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Let $f:[0, 2\pi) \to S^1 = \{(x, y): x^2 + y^2 = 1\}$ be such that

$f(t) \to (\cos t, \sin t)$

$f$ is a continuous bijection but it is NOT a homeomorphism.

I suppose the only point of contention is $(1, 0)$ in $S^1$.

Is it because there is no open $U \in \mathbb{R^2}$ such that $U \bigcap S^1 = f([0, 1))$? Because $f([0, 1))$ contains the point $(1, 0) \in \mathbb{R^2}$?

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  • $\begingroup$ Edited my original post, I made a mistake in the domain. $\endgroup$ – sonicboom Jul 8 '14 at 15:59
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To be a homeomorphism, $f$ must be an open map. The subset $[0,a)$ of $[0,2\pi)$ ($a<2\pi$ positive) is open in $[0,2\pi)$, but its image is not open in $S^1$.

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  • $\begingroup$ Your answer and Turion's say the same thing in different ways (and both are good). Yours explicitly shows where $f^{-1}$ is not continuous. $\endgroup$ – robjohn Jul 8 '14 at 16:06
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The other two answers demonstrate why this particular choice $f(t) = (\cos t, \sin t)$ is not a homeomorphism. More generally, you can show that no choice of $f:[0,1)\to S^1$ will be a homeomorphism.

One way to see this: suppose $f:[0,1)\to S^1$ is a continuous bijection. Then its inverse map $f^{-1}:S^1\to [0,1)$ is well-defined and surjective. But $S^1$ in the subspace topology on $\mathbb{R}^2$ is compact, while $[0,1)$ in the subspace topology on $\mathbb{R}^1$ is not compact. Since continuous maps send compact sets to compact sets, the inverse cannot be continuous.

There are other proofs that function in a similar way, namely by demonstrating that $[0,1)$ and $S^1$ differ in some sort of topological invariant. If you learn about algebraic topology, then you can try using the fundamental group or the first homology to argue essentially as above.

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Another way you can consider is that the numbers of connected components of two homeomorphic topological spaces have the same amount.

With that fact you can show that $[0,1)$ and $S^1$ are not homeomorphic:

If you remove $\frac{1}{2}$ from $[0,1)$ then we obtain $[0,\frac{1}{2}) \cup (\frac{1}{2},1)$ which is obviously not connected.

But if you remove an arbitrary $z\in S^1$ then we obtain $S^1$\ $\{z\}$, which is connected.

So $[0,1)$ and $S^1$ can not be homeomorphic.

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No, your explanation is wrong. Take $U = \mathbb{R}^2$, then $U\cap S^1 = S^1 = f([0,1))$ since it's a bijection. The point is that $f^{-1}$ is not continuous.

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