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Cell-like maps that are shape equivalences


Author: Jung-In K. Choi
Journal: Proc. Amer. Math. Soc. 108 (1990), 1011-1018
MSC: Primary 54C56; Secondary 54C55, 57N25
DOI: https://doi.org/10.1090/S0002-9939-1990-1038759-9
MathSciNet review: 1038759
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Abstract: Let $ f:X' \to X$ be a cell-like map between metric spaces and set $ {N_f} = \{ x \in X:{f^{ - 1}}(x) \ne {\text{point\} }}$. Even if $ {N_f} \subset \bigcup\nolimits_{n = 1}^\infty {{B_n}} $, where each $ {B_n}$ is closed and each $ f\vert{f^{ - 1}}({B_n}):{f^{ - 1}}({B_n}) \to {B_n}$ is hereditary shape equivalence, $ f$ may not be a hereditary shape equivalence. Conditions are placed on the $ {B_n}$ 's to assure that $ f$ is a hereditary shape equivalence. For example, if $ {N_f} \subset \bigcup\nolimits_{n = 1}^\infty {{B_n}} $, where $ {B_n}$ is closed for each $ n = 1,2, \ldots ,f\vert{f^{ - 1}}({B_n}):{f^{ - 1}}({B_n}) \to {B_n}$ is a hereditary shape equivalence, and $ {B_n}$ has arbitrary small neighborhoods whose boundaries miss $ \bigcup\nolimits_{i = 1}^\infty {{B_i}} $ then $ f$ is a hereditary shape equivalence. An immediate consequence is that if $ \{ {B_n}\} _{n = 1}^\infty $ is a pairwise disjoint null-sequence and each $ f\vert{f^{ - 1}}({B_n})$ is a hereditary shape equivalence, then $ f$ is a hereditary shape equivalence. Previously G. Kozlowski showed that if $ \{ {f^{ - 1}}({B_n})\} _{n = 1}^\infty $ is a pairwise disjoint null-sequence and each $ f\vert{f^{ - 1}}({B_n})$ is a hereditary shape equivalence, then $ f$ is a hereditary shape equivalence, which can be obtained as an immediate corollary of one of our results.


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Additional Information

DOI: https://doi.org/10.1090/S0002-9939-1990-1038759-9
Keywords: Cell-like map, hereditary shape equivalence, ANR, slice-trivial, nonde-generacy set, null-sequence
Article copyright: © Copyright 1990 American Mathematical Society

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