Find a counterexample to topology

Question

The following conjecture is rather attractive: if $\mathbb{D}$ is a descompostition of $X$ into homeomorphic sets, say all homemorphic to $Y$, then $X$ is homeomorphic to $\mathbb{D}$ x $Y$.

I tried to prove it but I suppose that is false, so could you help me to find a counterexample or prove it?

Answer 1

A good counterexample is the Klein bottle. One can describe it as the quotient space $K$ of the cylinder $S^1 \times [0,1]$ modulo the equivalence relation generated by $(p,1) \sim (-p,0)$, $p \in S^1$. For each $t \in [0,1]$ the quotient map $S^1 \times [0,1] \mapsto K$ restricts to an embedding of $S^1 \times \{t\}$ into $K$, and the images give a decomposition $\mathbb D$ of $K$ into subspaces homeomorphic to the circle $S^1$. Also, the quotient topology on $\mathbb D$ makes $\mathbb D$ itself into a circle. Thus $\mathbb D \times S^1$ is homeomorphic to the torus $S^1 \times S^1$.

However, it is well know that the Klein bottle and the torus are not homeomorphic; for example, their fundamental groups are not isomorphic.

Another counterexample, closely related to the Klein bottle but perhaps even easier, is the Möbius strip. It has a decomposition into subsets homeomorphic to the closed interval $[0,1]$, and the quotient space $\mathbb D$ is homeomorphic to the circle.

But $\mathbb D \times [0,1]$ is homeomorphic to $S^1 \times [0,1]$ which is a 2-dimensional manifold-with-boundary whose boundary is disconnected, whereas the Möbius strip has connected boundary. Since any homeomorphism between two manifolds-with-boundary restricts to a homeomorphism between the boundaries, it follows that the Möbius strip is not homeomorphic to $\mathbb D \times [0,1]$.

Answer 2

$$\Bbb R = \bigcup_{n \in \Bbb Z} [n,n+1)$$ is a decomposition of $\Bbb R$ in sets homeomorphic to $[0,1)$, indexed by $\Bbb Z$, but $\Bbb Z \times [0,1)$ is disconnected and not homeomorphic to $\Bbb R$. If $\Bbb Z$ (as a set of represntatives for $\Bbb D$ gets the quotient topology, then it has the indiscrete topology (!) and $\Bbb D \times [0,1)$ is not $T_0$ while $\Bbb R$ is.

If all parts of the decomposition are open (and thus clopen), then we do have such a theorem, provided $\Bbb D$ gets the discrete topology. In the above example, had we given $\Bbb R$ the Sorgenfrey topology it would have worked, but not in the usual topology..