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On the Product and Composition of Universal Mappings of Manifolds Into Cubes
W. Holsztyński
Proceedings of the American Mathematical Society
Vol. 58, No. 1 (Jul., 1976), pp. 311314
Published by: American Mathematical Society
DOI: 10.2307/2041406
Stable URL: http://www.jstor.org/stable/2041406
Page Count: 4
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Topics: Mathematical theorems, Cubes, Mathematical manifolds, Polyhedrons, Fixed point property, Universality, Topological theorems, Astronomy
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Abstract
A map $f: X \rightarrow Y$ is said to be universal $\operatorname{iff}$ for every $g: X \rightarrow Y$ 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 $f_t: M_t \rightarrow Q_t, f_0: M^n \rightarrow I^n$ and $f_k: I^n \rightarrow I^n$ be universal mappings for $t \in T$ and $k = 1, 2, \ldots$. Then (1.8) THEOREM. The product map $\Pi_{t \in T}f_t: M_t \rightarrow \Pi_{t \in T} Q_t$ is universal. (2.1) THEOREM. The composition $f_s \circ f_{s  1} \circ \cdots \circ f_1: M^n \rightarrow I^n$ is a universal map for $s = 1, 2, \ldots$. (2.2) THEOREM. The limit $X$ of the inverse sequence $$I^n \overset{f_1}{\leftarrow} I^n \overset{f_2}{\leftarrow} I^n \overset{f_3}{\leftarrow} \cdots$$ 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 $\leqslant n$. Then $f: X \rightarrow I^n$ is a universal mapping $\operatorname{iff}$ the element $f^\ast(e^n)$ of the $n$th Cech 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 $(S^{n  1} = \partial I^n)$.
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Proceedings of the American Mathematical Society © 1976 American Mathematical Society