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I would to like to know if there is an homeomorphism between the unit disk $D^2$ and $S^1\times I$, where $S^1$ is the unit circle. If I prove this homeomorphism I will be able to solve a question related which I'm struggling with.


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But! $D^2\setminus\{0\}$ is homeomorphic to $S^1\times (0,1]$. – Berci Nov 11 '12 at 19:51
up vote 5 down vote accepted

No they are not: $S^1 \times I$ is a cylinder and is homotopy equivalent to $S^1$. (To see this, stamp on it harshly once with your left foot.)

But $S^1$ is not homotopy equivalent to $D^2$ hence also not homeomorphic to $D^2$. (Every homeomorphism is also a homotopy equivalence.)

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See here, also. – Rudy the Reindeer Nov 11 '12 at 19:42
Can I say $S^1\times I/S^1\times \{1\}$ is homeomorphic to D^2? – user42912 Nov 11 '12 at 20:01
@user42912 Yes, you can! – Rudy the Reindeer Nov 11 '12 at 20:02
@user42912 An easy way to do so is to send the point $(\theta, x)$ ( in product coordinates) to the point $(\theta, 1-x)$ (in polar coordinates). – MartianInvader Nov 11 '12 at 20:11
An interesting follow-up question would be what happens if you stamp on $S^1\times I$ more than once. And what happens if you use your right foot? Is there a chirality phenomenon? – Miha Habič Nov 11 '12 at 20:38

No, they are not: $D^2$ is contractible, and $S^1\times I$ is not.

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Can I say $S^1\times I/S^1\times \{1\}$ is homeomorphic to D^2? – user42912 Nov 11 '12 at 20:02

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