Obtaining $\mathbb CP^n$ as an identification space of $D^{2n}.$ 
Prove that $\mathbb C P^n$ is the identification space $D^{2n}/\sim$ where $x \sim y$ if either $x = y$ or $q(x) = q(y),$ where $q$ is the usual quotient map $S^{2n+1} \xrightarrow {q} \mathbb CP^n.$

My Attempt $:$ There is an obvious natural embedding $\iota : D^{2n} \longrightarrow S^{2n+1}$ given by $$(z_1,\cdots,z_n) \longmapsto \left (z_1,\cdots,z_n,\sqrt {1 - \sum\limits_{i = 1}^{n} |z_i|^2} \right ).$$ So the image of $D^{2n}$ under this embedding is the upper hemisphere of $S^{2n+1}.$ Since the quotient map is the $S^1$ action on $S^{2n+1}$ and the action of $S^1$ identifies the all the points in $S^{2n+1}$ which are $S^1$ multiples it identifies antipodal points of $S^{2n+1}$ in particular. So if we restrict $q$ to the upper hemisphere it will still be an onto continuous map. So by universal property of quotient topology it will induce a bijective continuous map $\widetilde {q} : D^{2n}/\sim \longrightarrow \mathbb C P^n,$ where $D^{2n}/\sim$ is the quotient space consisting of fibres of $q \big \rvert_{D^{2n}},$ where $D^{2n}$ is viewed as the upper hemisphere of $S^{2n+1}$ by means of the embedding $\iota.$ Now we use the well-known trick. The domain of $\widetilde {q}$ is compact and the range is Hausdorff and hence $\widetilde {q}$ is a homeomorphism.
Is my above reasoning correct? Can anyone kindly check it?
Thanks in advance.
 A: Your proof has a flaw.
Let us first have a look at the equivalence relation on $D^{2n}$. This is not defined very accurately. Writing $q(x) = q(y)$ requires an explanation which map $q$ is. I think we must interpret it as $q : S^{2n-1} = \partial D^{2n} \to \mathbb CP^{n-1}$ and not as $q : S^{2n+1} \to \mathbb CP^{n}$. The second interpretation would require a canonical identification of points $\xi  \in D^{2n}$ with points $\xi' \in S^{2n+1}$. This can be done via your embedding $\iota : D^{2n} \to S^{2n+1}$; then we can define $x \sim y$ if $q(\iota(x)) = q(\iota(y))$. But doing so would make it obsolete to define $x \sim y$ if either $x = y$ or $q(x) = q(y)$. This is why we need to use $q : S^{2n-1} \to \mathbb CP^{n-1}$; this map can only be applied to boundary points of $D^{2n}$ which explains the $x = y$ part in  the definition of $x \sim y$.
For the sake of notational transparency let us write $q_m : S^{2m+1} \to CP^{m}$ for the quotient map. Let us also recall what the action of $S^1$ on $S^{2m+1}$ looks like. We have $S^{2m+1} = \{ (z_1,\ldots,z_{m+1}) \in \mathbb C^{m+1} \mid \sum_{i=1}^{m+1} \lvert z_i \rvert^2 = 1 \}$ and for $\lambda \in S^1$ we have $\lambda (z_1,\ldots,z_{m+1}) = (\lambda z_1,\ldots,\lambda z_{m+1})$.
Claim. $x \sim y$ (in the sense $x = y$ or $q_{n-1}(x) = q_{n-1}(y)$) if and only if $q_n(\iota(x)) = q_n(\iota(y))$.
Proof. We consider $ x= (u_1,\ldots,u_n) \in S^{2n-1}$ and  $y= (v_1,\ldots,v_n) \in S^{2n-1}$.
Let $x \sim y$. If $x=y$, then trivially $q_n(\iota(x)) = q_n(\iota(y))$. So assume that $q_{n-1}(x) = q_{n-1}(y)$. This means that $x,y \in S^{2n-1}$ and $(\lambda u_1,\ldots,\lambda u_n) = (v_1,\ldots, v_n)$ for some $\lambda \in S^1$. But $x,y \in S^{2n-1}$ implies that $\iota(x)  = (u_1,\ldots,u_n,0)$ and $\iota(y)  = (v_1,\ldots,v_n,0)$. This gives us
$$\lambda \iota(x)  = \left(\lambda u_1,\ldots,\lambda u_n, 0 \right ) = \left(v_1,\ldots,v_n, 0 \right ) = \iota(y)$$
and therefore $q_n(\iota(x)) = q_n(\iota(y))$.
Conversely, let $q_n(\iota(x)) = q_n(\iota(y))$ which means that $\lambda \iota(x) =  \iota(y)$ for some $\lambda \in S^1$, i.e.  $\lambda x =  y$ and
$$\lambda \cdot \sqrt {1 - \sum\limits_{i = 1}^{n} |u_i|^2} = \sqrt {1 - \sum\limits_{i = 1}^{n} |v_i|^2} .$$
Since the square roots are real and nonnegative, this is only possible if $\lambda = 1$ or $\sqrt {1 - \sum\limits_{i = 1}^{n} |u_i|^2} = \sqrt {1 - \sum\limits_{i = 1}^{n} |v_i|^2} = 0$. In the first case we have $\iota(x) =  \iota(y)$, thus $x = y$. In the second case we have $x,y \in S^{2n-1}$ and $\lambda x = y$ which means $q_{n-1}(x) = q_{n-1}(y)$.
Now you have an error: The image of $D^{2n}$ under $\iota$ is not the upper hemisphere of $S^{2n+1}$. In fact, the upper hemisphere of $S^{2n+1}$ is a copy of $D^{2n+1}$. But what we can say is that
$$\iota(D^{2n}) =  \bar S^{2n+1} = \{ (z_1,\ldots,z_{n+1}) \in S^{2n+1} \mid z_{n+1} \text{ is real and nonnegative} \} .$$
The inclusion $\iota(D^{2n}) \subset \bar S^{2n+1}$ is obvious. Conversely, if $(z_1,\ldots,z_{n+1}) \in \bar S^{2n+1}$, then $z_{n+1} = \lvert z_{n+1} \rvert = \sqrt{1 - \sum_{i=1}^n \lvert z_i \rvert^2}$, thus $(z_1,\ldots,z_{n+1}) = \iota((z_1,\ldots,z_n))$.
The good news is that nevertheless $q_n(\iota(D^{2n}) = \mathbb CP^{n}$. In fact, consider $\xi = (z_1,\ldots, z_{n+1}) \in S^{2n+1}$. If $z_{n+1} = 0$, then $x= (z_1,\ldots, z_n) \in S^{2n-1}$ and $\xi = \iota(x)$. If $z_{n+1} \ne 0$, then $\lambda = \frac{\overline{z_{n+1}}}{\lvert z_{n+1}\rvert} \in S^1$ and $\lambda \xi = (z'_1,\ldots,z'_n, \lvert z_{n+1} \rvert)$. We conclude that $\lambda \xi = \iota(x)$ with $x = (z'_1,\ldots,z'_n) \in D^{2n}$. Thus in both case $q_n(\xi) = q_n(\iota(x)) \in q_n(\iota(D^{2n})$.
From that point your arguments correctly apply.
Let us remark that the alternative representation $\mathbb CP^{n} = D^{2n}/\sim$ means nothing else than that $\mathbb CP^{n}$ is obtained from $\mathbb CP^{n-1}$ by attaching a $2n$-cell $D^{2n}$ via $q_{n-1} : S^{2n-1} \to CP^{n-1}$.
