Timeline for Can every manifold be represented as a quotient
Current License: CC BY-SA 4.0
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| Jun 5, 2021 at 2:37 | comment | added | Sam Nead | @MarcoGolla, yes, that is another description of $M$. I don't have a reference to hand for the universal cover. So here is a sketch. Let $S$ and $T$ be disjoint two-spheres in $M$ whose union is non-separating. Cut $M$ along these and denote the result by $X$. So $X$ is (homeomorphic to) $S^3$ with four small, open, round three-balls removed (also, they have disjoint closures). Note that $X$ is simply connected and so lifts to the universal cover. In fact, copies of $X$ tile the universal cover in a tree-like fashion; the result follows. | |
| Jun 1, 2021 at 13:47 | comment | added | Marco Golla | @SamNead, isn't $M = S^1\times S^2 \# S^1\times S^2$ in your example? Do you happen to have a reference for the description you give of its universal cover? | |
| Jun 1, 2021 at 12:17 | history | edited | Sam Nead | CC BY-SA 4.0 |
added 191 characters in body
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| Jun 1, 2021 at 12:15 | comment | added | ABIM | cool haha thanks Sam :) | |
| Jun 1, 2021 at 12:14 | comment | added | Sam Nead | Sure. Connect sums of simply connected four-manifolds... doubles of higher genus handlebodies... And there are more construction than these... for example torus bundles over hyperbolic surfaces will have nice universal covers but their fundamental groups will not “factor” as a product - instead they will be semi-direct products... I am sure that there are as many examples as there are topologists. :) | |
| Jun 1, 2021 at 12:06 | history | edited | Sam Nead | CC BY-SA 4.0 |
More examples
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| Jun 1, 2021 at 12:03 | comment | added | ABIM | Does the class of manifolds which we can describe in this way include any "interesting" family? | |
| Jun 1, 2021 at 12:02 | vote | accept | ABIM | ||
| Jun 1, 2021 at 11:59 | history | answered | Sam Nead | CC BY-SA 4.0 |