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I'm trying to prove the homeomorphism

$$\mathbb{R}^2 \backslash \{(0,0)\} \cong \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 = z^2, z>0\}$$

so I need to find a bijective, continuous function and it's inverse that maps the plane (minus the origin) onto the positive cone and vice-versa.

One possible function is

$$f(x,y) \rightarrow (x,y,\sqrt{x^2+y^2})$$

which has inverse

$$g(x,y,z) \rightarrow (x,y)$$

So now I need to show that $f$ and $g$ are continuous, bijective and open. I can gather the following intuitively but I'm not sure it's thorough enough to prove the conditions.

  • $f$ is continuous since $x^2 + y^2 > 0$ for all $x,y\in\mathbb{R}^2$ \ $\{(0,0)\}$, and $g$ is continuous for all $(x,y,z)\in\mathbb{R}^3, z>0$.

  • $f$ is bijective as each $(x,y)$ corresponds to a unique $(x,y,\sqrt{x^2+y^2})$ and the same clearly applies for $g$.

  • $f$ and $g$ are both open as they map open sets onto open sets.

Would this be an adequate proof?

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1 Answer 1

Why do you need to show that $f$ and $g$ are open maps? A homeomorphism just needs that $f$ and $g$ are continuous and mutually inverse.

As for the other points

  • you need to show that $f$ and $g$ are well defined (this means showing that the image of $f$ as a function on to $\mathbb{R}^3$ is a subset of the RHS and similarly for $g$ and the LHS),

  • that $f$ is surjective (you didn't mention this, only injectivity),

  • and continuity of $f$ and $g$ will depend on what is expected of you.

Many people would find it acceptable to just note that the functions are the restriction of a common continuous function without necessarily proving that the functions themselves are continuous.

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