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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?

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    Why was this downvoted?2011-12-04
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    I'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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    If 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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    No, the circle (as I thought of it anyway) doesn't work. Interior within what space?2011-12-05
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    The 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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    Another 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

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