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Let the map $f:[0,2\pi)\to S^1; t\mapsto e^{it}$. This map is clearly continuous and bijective. The inverse map associates to each point $z$ on the circle, the argument of $z$ modulo $2\pi$. it seems like $f$ is an homeomorphism, i know it is not but i don't see why? Is there any topological invariant that indicates that $[0,2\pi)$ and $S^1$ are not homeomorphic ?

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The interval can be disconnected by removal of a point; the circle, not. – Gerry Myerson Mar 19 '12 at 10:07
@Gerry That's really the simplest way to prove it. There are more direct ways to prove it,but they're a lot more complicated. – Mathemagician1234 Mar 19 '12 at 16:25
@palio : I tend not to post answers to questions that already have several answers including one that's "accepted", but there was a point worth noticing that I didn't see the others mentioning. Please have a look at it. – Michael Hardy Mar 19 '12 at 18:48
up vote 6 down vote accepted

Consider the sequence $z_n:=e^{i(-1)^n/n}$ in $S^1$. Then $(z_n)$ converges to $e^{0}=1$. But if you apply the inverse map $f^{-1}$, you obtain the sequence $(x_n)$ given by $x_{2n}=1/(2n)$ and $x_{2n+1}=2\pi-1/(2n+1)$. Since the latter does not converge, the inverse map $f^{-1}$ is not sequentially continuous. Hence it is not continuous. Of course, this elementary answer does not highlight the topological invariants you are looking for.

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VERY creative proof,Boston! – Mathemagician1234 Mar 19 '12 at 16:25

One way to see why the inverse mapping cannot be continuous is to note that $S^1$ is compact while $[0,2\pi)$ is not.

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This is essentially the same logic as the connectedness proof given by Gerry above.In each case,the proof is simple and decisive-which makes them very handy for exam answers. : ) – Mathemagician1234 Mar 19 '12 at 16:27

The inverse map is not continuous.

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Do we have this result: If $X$ is homeomorphic to a manifold then $X$ itself is a manifold? also, do we have that any submanifold has to be open? in this case $[0,2\pi($ is not open in $\mathbb R$ and so it is not a submanifold while $S^1$ is a $1$-manifold? i don't think so since $S^1$ itself is closed in $\mathbb R^2$!! – palio Mar 19 '12 at 10:13
@Palio: To flesh out Gerry's answer ever so slightly, the inverse mapping is continuous everywhere except at one point - at $z=1$ in $\mathbb{C}$. Why don't you try showing that? – Jason DeVito Mar 19 '12 at 12:34

Homeomorphisms preserve the structure of topologies...i.e. the open sets.

Continuous maps preserve the structure one way since a function of topological spaces is continuous if and only if the preimage of open sets is open.

However (bijective) homeomorphisms do this AND preserve open sets the other way, i.e. the image of open sets is now required to be open too.

Anyway, if you draw some pictures you will convince yourself that the open sets are not preserved.

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Careful. The images of open sets are not required to be open. If $f:X\to Y$ is a homeomorphism, then the preimage (under $f$) of open sets in Y must be open in X, and the preimage (under $f^{-1}$) of open sets in X must be open in Y. This is subtly different to what you said. Maps which send open sets to open sets are called Open Mappings. – Ragib Zaman Mar 19 '12 at 11:29
@Ragib Are you sure that's correct? If $x$ is a subset of $X$, then its preimage under $f^{-1}$, $(f^{-1})^{-1}(x)$, must be the same as $f(x)$, since $f$ is a bijection. Mustn't it? – Tanner Swett Mar 19 '12 at 11:52
@TannerL.Swett O Gosh Now I realized I remembered this the wrong way around - Homeomorphisms are always open maps, open maps aren't always homeomorphisms. Please excuse this mental lapse. – Ragib Zaman Mar 19 '12 at 12:37

Lots of answers, but this should also be mentioned: Look at the space $[0,2\pi)$ and ask what the open neighborhoods of $0$ are in that space. They are of the form $[0,\varepsilon)$. If every open neighborhood of $0$ were to reach into both that left end of the interval and also into the right end, looking like $[0,\varepsilon_1)\cup (1-\varepsilon_2,1)$, then $[0,2\pi)$ would be homeomorphic to the circle.

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