On my book I found the following function:
$$\phi:\mathbb{R}^2\setminus\{0\}\to [0,2\pi)\times(-1,\infty), \quad r(\sin\theta,\cos\theta)\to(\theta,r-1) $$
I think this function is an homeomorpshism, but I have trouble to prove it formally.
I think that $$\phi^{-1}: [0,2\pi)\times(-1,\infty)\to \mathbb{R}^2\setminus\{0\}, \quad (x,y)\to (y+1)(\sin x, \cos x) $$ is its inverse, and it is continuous.
I have trouble to prove formally that also $\phi$ is continuous. I know that if I do the pre-image of an open rectangle I find an "open" piece of circular crown, but I don't feel comfortable with a function defined in this manner.
In particular I was looking for a function $f:\mathbb{R}^2\to \mathbb{R}^2\setminus\{0\} $ which is an indentification and such that $\phi\circ f$ is continuous, and so from this i can deduce that also $\phi$ is continuous; or something similar. Is possibile to prove that $\phi$ is continuous in this way? Are there other smart and elegant way?
This can't be a homeomorphism, for example since the target space is contractible.
More explicitly it is homeomorphic to $[0, \infty) \times \mathbb R$, which does not really have the same "shape" as $\mathbb R^2 \setminus \{0\}$, since the latter has "nontrivial loops."
If you've seen why $[0,2 \pi)$ and $S^1$ are not homeomorphic, the reason is similar.
You can see that $\mathbb R^2 \setminus \{0\}$ is homeomorphic to $S^1 \times \mathbb R$ (a cyllinder) with basically that map though!
Take $(r \cos \theta,r \sin \theta) \mapsto (\theta,r)$. Note that it is important that $0$ is removed since it doesn't make sense to talk about $\theta$ there. Note that it is bijective and that open maps sets map to open sets in the product topology for example.