cartesian product rigidity for the punctured open disc

Question

Q1: Let $D^n$ ($n\geq 1$) be the n-dimensional open disk. If $D^n-\{0\}$ is homeomorphic to $X\times (0,1)$, for some topological space $X$, does it necessarily follow that $X$ is homeomorphic to $S^{n-1}$?

Q2 Let $Z$ be a topological space. What kind of assumptions can you impose on $Z$ in order to have the cancellation property for all (may be here one should impose some additional conditions on the cartesian decomposition) cartesian decompositions of $Z$, i.e., if $Z\simeq X\times Y$ and $Z\simeq X\times Y'$ then $Y\simeq Y'$

Answer 1

Q1: No, see the third-from-last theorem of

R. H. Bing, The cartesian product of a certain nonmanifold and a line is $E^4$.