The dunce cap results from a triangle with edge word $aaa^{-1}$. At the edge, a small neighborhood is homeomorphic to three half-disks glued together along their diameters. How do you prove this is not homeomorphic to a single disk?
How do you prove that the dunce cap is not a surface?
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$\begingroup$
general-topology
algebraic-topology
manifolds
surfaces
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2Why was this downvoted? – 2011-12-04
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0I'm thinking maybe the trick is that (if I'm not mistaken) removing a set homeomorphic to the circle from a disk separates it into no more than two pieces, but can separate the the three glued half-disks into three pieces. Is this a good direction? – 2011-12-04
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0If that works, you should be able to do it with a segment instead, which might be easier. A similar idea would be to remove a circle from the interior: with the three glued half-disks that can leave a connected set, but by the Jordan curve theorem it has to split a disk. – 2011-12-04
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0No, the circle (as I thought of it anyway) doesn't work. Interior within what space? – 2011-12-05
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1The interior of the closed nbhd consisting of three closed half-disks glued together. You can embed the circle so that it snakes into each of the ‘pages’, meeting the ‘spine’ three times, and has a path-connected complement. – 2011-12-05
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1Another way of putting this argument: the three-halfdiscs-glued contains an open disc as a nonopen subspace. By [invariance of domain](http://en.wikipedia.org/wiki/Invariance_of_domain), anything in $\mathbb{R}^2$ which is homeomorphic to an open disc is in fact open in $\mathbb{R}^2$. Thus the three-halfdiscs-glued cannot be homeomorphic to $\mathbb{R}^2$. – 2011-12-07