The dunce cap results from a triangle with edge word $aaa^{1}$. At the edge, a small neighborhood is homeomorphic to three halfdisks glued together along their diameters. How do you prove this is not homeomorphic to a single disk?

2Why was this downvoted? – J. M. ain't a mathematician Dec 04 '11 at 05:50

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 halfdisks into three pieces. Is this a good direction? – dfeuer Dec 04 '11 at 19:26

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 halfdisks that can leave a connected set, but by the Jordan curve theorem it has to split a disk. – Brian M. Scott Dec 04 '11 at 20:31

No, the circle (as I thought of it anyway) doesn't work. Interior within what space? – dfeuer Dec 05 '11 at 04:54

1The interior of the closed nbhd consisting of three closed halfdisks glued together. You can embed the circle so that it snakes into each of the ‘pages’, meeting the ‘spine’ three times, and has a pathconnected complement. – Brian M. Scott Dec 05 '11 at 16:30

1Another way of putting this argument: the threehalfdiscsglued 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 threehalfdiscsglued cannot be homeomorphic to $\mathbb{R}^2$. – Chris Eagle Dec 07 '11 at 18:59
2 Answers
If you have two homotopic maps $f,g\colon S^1 \to X$, then $X \cup_f D^2$ is homotopy equivalent to $X \cup_g D^2$.
You can use this to show that the dunce cap is homotopy equivalent to $D^2$, and thus contractible. Since no closed surface is contractible (using classification of surfaces), the dunce cap is not a surface.
$D^2$ is the closed unit disk. By $X \cup_f D^2$, I mean gluing $D^2$ via the map $f\colon S^1 = \partial D^2 \to X$. This is the quotient space of $X \sqcup D^2$ identifying each point of $\partial D^2$ with its image under $f$ in $X$. So in our specific case, $D^2$ is homeomorphic to $S^1$ glued to $D^2$ under the identity map $S^1 \to S^1$. On the other hand, we have that the dunce cap is constructed by gluing $D^2$ to $S^1$ under the map $g\colon S^1 \to S^1$ given by $$ g(e^{i\theta}) = \begin{cases} \exp(4 i \theta) & 0 \leq \theta \leq \pi/2\\ \exp(4 i (2 \theta  \pi)) & \pi/2 \leq \theta \leq 3\pi/2\\ \exp(8 i(\pi  \theta)) & 3\pi/2 \leq \theta \leq 2\pi \end{cases}$$
It is not hard to show that $g$ is homotopic to to the identity map, and so (using the result I mentioned above), $D^2$ is homotopy equivalent to the dunce cap. So the dunce cap must be contractible.
Edit: I have now realized that the above answers the question in the title, which is not the question posed by the OP. To see that the dunce cap is not homeomorphic to $D^2$, you can simply note that the dunce cap is a disk glued along its boundary (albeit in a strange way), and thus has no 2dimensional boundary, while $D^2$ does.
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I haven't yet studied algebraic topology, though I do know what homotopic means. Could you explain what you mean by \cup_f and \cup_g? Is D^2 the open or closed unit disk? – dfeuer Dec 07 '11 at 18:22



@BrandonCarter: I casted the downvote, because I feel this proof is ugly and may be wrong. I do not see the need to ask others explain downvoting. Sorry this may be provocative from the perspective that you believe your proof is right, and I totally understood that. – Kerry Dec 07 '11 at 22:48

1@Changwei: I know that the above proof is correct, especially since it is given as a sequence of exercises in Armstrong's *Basic Topology*. In general, explaining your downvote helps to improve the quality of answers. I afforded you the courtesy of not downvoting your answer, despite the contractibility of the dunce cap being wellknown (and thus incorrectness of your answer). That is a courtesy that I have since rescinded. – Brandon Carter Dec 07 '11 at 22:58

@BrandonCarter The question I intended to pose, which I thought I did, is how to show that there are points of the dunce cap that do not have neighborhoods homeomorphic to the closed disk, which as I understand it is a requirement for something to be a "surface". – dfeuer Dec 08 '11 at 04:45

@dfeuer: That is a requirement for something to be a surface. In fact, every point must lie in some neighborhood homeomorphic to the disk. However, if $X$ is what is called a "closed surface", which means a compact surface (without boundary), then we know by the classification of surface what all of the possibilities of $X$ are. None of those possibilities is contractible, while the dunce cap is compact and has no 2dimensional boundary, hence my answer above. – Brandon Carter Dec 08 '11 at 05:31

@BrandonCarter I see. I wasn't excluding the possibility of a boundary, although the dunce cap certainly does not appear to have one. I guess my question could have been phrased better. – dfeuer Dec 08 '11 at 05:46

@BrandonCarter: I disagree. I cannot assert your various proofs are correct (though I am not familiar enough with fundamental groups to say it is wrong)and the fact that dunce hat appeared to be contractible in standard textbooks, etc does not convince me either. While thanks for your courtesy I believe every member of MSE should be able to cast a vote for their opinion on the quality of the answer mattered to their interest. I do not think explaining downvoting helps to improving the quality of the answer, because people's opinion can be complicated and one has one's own unique writing style – Kerry Dec 09 '11 at 00:03
The question is essentially answered in the way that it is posed. No point of a disc has a neighbourhood homeomorphic to three half discs glued along their diameters.
The question has been changed since I wrote this reply.
In the talk about the topological dunce hat, I have described a simple method of contracting it.
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