The projection $\pi(x,y,z) = (x,y)$ defines a homeomorphism between $\mathbb{R}^{2}$ and $C = \{(x,y,z) \in \mathbb{R}^{3} \mid x^{2} + y^{2} - z = 0, z \geq 0\}$. This is the classic approach to show that the cone and $\mathbb{R}^{2}$ are homeomorphic. My question is just curiosity:
- Is there a more elegant way to show this homeomorphism? I mean, without having to check each condition, maybe using some stronger result.
Edit. I know that the projection give a simple proof. I wanted to know if this homeomorphism can be obtained as a consequence of some stronger result, without the need to use an explicit function.
It as an obvious inverse: $(x,y)\mapsto(x,y,-x^2-y^2)$. Furthermore, this functions is clearly continuous. How more elegant than this can you get?