Euclidean $(q+r)$-space modulo an $r$-plane of collapsible $p$-complexes

Abstract:

The following general decomposition result is obtained: Suppose ${K^p}(p \geqq 1)$ is a finite collapsible $p$-complex topologically embedded as a subset of a separable metric space ${X^q}$ where, for some $r \geqq 1,{X^q} \times {E^r}$ is homeomorphic to Euclidean $(q + r)$-space ${E^{q + r}}$. Then the Cartesian product of the quotient space ${X^q}/{K^p}$ with ${E^r}$ is topologically ${E^{q + r}}$ provided that $q \geqq 3$ and, for each simplex ${\Delta ^k} \in {K^p},({X^q} \times {E^r},{\Delta ^k} \times ({[0,1]^{r - 1}} \times 0))$ is homeomorphic, as pairs, to \[ ({E^{q + r}},{[0,1]^{k + r - 1}} \times (0, \ldots ,0)).\] It is known that this condition is satisfied if $q - p \geqq 2$ and $q + r \geqq 5$. This result implies that if ${K^k}$ is a finite collapsible $k$-complex topologically embedded as a subset of Euclidean $n$-space ${E^n}$, then the Cartesian product of the quotient space ${E^n}/{K^k}$ with ${E^1}$ is topologically ${E^{n + 1}}$ provided either (i) $n \leqq 3$, (ii) $n - k \geqq 2$, or (iii) each simplex of ${K^k}$ is flat in ${E^{n + 1}}$.