I am reading Hatcher's book and I have a problem understading how the complex projective space $\mathbb CP^n$ can be realised as a quotient of $D^{2n}$ (page 7)
Let me briefly outline his arguments (or what I make of them):
$\mathbb CP^n=S^{2n+1}/\sim$ where $x\sim \lambda x$, $\left|\lambda \right|=1$.
$S^{2n+1}=\left\{(w,a\sqrt {1-\left|w\right|^2})\in \mathbb C^n\times \mathbb C: \left|w\right|\le 1, \left|a\right|=1\right\}$, $S^{2n-1}=\left\{(w,0)\in \mathbb C^n\times \left\{0\right\}: \left|w\right|=1\right\}$ $D^{2n}_+=\left\{(w,\sqrt {1-\left|w\right|^2})\in \mathbb C^n\times [0,+\infty):\left|w\right|\le 1\right\}$.
"Each vector in $S^{2n+1}$ is equivalent under the identifications $v ∼ λv$ to a vector in $D^{2n}_+$ , and the latter vector is unique if its last coordinate is nonzero. If the last coordinate is zero, we have just the identifications $v ∼ λv$ for $v ∈ S^{2n−1}$"
My problem is with Step 3, I do not see how this implies that the two quotient spaces are homeomorphic.
Here is my idea: Consider the map $S^{2n+1}-S^{2n-1}\to D^{2n}_+$, \begin{equation*}\left(w, a\sqrt{1-\left|w\right|^2}\right)\mapsto \left(\frac wa,\sqrt{1-\left|w\right|^2}\right)\end{equation*} Then, starting from $S^{2n+1}-S^{2n-1}\coprod S^{2n-1}$ we can first go to $S^{2n+1}$ (via the identity) and then to $\mathbb CP^n$ (call this composition of maps $f$) or first to $D^{2n}_+-S^{2n-1}\coprod S^{2n-1}$ and then to $D^{2n}_+$ (by the identity) and finally to $D^{2n}_+/\sim$ (call this composition of maps $g$) . $f,g$ are both continuous, have the same domain and make the same identifications. If I could show that they are open then by a general Theorem, the quotient spaces would be homeomorphic.
However, I do not think it should be that complicated; Hatcher must have something else in mind. Is there a general principle for quotient spaces that I am forgetting?