Topology of complex projective plane It is well known there are two ways to construct topology of $\mathbb{C}P^n$:


*

*quotient space of $S^{2n+1}$ by identifying $x$ with $\lambda x$, where $\lambda$ is complex nonzero constant.

*According to cell complex, quotient space of $D^{2n}$ by identifying $x \in \partial D^{2n}$ with $\lambda x \in \partial D^{2n}$, where $\lambda$ is complex nonzero constant. 
Well, of course, topologies of above two spaces are same, but I want to prove this fact implicitly. I tried an analytic method, but too absurd: Can anyone give me a hint? 
 A: The key theorem you need from elementary topology is that any surjective map from a compact space to a Hausdorff space is a quotient map. To apply that theorem you must set up the appropriate quotient maps. This is a purely topological problem, requiring you to guess the correct formulas for the appropriate quotient maps.
You have described constructions 1 and 2, but let me also describe a construction which which is closest to the definition of $\mathbb{C}P^n$: 


*

*Construction 0: $\mathbb{C}P^n$ is the quotient of $\mathbb{C}^{n+1}-\{0\}$ defined by the equivalence relation $x \sim zx$ for each $x \in \mathbb{C}^{n+1}-\{0\}$ and each $\lambda \in \mathbb{C} - \{0\}$. 


Let $f : \mathbb{C}^{n+1}-\{0\} \to \mathbb{C}P^n$ be the function such that $f(x)$ equals the equivalence class of $x$. The topology on $\mathbb{C}P^n$ is defined to be the unique one such that $f$ is a quotient map.
Next let me compare construction 0 and construction 1. Let $g : S^{2n+1} \to \mathbb{C}P^n$ be the restriction of the function $f$ to the unit sphere $S^{2n+1} \in \mathbb{C}^{n+1}-\{0\} = \mathbb{R}^{2n+2}-\{0\}$. According to the definition of $f$, two points $x,y \in S^{2n+1}$ satisfy $g(x)=g(y)$ if and only if there exists $\lambda \in \mathbb{C}-\{0\}$ such that $x=\lambda y$ (notice that $\lambda$ must also have norm $1$ since $x$ and $y$ have norm $1$). Since the domain of $g$ is compact and its range is Hausdorff, the above theorem applies, and therefore $g$ is a quotient map. It follows that $\mathbb{C}P^n$ is homeomorphic to the quotient space obtained from $S^{2n+1}$ by identifying $x,y$ if and only if there exists $\lambda \ne 0$ such that $x=\lambda y$ (again the $\lambda$ must have norm $1$).
Now, to get to the heart of your question, let's compare construction 1 and construction 2. The method is similar: construct a quotient map $h : \mathbb{D}^{2n} \to \mathbb{C}P^n$ such that $h(x)=h(y)$ if and only if $x,y \in \partial\mathbb{D}^{2n}$ and $x=\lambda y$ for a nonzero complex constant (which, again, must have norm $1$). We will construct $h$ using $g$. To do this, consider the inclusion $\mathbb{D}^{2n} \subset \mathbb{C}^{n} \subset \mathbb{C}^{n+1}$. Map $\mathbb{D}^{2n}$ to $S^{2n+1}$ using the function $$p(a_1,b_1,...,a_n,b_n,0,0) = (a_1,b_1,...,a_n,b_n,\,\sqrt{1 - (a_1^2+b_1^2+...+a_n^2+b_n^2)}\,\,,\,\,0)
$$
We then have a map
$$h = g \circ p : \mathbb{D}^{2n} \to \mathbb{C}P^n
$$
Now check that this map $h$ is surjective, its domain is compact, and its range is Hausdorff. Applying the key theorem, $h$ is a quotient map. Also check that $h(x)=h(y)$ if and only if $x=y$ or $x,y \in \partial \mathbb{D}^{2n}$ and $x = \lambda y$ for some $\lambda \in \mathbb{C}$ (which as said must have norm $1$). It follows that $\mathbb{C}P^n$ has the description in your construction 2.
A: I am not sure that you look for this type of argument, but anyway:
Try to look at the sphere $S^{2n+1}$ embedded in the usual way in $\mathbb{C}^{n+1}$. By definition elements in $\mathbb{C}P^n$ are lines in $\mathbb{C}^{n+1}$ through the origin. Now look at the sphere and examine the intersection of each line with the sphere. You will then receive the object from your first definition.
The second one follows from the first. This disk has to be viewed as one half of the sphere from the first definition. Imagine that you glue all the upper half of $S^{2n+1}$ with the bottom half according to your quotient. Then you have not glued only the border. Voilà!
All these arguments are in some way analytic, as all these equivalences can be written as maps.
I hope my answer clarifies a bit, I will try to ameliorate it otherwise.
