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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.

Thanks

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    But! $D^2\setminus\{0\}$ *is* homeomorphic to $S^1\times (0,1]$.2012-11-11

2 Answers 2

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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](http://books.google.com/books?id=O319SgIPsJ8C&lpg=PA15&ots=EWUOFOPhHQ&dq=cylinder%20is%20homotopy%20equivalent%20to%20S%5E1&pg=PA15#v=onepage&q&f=false), also.2012-11-11
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    Can I say $S^1\times I/S^1\times \{1\}$ is homeomorphic to D^2?2012-11-11
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    @user42912 Yes, you can!2012-11-11
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    @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).2012-11-11
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    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?2012-11-11
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    @MihaHabič Well, I am not sure whether you are taking up the train of humorous writing or whether you are complaining about it. If the latter I will gladly edit my answer. If the former: once you stamped it's infinitely thin and flat hence cannot be compressed any more and hence nothing happens. As for the right foot: I am left-footed so I cannot try.2012-11-11
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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?2012-11-11