How to prove that $\overline{\Bbb H^1} \times \overline{\Bbb H^1} \cong \overline{\Bbb H^2}$? Let  $\overline{\Bbb H^n} = \{(x^1, ..., x^n) \in \Bbb R^n \mid x^n \geq 0\}$, equipped with the usual subspace topology of $\Bbb R^n$.
I'm trying to prove that $\overline{\Bbb H^1} \times \overline{\Bbb H^1}$ is homeomorphic to $\overline{\Bbb H^2}$.
I can see it "visually" (it's like opening a book) but I don't know how to find an explicit map...
 A: So we are actually trying to show $[0, +\infty\rangle \times [0, +\infty\rangle \cong \mathbb{R} \times [0, +\infty\rangle$.
As @Moishe Coher suggests, you can use polar coordinates. Notice that:
$$[0, +\infty\rangle \times [0, +\infty\rangle = \left\{(r,\phi) : r \ge 0, \phi\in \left[0, \frac\pi2\right]\right\}$$
$$\mathbb{R} \times [0, +\infty\rangle = \left\{(r,\phi) : r \ge 0, \phi\in \left[0, \pi\right]\right\}$$
Obviously we should consider the map $f : [0, +\infty\rangle \times [0, +\infty\rangle \to \mathbb{R} \times [0, +\infty\rangle$ given by $$(r,\phi) \stackrel{f}{\mapsto} (r,2\phi)$$ It is indeed a homeomorphism of the two spaces.

You can express $f$ explicitly in Cartesian coordinates. After a bit of computation we arrive at the formula:
$$f(x,y) = \begin{cases}
\frac1{\sqrt{x^2+y^2}}\left(x^2-y^2, 2xy\right),  & \text{if $(x,y) \ne (0,0)$} \\
(0,0), & \text{if $(x,y) = (0,0)$}
\end{cases}$$
$f$ is obviously continuous for $(x,y) \ne (0,0)$. To verify continuity of $(0,0)$ note that
$$\|f(x,y)\|^2 = \frac{(x^2-y^2)+4x^2y^2}{x^2+y^2} = x^2 + y^2 = \|(x,y)\|^2$$
The inverse function is given explicitly as:
$$f^{-1}(x,y) = \begin{cases}
\frac1{\sqrt{2}}\left( w(x,y), \frac{-2x\sqrt{x^2+y^2}w(x,y) + w(x,y)^{3}}{y\sqrt{x^2+y^2}} \right),  & \text{if $(x,y) \ne (0,0)$} \\
(0,0), & \text{if $(x,y) = (0,0)$}
\end{cases}$$
where $w(x,y) = \sqrt{x\sqrt{x^2+y^2} + x^2 + y^2}$.
Again, continuity of  $f^{-1}$ at $(x,y) \ne (0,0)$ is evident from the formula, while continuity at $(0,0)$ follows from $\|f^{-1}(x,y)\| = \|(x,y)\|$.
We conclude that $f$ is a homeomorphism.
