Let $H^2$ be the closed upper hemisphere, that is $$H^2=\big\{(x,y,z\in\mathbb{R}^3)\;|\;x^2+y^2+z^2=1, z\ge0\big\},$$ and let $D^2$ be the closed unit disk $$D^2=\big\{(x,y)\in\mathbb{R}^2)\;|\;x^2+y^2=1\big\}.$$
On $H^2$ , define an equivalence relation $\sim$ by identifying the antipodal points on the equator: $$(x, y, 0)\sim(−x, −y, 0),\quad x^2 + y^2 = 1.$$ On $D^2$ , define an equivalence relation $\sim$ by identifying the antipodal points on the boundary circle: $$(x, y) \sim (−x, −y),\quad x^2 + y^2 = 1.$$
I must prove that $$H^2/\sim\;\simeq D^2/\sim.$$
The space $D^2$ and $H^2$ are homeomorphic to each other via the continuous map $\varphi\colon H^2\to D^2$, $\varphi(x, y, z)=(x, y)$, and its inverse $\psi\colon D^2\to H^2$ , $\psi(x, y) = \big(x, y,\sqrt{1-x^2-y^2}\big)$. Note that if $\pi_1\colon H^2\to H^2/\sim$ and $\pi_2\colon D^2\to D^2/\sim$ are the projection map, then exists the following commutative diagram:
$\require{AMScd}$ \begin{CD} H^2 @>\varphi>> D^2\\ @V \pi_1 V V @VV \pi_2 V\\ H^2/\sim @>>\overline{\varphi}> D^2/\sim \end{CD}
indeed the map $(\pi_2\circ\varphi)\colon H^2\to D^2/\sim$ is constant on each equivalence class of $H^2/\sim$, therefore it induces a map $\overline{\varphi}\colon H^2/\sim\to D^2/\sim$ by $$\overline{\varphi}\big([(x,y,z)]\big)=\pi_2\circ\varphi(x,y,z)=\pi_2(x,y)=[(x,y)],\;\text{for}\; (x,y,z)\in H^2.$$ The map $\overline{\varphi}$ is continuos, since $\pi_2\circ \varphi$ is continuos.
Similarly, exists the following commutative diagram:
$\require{AMScd}$ \begin{CD} D^2 @>\psi>> H^2\\ @V \pi_2 V V @VV \pi_1 V\\ D^2/\sim @>>\overline{\psi}> H^2/\sim \end{CD}
indeed the map $(\pi_1\circ\psi)\colon D^2\to H^2/\sim$ is constant on each equivalence class of $D^2/\sim$, therefore it induces a map $\overline{\psi}\colon D^2/\sim\to H^2/\sim$ by $$\overline{\psi}\big([(x,y)]\big)=\pi_1\circ\psi(x,y)=\pi_1(x,y,z)=[(x,y,z)],\;\text{for}\; (x,y)\in D^2.$$ The map $\overline{\psi}$ is continuos, since $\pi_1\circ \psi$ is continuos. It remains to show that $\overline{\varphi}$ and $\overline{\psi}$ are the inverse of the other. Therefore, $$\overline{\varphi}\circ\overline{\psi}\big([(x,y)]\big)=\overline{\varphi}\big([(x,y,z)]\big)=[(x,y)],$$ and
$$\overline{\psi}\circ\overline{\varphi}\big([(x,y,z)]\big)=\overline{\psi}\big([(x,y)]\big)=[(x,y,z)].$$
Question. Is this procedure correct? If I could have shown the result more immediately?
Thanks!
As you know, the projection $\varphi : H^2 \to D^2, \varphi (x,y,z) = (x,y)$, is a homeomorphism. You have $p \sim p'$ in $H^2$ if and only if $\varphi (p) \sim \varphi (p')$ in $D^2$. It is therefore obvious that $\varphi$ induces a homeomorphism $\tilde{\varphi} : H^2/\sim \quad \to D^2/\sim, \tilde{\varphi}[p]) = [\varphi(p)]$.
But of course, your proof is correct.