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For calculations in homology we often need to know how to identify some quotient spaces.

Let $\mathbb{C}P^n:=\mathbb{C}^{n+1}\setminus \{0\} /\sim$, the complex projective space where $x,y\in \mathbb{C}^{n+1}\setminus \{0\}$ satisfy $x\sim y$ $:\iff \exists \lambda\in \mathbb{C}\setminus \{0\}$ such that $x=\lambda y$. I also know that $\mathbb{C}P^n\cong S^{2n+1}/x\sim \lambda x,\;\lambda\in S^1$ and $\mathbb{C}P^n\cong D^{2n}/v\sim \lambda v, v\in S^{2n}$.

Question: How to see that $\mathbb{C}P^{n+1}/\mathbb{C}P^n\cong D^{2n+2}/ S^{2n+1}$?

We have $$\mathbb{C}P^{n+1}/\mathbb{C}P^n\cong \bigg(D^{2(n+1)}/v\sim \lambda v, v\in S^{2(n+1)}\bigg)/\bigg(D^{2n}/v\sim \lambda v, v\in S^{2n}\bigg)$$. But how to proceed or how to construct that homeomorphism :$\mathbb{C}P^{n+1}/\mathbb{C}P^n\to D^{2n+2}/ S^{2n+1}$?

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    When you write $\mathbb{C}P^{n+1}/\mathbb{C}P^n$, what exactly do you mean? Also: to construct homeomorphisms of quotients like this, it's not a bad idea to start with a map of the total spaces, i.e., a map from $D^{2n+2}$ to $CP^{n+1}$, and then see whether it can be made to pass to the quotient nicely. – 2017-01-08
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    I mean the quotient space of $\mathbb{C}P^{n+1}$ with $\mathbb{C}P^{n}$. $\mathbb{C}P^{n}$ is considered as a subspace of $\mathbb{C}P^{n+1}$ by using the above identifications of the complex projective space and by considering $S^{2n+1}$ as the boundary of $D^{2n+2}$. Or do you mean something else? – 2017-01-08
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    @Ryan: The natural approach given what you've stated (and following John Hughes' good advice) is to start with the map $D^{2n+2} \to \mathbf{CP}^{n+1}$ obtained by (partial) boundary identification, noting that $S^{2n+1} \to \mathbf{CP}^{n}$; then collapse $\mathbf{CP}^{n}$ to a point and consider what happens to $S^{2n+1}$, the boundary of $D^{2n+2}$. – 2017-01-08

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The first step is that $\mathbb{C}P(n)-\mathbb{C}P(n-1)$ is just the result of removing the points $[z_0,\ldots,z_n]$ so that $z_n=0$. This is an open set that we denote $U_n\subset \mathbb{C}P(n)$. In any standard development of $\mathbb{C}P(n)$ you will see that there is a homeomorphism $$\phi_n:U_n\rightarrow \mathbb{C}^n$$ given by $\phi_n([z_0,\ldots,z_n])=(z_0/z_n,z_1/z_n,\ldots,z_{n-1}/z_n)$. Hence $\mathbb{C}P(n)-\mathbb{C}P(n-1)$ is homeomorhic to $\mathbb{C}^n=\mathbb{R}^{2n}$. Notice that $\mathbb{C}P(n)/\mathbb{C}P(n-1)$ is a compact Hausdorff space that has one more point than $U_n$, therefore, it is the one point compactification of $\mathbb{R}^{2n}$, which is $S^{2n}$. On the other hand $D^{2n}/S^{2n-1}$ is also the one point compactification of $\mathbb{R}^{2n}$, so they are homeomorphic.