Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have two topological spaces $X,Y$ and I know that $X\times Y$ is homeomorphic to a manifold with boundary. Can I conclude that $X$ and $Y$ are manifolds (maybe with boundary)?

If not, suppose that $Y=[0,1]$. Is it then true? My intuition states that this is true, but I cannot see directly an elementary proof.

share|cite|improve this question
up vote 7 down vote accepted

Bing's dogbone space is a non-manifold $X$ such that $X\times\mathbb R\cong\mathbb R^4$. It is constructed as a quotient of $\mathbb R^3$ which is the identity outside of a ball. Hence we can do the construction inside a ball, to get a modified dogbone space $W$ with $\partial W=S^2$. Then I think that $W\times\mathbb R\cong D^3\times \mathbb R,$ which has boundary $S^2\times\mathbb R$. The basic idea as I understand it is that the nested tangle of genus 2 handlebodies unknots itself in $4$ dimensions, and this doesn't appear to use anything outside of a ball. However I haven't ever gone through the proof in detail, so I might be missing something.

If you take $Y=[0,1]$ I would guess that $X$ is probably a manifold with boundary. I recall hearing in a lecture many years ago that if $X\times S^1$ is a manifold, then so is $X$, which strikes me as a very similar problem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.