How to show that $p(x)=(\cos2\pi x,\sin2\pi x)$ is a homeomorphism?

Question

Let $p:(0,1)\to S^1-{(1,0)}$ Defined by $p(x)=(\cos2\pi x,\sin2\pi x)$

How to show that this map is a homeomorphism? Note that by $S^1$ I mean a unit circle in $R^2$ and ${(1,0)}$ it's North pole.

Answer 1

Use $p(x)=e^{2\pi i x}$. Note that $p(0)=p(1)=1$.

Answer 2

It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $\sqrt{(cos(2\pi y)-cos(2\pi x))^2+(sin(2\pi y)-sin(2\pi x))^2}=\sqrt{2(1-cos(2\pi x-2\pi y)}$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0\Leftrightarrow x=y$ and $|x-y|<\epsilon\Leftrightarrow d(p(x),p(y))<\sqrt{2(1-cos(2\pi\epsilon)}$ We know that $p$ is a diffeomorphism.

Answer 3

We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.

1) Note that for $x,y \in (0,1)$ $$P(x)=P(y) \implies (\cos ( 2\pi x, \sin ( 2\pi x) )= (\cos ( 2\pi y, \sin ( 2\pi y) ) \implies x=y $$ $ 2) If $$(\cos ( 2\pi x), \sin ( 2\pi x) ) \in S^1-(0,1)$ then $0<x<1$

3) Both function $P$ and its inverse function are continuous.

Thus $P$ it is a homeomorphism.