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Problem: Let $X$ the half open interval $[0,1)\subseteq \mathbb{R}$ and $\mathbb{S}^1$ be the unit circle in $\mathbb{C}$. Define a map $\phi: [0,1) \rightarrow \mathbb{S}^1$ by $\phi(x)= \cos(2\pi x)+i\sin(2\pi x)$. Show that it is continuous and bijection but not a homeomorphism.

My attempt:$\phi(x)=\phi(y)$ $\implies$ $\cos(2\pi(x-y))=1$ $\implies$ $x=y$. So the map is injective. The map is also surjective and thus the map is bijective. Let $\epsilon>0$ and set $\delta = \frac{\epsilon}{4\pi}$. if $y\in [0,1)$ such that $|x-y|<\delta$ then $|f(x)-f(y)|\leq 4\pi |x-y|<\epsilon$. Thus the map is continuous. It suffices to show that the map is not open. Observe, since $[0,\frac{1}{2})= (-\frac{1}{2},\frac{1}{2}) \cap [0,1)$, it is thus open in $[0,1)$.

How do I show that $[0,\frac{1}{2})$ is not open in the image?

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  • $\begingroup$ What is open subsets in $\Bbb{S}^1$? $\endgroup$ Commented Dec 27, 2019 at 7:55
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    $\begingroup$ Show that the image of $0$ is not an interior point of the image of $[0,\frac 1 2)$. Make sure to see the image geometrically. $\endgroup$ Commented Dec 27, 2019 at 7:55
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    $\begingroup$ Note: To get the correct typesetting of $\sin$, put a backslash in front of the name: \sin. This also automatically takes care of the spacing. The same is true of $\cos$. $\endgroup$
    – celtschk
    Commented Dec 27, 2019 at 7:57

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By contraddiction, if $\phi$ would be an homeomorphism, then $[0,1)\setminus \{\frac{1}{2}\}$ is homeomorphic to

$\mathbb{S}^1\setminus \{\phi(\frac{1}{2})\}$

but the first space is not connected while the second is connected.

Another way can be to observe that $[0,1)$ is not compact while $\mathbb{S}^1$ is a compact Space because is a closed and limited subset of $\mathbb{R}^2$

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    $\begingroup$ I had to read several times to get that where you write “omeomorphic” you probably mean “homeomorphic” (my brain consistently parsed it as “onemorphic”, which looked like a term that might exist). $\endgroup$
    – celtschk
    Commented Dec 27, 2019 at 8:04
  • $\begingroup$ @celtschk thanks you very much, i’m writing with the telephone $\endgroup$ Commented Dec 27, 2019 at 8:07
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You can also show that $\phi$ is not a homeomorphism because $\phi^{-1}$ is not continuous at $1 \in \Bbb C$: $y_n = \cos(2 \pi \frac{-1}{n}) + i \sin(2 \pi \frac{-1}{n})$ converges to $1$, but $\phi^{-1}(y_n)= 1-\frac{1}{n}$ for all $n$ (note that $$\cos(2 \pi (1-\frac{1}{n}))=\cos(2\pi \frac{-1}{n})$$ because the inputs differ by $2\pi$, and likewise for the sine values) and $\phi^{-1}(y_n)$ does not converge to $0 = \phi^{-1}(1)$ in $[0,1)$.

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Since $[0,1)$ is connected and $\varphi$ is continuous, $\varphi([0,1))$ is connected. We see then that $$\varphi([0,1)) = \{\cos(2\pi x)+i\sin(2\pi x):x\in[0,1)\} = \{\cos(\theta) + i\sin(\theta):\theta\in[0,\pi/4)\}$$ is not open, as $0$ is not an interior point. It follows that $\varphi$ is not a homeomorphism.

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