Consider the quotient space $Q = D^{2n}/ (\Bbb S^1 \circlearrowleft \partial D^{2n})$ obtained from $D^{2n}$ by letting $\Bbb S^1$ act on the boundary $\partial D^{2n}$ by left (or right) multiplication. Show that $Q \approx \Bbb CP^n.$
I know that $\Bbb CP^n : = \Bbb S^{2n+1}/ \Bbb S^1.$ But from here how do I deduce that? For that I need to get a continuous surjective map $p : D^{2n} \longrightarrow \Bbb CP^n$ such that fibres of $p$ are the orbits under the action. But how do I find that?
Any help in this regard will be appreciated. Thanks in advance.
EDIT $:$ I have one idea. Let $w = (w_1,w_1,\cdots,w_n) \in D^{2n}$ where $w_i \in \Bbb C,$ for $i = 1,2, \cdots, n$ then we send it to $\left (w_1,w_2, \cdots, w_n, \sqrt {1 - |w|^2} \right) \in \Bbb S^{2n+1}$ via some map $\phi$ and then compose it with the quotient map $p : \Bbb S^{2n+1} \longrightarrow \Bbb C P^n.$ So I only need to verify whether $p \circ \phi$ is surjective and the fibres of the map $p \circ \phi$ are the orbits of the given action.