]) (X 0}) U. There exists a continuous map f : X [0, 1] U such that for each (x, t) (A [0, 1]) (X 0}), f(x, t) = (x, t), see [1, (8.2), p. 94]. Let H : X [0, 1] Y be a map given by H(x, t) = χ(f(x, t)) for each (x, t) X [0, 1]. From the map H is the TSA-extension of H. If (x, t) A [0, 1], then H(x, t) = χ(f(x, t)) = χ(x, t) = H(x, t) and if (x, 0) X 0}, then H(x, 0) = χ(f(x, 0)) = χ(x, 0) = φ(x). We observe that if a map H : X [0, 1] Y is a TSA-homotopy such that the conditions (8) are satisfied, then from Proposition 3.7, (H φ) TSA. Homotopy is a very good tool for examining the acyclicity of a space. The following definition will be introduced. 2195

10 The homotopies of admissible multivalued mappings Definition Let X be a metric space and let A X be a non-empty and compact set. We say that the space X is TSA-contractible to A if there exists a TSA-homotopy χ : X [0, 1] X such that χ(x, 0) = x and χ(x, 1) = A for each x X. If the set A = x 0 }, where x 0 X, is a stationary point, then we say that the space X is TSA-contractible to x 0 (TSA-contractible). Certainly every set X is TSA-contractible to itself. Proposition Let X be a metric space and A X non-empty and compact. If the space X is TSA-contractible to A, then there exists a monomorphism H (X) H (A). From Theorem 3.2 there exists (p, q) χ X 1} such that Id = q p 1, where p: Z X [0, 1] is a cell-like map, χ X 1} : X 1} X, χ X 1} (x, 1) = χ(x, 1) for each x X, and Id: X X is the identity map. We have the following diagram: q p 1 H (X) H (A) H (X), i where i: A X is an inclusion and q: Z A is given by q(z) = q(z) for each z Z. We get Id = q p 1 = ( i qp 1) = i q p 1. Hence a map q p 1 : H (X) H (A) is a monomorphism and the proof is complete. It follows from Proposition 3.13 that a set which is non-acyclic, is not TSA-contractible to an acyclic subset. In particular, the sphere is not TSA-contractible to a point, that is, the set A that consists of one point. From the last proposition it also follows that the sphere S n is not TSA-contractible to the set A = S m, m < n. It should be noticed that the compact set X of trivial shape is TSA-contractible. As a homotopy χ : X [0, 1] X, the mapping given by the following formula can be taken: x for t < 1 2, χ(x, t) = X for t = 1 2, x 0 for t > 1 2, where x 0 X is a stationary point. A more general fact can be proven. Proposition Let Z be a continuum of trivial shape and let X be a continuum. Let φ : X Z be a map of trivial shape. If there exists a continuous map f : Z X such that f(φ(x)) = x} for each x X, (9) then X is TSA-contractible. 2196

11 M. Ślosarski The map χ : X [0, 1] X given by x for t < 1 2, χ(x, t) = X for t = 1 2, x 0 for t > 1 2, is a TSA-homotopy, where x 0 X is a stationary point. Let ψ : X [0, 1] Z be φ(x) for t < 1 2, ψ(x, t) = Z for t = 1 2, φ(x 0 ) for t > 1 2. Then for each (x, t) X [0, 1], f(ψ(x, t)) = χ(x, t) and from the proof is complete. Further some facts will be proven and examples presented, which show that TSA-contractibility is more general than contractibility. Proposition Let Z be a continuum, contractible space and let X be a continuum. If there exist a continuous map f : Z X, a TSA-map φ : X Z such that (9) holds, then X is TSA-contractible. Let x 0 X be a stationary point. We define a TSA-homotopy χ : X [0, 1] X as χ(x, t) = f(h(φ(x), t)) for each (x, t) X [0, 1], where the map H : Z [0, 1] Z is a homotopy such that H(z, 0) = z, H(z, 1) = z 0 for each z Z and z 0 φ(x 0 ). The map χ TSA, since χ = f H ψ, see 2.4.1, where X [0, 1] ψ Z [0, 1] H Z f X, ψ(x, t) = φ(x) t} for each (x, t) X [0, 1]. It is known, see [1], that every compact space of trivial shape is movable, see Definition 2.8. Example Let X Q be a compact and non-movable space such that there exists a cell-like map p: Q X, where Q is the Hilbert cube, see [4]. From Propositions 3.15 or 3.14 the space X is TSA-contractible, where f = p and φ = p 1. The space X is not movable so it cannot be contractible to a point, see [1]. Proposition Let Z be a contractible continuum and let X be a continuum. Let φ : Z X be a map of trivial shape, see (6). If there exists a continuous map f : X Z such that x φ(f(x)) for each x X, then X is TSA-contractible. 2197

12 The homotopies of admissible multivalued mappings Fix some point x 0 X. We define a TSA-homotopy χ : X [0, 1] X as x for t < 1 3, χ(x, t) = φ ( H(f(x), 3t 1) ) 1 for 3 t 2 3, x 0 for t > 2 3, where the map H : Z [0, 1] Z is a homotopy such that H(z, 0) = z, H(z, 1) = z 0 for each z Z, z 0 = f(x 0 ). Example Let X Q be a compact and non-movable space such that there exists a cell-like map p: X Q, where Q is the Hilbert cube, see [7]. From Proposition 3.17 the space X is TSA-contractible, where f = p and φ = p 1. The space X is not movable so it cannot be contractible to a point, see [1]. Another fact shows that a TSA-homotopy is a more flexible tool than the Suszycki multi-homotopy, see [6]. Proposition Let f : X Y be a continuous map. The metric space X is TSA-contractible if and only if the graph Γ f of f is TSA-contractible. Assume that X is TSA-contractible. We have the following diagram: Γ f [0, 1] h X [0, 1] H X g Γ f, where h(x, y, t) = (x, t) for each (x, y, t) Γ f [0, 1], H is a TSA-homotopy such that H(x, 0) = x, H(x, 1) = x 0 X and g(x) = (x, f(x)) for each x X. We define a TSA-homotopy χ : Γ f [0, 1] Γ f as χ(x, y, t) = g(h(h(x, y, t))) for each (x, y, t) Γ f [0, 1]. From Proposition 2.4, χ is a TSA-map. Assume now that Γ f is TSA-contractible. We have the following diagram: X [0, 1] h Γ f [0, 1] H Γ f g X, where h(x, t) = (x, f(x), t) for each (x, t) X [0, 1], H is a TSA-homotopy such that H(x, y, 0) = (x, y), H(x, y, 1) = (x 0, y 0 ), y 0 = f(x 0 ) and g(x, y) = x for each (x, y) Γ f. We define a TSA-homotopy χ : X [0, 1] X as χ(x, t) = g(h(h(x, t))) for each (x, t) X [0, 1]. From Proposition 2.4, χ is a TSA-map and the proof is complete. Open problem. Is there a compact set that is acyclic in the sense of Čech homology, but that is not TSA-contractible? 2198

13 M. Ślosarski References [1] Borsuk K., Theory of Retracts, Monogr. Mat., 44, PWN, Warsaw, 1967 [2] Borsuk K., Theory of Shape, Monogr. Mat., 59, PWN, Warsaw, 1975 [3] Górniewicz L., Topological Fixed Point Theory of Multivalued Mappings, Topol. Fixed Point Theory Appl., 4, Springer, Dordrecht, 2006 [4] Keesling J.E., A non-movable trivial-shape decomposition of the Hilbert cube, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., 1975, 23(9), [5] Suszycki A., On extensions of multivalued maps, Bull. Acad. Polon. Sci. Sér. Sci. Math., 1979, 27(2), [6] Suszycki A., Retracts and homotopies for multimaps, Fund. Math., 1983, 115(1), 9 26 [7] Taylor J.L., A counterexample in shape theory, Bull. Amer. Math. Soc., 1975, 81,