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

**Attempt at solution:**

Injective: Suppose $\displaystyle f(x)= f(y)$. Then $\displaystyle (\cos x, \sin x) = (\cos y, \sin y)$

This holds if and only if $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = \sin y$.

From the first equality we have $\displaystyle x= y$ or $\displaystyle x = 2 \pi - y$.

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 $\displaystyle (x,y) \in S^1$ then would it be to say as the next logical step that $\displaystyle x = \cos \theta$ and $\displaystyle y = \sin \theta$ for some $\displaystyle \theta \in [0, 2 \pi)$ or do I have to justify this further?

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

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

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

Quote:

Originally Posted by

**MrMeth** **The problem:**

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

Injective: Suppose $\displaystyle f(x)= f(y)$. Then $\displaystyle (\cos x, \sin x) = (\cos y, \sin y)$

This holds if and only if $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = \sin y$.

From the first equality we have $\displaystyle x= y$ or $\displaystyle x = 2 \pi - y$.

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: $\displaystyle x=y\text{ or }y=2\pi-x. $

That proves it is one-to-one.

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

Thank you for your reply.

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

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

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.

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

Thanks for your help (Happy) .

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

**Injective**

Suppose $\displaystyle f(x) = f(y)$.

This holds if and only if $\displaystyle \cos x = \cos y$ and $\displaystyle \sin x = \sin y$.

From the first equality we have $\displaystyle x=y$ or $\displaystyle y = 2\pi - x$.

But $\displaystyle \sin x \neq \sin 2\pi - x$ so we must have $\displaystyle x=y$.

**Surjective**

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

Then $\displaystyle f(\theta) = (\cos \theta, \sin \theta) = (x,y) $.

**Continuity of f**

This is clear from the continuity of $\displaystyle \sin $ and $\displaystyle \cos $ on $\displaystyle [0, 2 \pi)$ in $\displaystyle \mathbb{R}$.

(we can use the fact that $\displaystyle \sqrt{ (\cos x - \cos y)^2 + (\sin x - \sin y)^2} \leq |\cos x - \cos y| + | \sin x - \sin y|$ in an $\displaystyle \varepsilon - \delta$ statement to make this precise)

**Discontinuity of $\displaystyle f^{-1}$**

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

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

Then for $\displaystyle \varepsilon_1 = 1$ we have for all $\displaystyle (\cos x, \sin x ) \in S^1$ where $\displaystyle x \in (2\pi - \varepsilon, 2\pi)$

and for all $\displaystyle \delta>0$

$\displaystyle |(\cos x, \sin x ) - (1,0)| < \delta \Rightarrow |f^{-1}(\cos x, \sin x ) - f^{-1} (1,0)|> 2 \pi - \varepsilon > 1 $

provided $\displaystyle \varepsilon$ is small.

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

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.

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

Quote:

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!

Thanks for your help.

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