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On the product and composition of universal mappings of manifolds into cubes


Author: W. Holsztyński
Journal: Proc. Amer. Math. Soc. 58 (1976), 311-314
MSC: Primary 55C20; Secondary 57A15
MathSciNet review: 0407832
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Abstract: A map $ f:X \to Y$ is said to be universal iff for every $ g:X \to Y$ there exists $ x \in X$ such that $ f(x) = g(x)$.

Let $ {M_t},t \in T$, and $ {M^n}$ be orientable compact manifolds (in general with boundary). Let $ \dim {M^n} = n$ and let $ {Q_t}$, be a cube with $ \dim {Q_t} = \dim {M_t}$. Let $ {f_t}:{M_t} \to {Q_t},{f_0}:{M^n} \to {I^n}$ and $ {f_k}:{I^n} \to {I^n}$ be universal mappings for $ t \in T$ and $ k = 1,2, \ldots $ Then

(1.8) Theorem. The product map $ \prod\nolimits_{t \in T} {{f_t}:{M_t} \to \prod\nolimits_{t \in T} {{Q_t}} } $ is universal.

(2.1) Theorem. The composition $ {f_s} \circ {f_{s - 1}} \circ \cdots \circ {f_1}:{M^n} \to {I^n}$ is a universal map for $ s = 1,2, \ldots $

(2.2) Theorem. The limit $ X$ inverse sequence

$\displaystyle {I^n}\xleftarrow{{{f_1}}}{I^n}\xleftarrow{{{f_2}}}{I^n}\xleftarrow{{{f_3}}} \cdots $

is an $ n$-dimensional space with the fixed point property.

Some ``counterexamples'' are furnished. Also the following variant of Proposition (1.5) from [3] is given:

Theorem A (Proposition (1.5) of [3]). Let $ X$ be a compact space of (covering) dimension $ \leq n$. Then $ f:X \to {I^n}$ is a universal mapping iff the element $ {f^\ast }({e^n})$ of the $ n$th Čech cohomology group $ {H^n}(X,{f^{ - 1}}({S^{n - 1}});{\mathbf{Z}})$ is different from 0 for a generator $ {e^n}$ of $ {H^n}({I^n},{S^{n - 1}};{\mathbf{Z}})$ where $ ({S^{n - 1}} = \partial {I^n})$.


References [Enhancements On Off] (What's this?)

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Additional Information

DOI: http://dx.doi.org/10.1090/S0002-9939-1976-0407832-0
Article copyright: © Copyright 1976 American Mathematical Society