Given a compact Lie group G acting freely on a topological manifold M, is it true that the orbit space M/G is also a topological manifold? If so, why?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
The answer is no. Bing constructed a space $X$ (called the dogbone space) so that $X$ is not a manifold, but $X\times R$ is homeomorphic to $R^4$. In particular, $M^4=X\times S^1$ is a $4$manifold (since its universal cover is $R^4$) and $X$ is the quotient of $M^4$ by free $S^1$action. 

