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.
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?
Thanks for your help .
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! )? Is there a simpler way I could have deduced that is not continuous from the fact that is the union of disjoint subsets?
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.