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

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 \[ {I^n}\xleftarrow {{{f_1}}}{I^n}\xleftarrow {{{f_2}}}{I^n}\xleftarrow {{{f_3}}} \cdots \] 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 $({S^{n - 1}} = \partial {I^n})$.