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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? –  J. M. Dec 4 '11 at 5:50
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 half-disks into three pieces. Is this a good direction? –  dfeuer Dec 4 '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 half-disks that can leave a connected set, but by the Jordan curve theorem it has to split a disk. –  Brian M. Scott Dec 4 '11 at 20:31
No, the circle (as I thought of it anyway) doesn't work. Interior within what space? –  dfeuer Dec 5 '11 at 4:54
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. –  Brian M. Scott Dec 5 '11 at 16:30
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2 Answers

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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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 7 '11 at 18:22
@dfeuer: I have updated to answer your questions. –  Brandon Carter Dec 7 '11 at 18:48
To whomever downvoted, please explain your downvote. –  Brandon Carter Dec 7 '11 at 22:05
@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 7 '11 at 22:48
@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 well-known (and thus incorrectness of your answer). That is a courtesy that I have since rescinded. –  Brandon Carter Dec 7 '11 at 22:58
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I think the dunce cap has a homology group of $\mathbb{Z}/\mathbb{3Z}$ in $H_{1}$ while $H_{1}D_{2}=0$. So the two surfaces cannot be homeomorphic. I am not sure how calculate this via tools like fundamental group, covering space, etc. Intuitively the fundamental group should be $\mathbb{Z}/3\mathbb{Z}$.

The 1st chain group $(a,b,c)$ is the points, and the second chain group $(A,B,C)$ are the edges, with the third chain group $X$ being the surface itself. Then we have a map $0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0$ as the three edges and vertices are identified together. Now the top map from the surface to the edges is $A-B+C$, but $B=-A,C=A$ so the image is $3A$ instead. The boundary map from the edges to the vertices is $A\rightarrow a-b$, $B\rightarrow b-c$, $C\rightarrow c-a$, since $a=b=c$ this amounts to a zero map. So we have $H_{1}\mathbb{X}=\mathbb{Z}/3\mathbb{Z}$, $H_{2}\mathbb{X}=0$. But I have not done homology calculations for a while, so I am not totally confident the result is correct.

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This is not correct. The dunce cap is contractible, so $H_q = 0$ for $q > 0$. –  Brandon Carter Dec 7 '11 at 1:55
@Brandon Carter: I agree if it is contractible then it should have 0 homology group, but I could not find where the computation go wrong. –  Kerry Dec 7 '11 at 2:03
Intuitively, the fundamental group is generated by $a$, and has relation equal to the surface symbol, $aaa^{-1}$. Thus $\pi_1(X) = \langle a|a \rangle = 0$. This can, of course, be made precise with Seifert-van Kampen. –  Brandon Carter Dec 7 '11 at 2:09
@BrandonCarter: I feel intuitively the fundamental group is generated by $2a$ and the defining relation is $3a=0$. But as I wrote I am not sure if this is rigorous. –  Kerry Dec 7 '11 at 2:15
In general, if you have a 2-dimensional polygon with surface symbol $W$ composed of characters $r_1, \dots, r_n$, then the fundamental group is $\langle r_1, \dots, r_n | W \rangle$. You can use Seifert-van Kampen ($U$ = interior of polygon, $V$ = polygon once punctured in the interior). $V$ deformation retracts to the wedge of $n$ circles, $U$ is contractible. But the nontrivial loop in $U \cap V$ is homotopic to the surface symbol in $V$. –  Brandon Carter Dec 7 '11 at 2:24
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