# Math Help - Continuous bijection from an interval to the circle but homeomorphic

1. ## Continuous bijection from an interval to the circle but homeomorphic

Hi,

This is my first mathematics related post. I'd be grateful for some suggestions on how to proceed.

The problem:
Let $f: [0, 2\pi) \rightarrow S^1$ be defined by $f(\theta) = (\cos \theta, \sin \theta)$. Show that this is a bijection but that its inverse is not continuous.

Attempt at solution:

Injective: Suppose $f(x)= f(y)$. Then $(\cos x, \sin x) = (\cos y, \sin y)$
This holds if and only if $\cos x = \cos y$ and $\sin x = \sin y$.
From the first equality we have $x= y$ or $x = 2 \pi - y$.
What property of sin can I use in the second equality to show that the only simultaneous solution is x=y?

Surjective: Not quite sure where to start from here. I'm thinking if I let $(x,y) \in S^1$ then would it be to say as the next logical step that $x = \cos \theta$ and $y = \sin \theta$ for some $\theta \in [0, 2 \pi)$ or do I have to justify this further?

Continuous: I know both $\cos$ and $\sin$ are continuous on $[0, 2\pi)$. How can I use this to justify that $f$ is continuous?

Inverse not continuous: The hint I'm given is that this is not continuous because the pre-image of a small open set containing 0 is not open. How can there be a small open set containing 0 in the set $[0,2\pi)$?

Any advice on how I can proceed would be much appreciated.

2. ## Re: Continuous bijection from an interval to the circle but homeomorphic

Originally Posted by MrMeth
The problem:
Let $f: [0, 2\pi) \rightarrow S^1$ be defined by $f(\theta) = (\cos \theta, \sin \theta)$. Show that this is a bijection but that its inverse is not continuous. Attempt at solution:
Injective: Suppose $f(x)= f(y)$. Then $(\cos x, \sin x) = (\cos y, \sin y)$
This holds if and only if $\cos x = \cos y$ and $\sin x = \sin y$.
From the first equality we have $x= y$ or $x = 2 \pi - y$.
What property of sin can I use in the second equality to show that the only simultaneous solution is x=y?
In both cases you have: $x=y\text{ or }y=2\pi-x.$
That proves it is one-to-one.

3. ## Re: Continuous bijection from an interval to the circle but homeomorphic

Okay, so $y = 2 \pi - x$ means that $\sin x \neq \sin y$ so the only possibility is $x=y$.

Have you any suggestions on how to proceed with the other parts of the proof?

4. ## Re: Continuous bijection from an interval to the circle but homeomorphic

The pre-image of a neighbourhood of (1,0) in S^1 is a set of the form [0,epsilon) union (epsilon,2.pi). The points of these sets are not arbitrarily close.

5. ## Re: Continuous bijection from an interval to the circle but homeomorphic

Okay here is my revised attempt at a solution; i've tried to make some things a bit more precise:

Injective

Suppose $f(x) = f(y)$.
This holds if and only if $\cos x = \cos y$ and $\sin x = \sin y$.
From the first equality we have $x=y$ or $y = 2\pi - x$.
But $\sin x \neq \sin 2\pi - x$ so we must have $x=y$.

Surjective
Let $(x,y) \in S^1$. Then the radius at point $(x,y)$ makes angle $\theta$ in the counter clockwise direction from the positive x-axis where $\theta \in [0, 2\pi)$, then we can set $x = \cos \theta$ and $y = \sin \theta$.
Then $f(\theta) = (\cos \theta, \sin \theta) = (x,y)$.

Continuity of f
This is clear from the continuity of $\sin$ and $\cos$ on $[0, 2 \pi)$ in $\mathbb{R}$.
(we can use the fact that $\sqrt{ (\cos x - \cos y)^2 + (\sin x - \sin y)^2} \leq |\cos x - \cos y| + | \sin x - \sin y|$ in an $\varepsilon - \delta$ statement to make this precise)

Discontinuity of $f^{-1}$
Consider the point $(1,0) = (\cos 0, \sin 0) \in S^1$ and an open set containing this point given by $U=\{ (x,y) \in S^1 : x = \cos t, y = \sin t,$ and $t \in ( - \varepsilon, \varepsilon) \}$ where $\varepsilon > 0$ is small. This set is open as it is homeomorphic to $(-\varepsilon, \varepsilon)$

But $f^{-1}(U) = [0,\varepsilon) \cup (2\pi - \varepsilon, 2 \pi)$.

Then for $\varepsilon_1 = 1$ we have for all $(\cos x, \sin x ) \in S^1$ where $x \in (2\pi - \varepsilon, 2\pi)$
and for all $\delta>0$
$|(\cos x, \sin x ) - (1,0)| < \delta \Rightarrow |f^{-1}(\cos x, \sin x ) - f^{-1} (1,0)|> 2 \pi - \varepsilon > 1$
provided $\varepsilon$ is small.

How does this look? Does anyone think I might have used any circular arguments here (pun not intended! )? Is there a simpler way I could have deduced that $f^{-1}$ is not continuous from the fact that $f^{-1}(U)$ is the union of disjoint subsets?

6. ## Re: Continuous bijection from an interval to the circle but homeomorphic

Looks fine. No circular arguments.

Regarding your last question, if you have talked about connected sets, you should probably know that a continuous function's image of a connected set is connected. f^-1(U) is not connected although U is, so f^-1 can't be continuous. If you haven't talked about connectedness, I don't see a shorter way than the one you described.

7. ## Re: Continuous bijection from an interval to the circle but homeomorphic

Originally Posted by ModusPonens

Regarding your last question, if you have talked about connected sets, you should probably know that a continuous function's image of a connected set is connected. f^-1(U) is not connected although U is, so f^-1 can't be continuous. If you haven't talked about connectedness, I don't see a shorter way than the one you described.

Can't believe I didn't think of connectedness. Of course!