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Timeline for free action on contractible spaces

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Apr 29, 2014 at 20:12 comment added Tom Goodwillie I don't think anybody is going to disprove any big conjectures on this thread! Here's a compact 1-dimensional non-manifold example. The graph that looks like the letter $\theta$ and the graph you get by connecting two circles with an arc have isomorphic fundamental groups and homeomorphic universal covers.
Apr 29, 2014 at 18:31 comment added Ilias A. I'm quite optimistic about Borel rigidity conjecture, I would say that a possible source of a counter example to the asked question ($E/G_{i}$ compact) should arise when $E$ is not a n-manifold. :)
Apr 29, 2014 at 18:12 comment added Craig Westerland I agree that it is a different question. I'm just pointing out that if Tom has a counterexample when the $E/G_i$ are compact manifolds, then he also has a counterexample to the Borel conjecture. I certainly would be excited to hear it. Tom?
Apr 29, 2014 at 18:00 comment added Ilias A. Dear Craig, I wanted to add this ad the begining of the question. But I think that the question is little bit different since I assume in the hypothesis that the universal cover is the same (or homemorphic) and I dorp the hypothesis to be a manifold. I just asked Tom what happens if $E/G_{1}$ and $E/G_{2}$ are assumed to be compact.
Apr 29, 2014 at 17:54 comment added Craig Westerland @Fedotov, if you assume that E is a contractible manifold, that is very nearly the (open) Borel conjecture: if two aspherical compact manifolds of the same dimension are homotopy equivalent, then they are homeomorphic. Here, asphericity implies that homotopy equivalence is the same as having the same fundamental group, which is true of your setup.
Apr 29, 2014 at 17:27 comment added Ilias A. Thank you Tom! I was wondering if we add the assumption that $E/G_{1}$ and $E/G_{2}$ are compact do you still have an other counter example?!
Apr 29, 2014 at 16:35 history answered Tom Goodwillie CC BY-SA 3.0