# Homeomorphism and Unit circle

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?

• What is open subsets in $\Bbb{S}^1$? Commented Dec 27, 2019 at 7:55
• 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. Commented Dec 27, 2019 at 7:55
• 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 $\cos$. Commented Dec 27, 2019 at 7:57

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 $$\mathbb{R}^2$$

• 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). Commented Dec 27, 2019 at 8:04
• @celtschk thanks you very much, i’m writing with the telephone Commented Dec 27, 2019 at 8:07

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 $$[0,1)$$.

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.