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.