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Assume that $A$ is compact, connected and contractible set in $\mathbb{R}^{2}$ (for example: simple square). If we contract this set to a point the space still will be homeomorphic to $\mathbb{R}^{2}$. Formally: the space $\mathbb{R}^{2}/{}_{\approx}$, where $\approx$ is equivalence relation which equivalence classes are $A$ and singletons, is homeomorphic to $\mathbb{R}^{2}$.

This is should be know, it is propably a folklore in topology, but I could not find a source. It bring to my mind Mosers name, but I couldn't find a right theory. Can you bring proper reference?

Grigory M
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M314
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    $S^1\subset\mathbb R^2$ is compact and connected but not contractible. – Grigory M Dec 29 '13 at 19:09
  • Thanks for pointing mistake. I have fixed question - I consider only contractible sets. – M314 Dec 29 '13 at 19:16
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    You do not need the hypothesis "$A$ is connected" , since contractible spaces are automatically path-connected, hence connected. – Amr Dec 29 '13 at 19:54
  • I'm not even sure if this is true — for example for $\mathbb R^3$ the statement seems to be false (take $A$ to be the horned ball...). – Grigory M Dec 30 '13 at 06:02
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    I do not see why 'horned ball' whould be a problem. – M314 Dec 30 '13 at 08:40
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    @M314 Complement to the horned ball is not simply connected. So in $\mathbb R^3/A$ we have a point $A/A$ with non-simply-connected complement. – Grigory M Dec 31 '13 at 20:50

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This is a special case of a more general theorem due to Moore, whose proof should be in the book R.Wilder, "Topology of manifolds". Dimension 2 is very different from higher dimensions (2-dimensional homology manifolds are topological manifolds). Bing (I think) proved that contracting a wild arc in $R^3$ results in a space which is not a manifold.

Moishe Kohan
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