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 be defined by . Show that this is a bijection but that its inverse is not continuous.

**Attempt at solution:**

Injective: Suppose . Then

This holds if and only if and .

From the first equality we have or .

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 then would it be to say as the next logical step that and for some or do I have to justify this further?

Continuous: I know both and are continuous on . How can I use this to justify that 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 ?

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

be defined by

. Show that this is a bijection but that its inverse is not continuous.

**Attempt at solution:**
Injective: Suppose

. Then

This holds if and only if

and

.

From the first equality we have

or

.

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:

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 means that so the only possibility is .

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 .

This holds if and only if and .

From the first equality we have or .

But so we must have .

**Surjective**

Let . Then the radius at point makes angle in the counter clockwise direction from the positive x-axis where , then we can set and .

Then .

**Continuity of f**

This is clear from the continuity of and on in .

(we can use the fact that in an statement to make this precise)

**Discontinuity of **

Consider the point and an open set containing this point given by and where is small. This set is open as it is homeomorphic to

But .

Then for we have for all where

and for all

provided 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 is not continuous from the fact that 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