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? 

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.
