Let $Y:=\mathbb{C}\cup\{\infty\}$ be the one-point compactification of $\mathbb{C}$. Define
Consider the map $$\Phi: SU(2)\to Y: \begin{pmatrix}a & b\\ c & d\end{pmatrix}\mapsto \frac{b}{a}$$ where we define $b/0:= \infty$ for every $b\in \mathbb{C}$.
I want to show that this map is continuous. I tried the following:
Indeed, if $W\subseteq Y$ is an open subset, we distinguish two cases. In the first case, $\infty \notin W$, but then $$\Phi^{-1}(W)= \Psi^{-1}(W)$$ where $$\Psi: \left\{\begin{pmatrix}a & b \\c & d\end{pmatrix}\in SU(2): a \ne 0\right\}\to \mathbb{C}: \begin{pmatrix}a & b \\c & d\end{pmatrix}\mapsto \frac{b}{a}$$ so by continuity of $\Psi$ we see that $\Psi^{-1}(W)$ is open in the domain of $\Psi$, which itself is open in $SU(2)$. Therefore, $\Psi^{-1}(W)$ is open in $SU(2)$.
In the second case, $\infty \in W$. Then I couldn't finish.
Any input is highly appreciated!
You need to exploit that $SU(2)$ has the topology of the three-sphere:
Since the determinant of each matrix in $SU(2)$ must be one we know that $ad-cb=1\,.$ Also, since each matrix is orthonormal, we can write it as $$ \begin{pmatrix}u+iv & -w+ix\\w+ix&u-iv \end{pmatrix} $$ with $u^2+v^2+w^2+x^2=1\,.$ This means that $u^2+v^2\le 1$ and $w^2+x^2\le 1\,.$ In other words, it is not possible that any of $a,b,c,d$ diverges to $\infty\,.$
Now let $a_n\to 0$ and $b_n\to b\in\mathbb C\,,c_n\to c\in\mathbb C\,,d_n\to d\in\mathbb C\,.$ Since $(d_n)$ cannot diverge it follows from $a_nd_n-c_nb_n=1$ that $$ c_nb_n\to 1\,. $$ Since $c_n$ cannot diverge it is not possible that $b_n\to 0\,.$ This implies $$ \frac{b_n}{a_n}\to\infty=\Phi\left(\begin{smallmatrix}0&b\\c&d\end{smallmatrix}\right)\,. $$