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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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    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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If you have two homotopic maps $f,g: S^1 \to X$, 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 $f: S^1 = \partial D^2 \to X$. 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 $g: S^1 \to S^1$ 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 2-dimensional boundary, while $D^2$ does.

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    @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 style2011-12-09
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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.