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Suppose $U$, $V$ and $W$ are connected topological spaces. Is it possible for $U \times V$ and $U \times W$ to be homeomorphic but for $V$ and $W$ not to be? It seems like it should not always be true but I am unable to find a counterexample.

user18063
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  • http://math.stackexchange.com/questions/396608/does-x-times-s1-cong-y-times-s1-imply-that-x-times-mathbb-r-cong-y-times-ma/410699#410699, http://math.stackexchange.com/questions/1057793/cancellation-in-topological-product – Moishe Kohan Feb 14 '17 at 11:06

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Let $U$, $V$, and $W$ all have the indiscrete topology with $0<|V|<|W|\leq |U|$ and $|U|$ infinite. Then $U\times V$ and $U\times W$ both have the indiscrete topology and have cardinality $|U|$, and so they are homeomorphic. But $V$ and $W$ have different cardinality and so are not homeomorphic.

For another example, let $U=V=[0,1]^\mathbb{N}$ and $W=[0,1]$. Then $U\times V\cong U\times W\cong [0,1]^\mathbb{N}$, but $V\not\cong W$. You can find many similar examples using infinite products like this.

For some further discussion and some more surprising examples, see this question on MO.

Community
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Eric Wofsey
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  • The example I find most amusing is the one of ${\mathbb Q}^2\cong {\mathbb Q}\times {0}$. http://mathoverflow.net/questions/26001/are-the-rationals-homeomorphic-to-any-power-of-the-rationals – Moishe Kohan Feb 14 '17 at 11:20