Showing that $\varphi: \overline{\mathbb{B}^2} \amalg \overline{\mathbb{B}^2} \to \mathbb{S}^2$ is a quotient map

Question

Let $\overline{\mathbb{B}^2}$ be the closed unit ball in $\mathbb{R}^2$, i.e. $$\overline{\mathbb{B}^2} := \{x \in \mathbb{R}^2 : \|x\| \leq 1\}$$

Consider the mapping $$\varphi: \overline{\mathbb{B}^2} \amalg \overline{\mathbb{B}^2} \to \mathbb{S}^2$$ which (informally) maps one $\overline{\mathbb{B}^2}$ to the lower and the other to the upper hemisphere of $\mathbb{S}^2$ (of course they will overlap on $\mathbb{S}^1$). Now I want to show that this is a quotient map. My question is, how should I approach this task?

First of all, it is somehow clear that $\varphi$ is continuous and surjective. I mean, we could construct a continuous map from $\overline{\mathbb{B}^2}$ to, say, the upper hemisphere. Hence it will be enough to show that $\varphi$ is open. To do so, it should be enough to show that it maps open balls to open balls since the open balls are a basis of any metric topology. But then I struggle,. Do I have to write out $\varphi$ explicitely? I do not see another way.

Edit. I think that a possible explicit formula for $\varphi$ is $$\varphi: \begin{cases} \overline{\mathbb{B}^2} \to \mathbb{S}^2\\ (x,y) \mapsto (x,y,\pm\sqrt{1 - x^2 - y^2})\end{cases}$$ Hence left to show is that $\varphi$ is open.

Answer 1

Your formula for $\varphi$ is correct (assuming the $\pm$ is supposed to become $+$ on one of the $\overline{\mathbb{B}^2}$s and $-$ on the other). So, having constructed a continuous surjection $\overline{\mathbb{B}^2} \amalg \overline{\mathbb{B}^2} \to \mathbb{S}^2$ you are done for free by the following general fact:

Let $f:X\to Y$ be a continuous surjection from a compact space to a Hausdorff space. Then $f$ is a quotient map.

To prove this, let $Z$ be the set $Y$ equipped with the quotient topology with respect to $f$, so that $f$ is a quotient map when considered as a map $X\to Z$. By the universal property of the quotient topology (or in other words, the fact that the quotient topology is the finest topology that makes $f$ continuous), the identity map $id:Z\to Y$ is continuous. But $Z$ is compact, being a continuous image of $X$, and $Y$ is Hausdorff. Since a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, $id$ is a homeomorphism. That is, $Y$ has the quotient topology and $f$ is a quotient map.