# How do you prove that the dunce cap is not a surface?

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?

• 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? Dec 4, 2011 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. Dec 4, 2011 at 20:31
• 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. Dec 5, 2011 at 16:30
• Another way of putting this argument: the three-halfdiscs-glued contains an open disc as a nonopen subspace. By 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 $\mathbb{R}^2$. Dec 7, 2011 at 18:59

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 2-dimensional boundary, while $$D^2$$ does.

• 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? Dec 7, 2011 at 18:22
– bzc
Dec 7, 2011 at 18:48