We have that f:X->Y is continuous iff for all closed subsets A in Y, the inverse image of A in X is also closed.
That's sound, and I have no problems to prove it.
However, I am asked to show that f:[0,2π[-> C defined by f(θ)= exp{iθ} don't have an inverse function, although it is continuous and bijective.
Here's my argument:
We have f:[0,2π[ --> Y={z element in C : |z|=1}, defined by f(θ)=exp{iθ}. As this is the exponential function, it is continuous. [0,2π[ is not closed, consequently not compact. As Y is the unit circle, it is compact.
Using argument from a), I can find a closed subset in Y, namely Y itself. But the inverse image of Y is [0,2π[, which is not closed. As I have a closed subset with an inverse image which is not closed, It should follow that f:[0,2π[->Y is not continous, which is clearly wrong!
Could anyone please point out where the mistake in my reasoning is?