Homeomorphism between compactification of real line and unit circle. I'm currently working on a topology assignment which is, unfortunately, due today.
As part of that, I need to show that one-point compactification of the real line, $\mathbb{R}\cup\infty$ is homeomorphic to the unit circle.
I've come so far to have defined a function $f$ with
$(x,y)=(\cos(2\arctan(t)),\sin(2\arctan(t)))$, which should map the reals onto the unit circle and be bijective and continuous. Now since my domain is compact, if I can show that the unit circle is Hausdorff, I can conclude that my function is a homeomorphism, correct?
However, which topology would I use for that?
Also, I am having a hard time trying to prove that f is surjective. Thought about dividing it up into two functions, from reals to $(-\pi,\pi)$ and then to the unit circle, but that doesn't really seem to work either.
Any thoughts?
Topology is confusing...
(Intuitively, I absolutely see why all this should be the case, however I am struggling with the formal stuff.)
Edit:
Okay I have tried the approach of defining an inverse function, suggested and then proving that it is continuous.
So far I've got:
$$g=\tan\left(\frac{1}{2}\arctan\left(\frac{y}{x}\right)\right)$$
Problem is, that this function is not bijective, so there is something missing (case distinction?) but I can't figure out what...
At least within $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ it seems to work, and it's easy to show that combining f and g gives the identity.
Any tips maybe?
Cheers
Tom

The unit circle is a subspace of the plane; thus if you already know the plane is Hausdorff, that carries over to the circle too.
Showing that
$$ \mathrm{R} \to (-\pi, \pi) \to \text{everything but $(-1,0)$} $$
is surjective and $f(\infty) = (-1,0)$ would be enough to show $f$ is surjective.
Sometimes, though, it's easier to just find the inverse of your function.
