Homeomorphism of the projective n-space

[InvisibleBlue]

Hi,

I'm trying to prove that the projective n-space is homeomorphic to identification space B^n / ~ where for x, x' \in B^n: x~x'~\Leftrightarrow~x=x' or x'=\pm x \in S^{n-1},
The way I have tried to solve this is, I introduced:
{H_{+}}^{n}=\{x\in S^n | x_n \geq 0\}

Then {H_{+}}^{n}\cong B^n by the function F(x)=(\frac{x}{|x|}sin\frac{\pi}{2}|x|,~cos\frac{\pi}{2}|x|) [here \frac{x}{|x|}sin\frac{\pi}{2}|x|\in \mathbb{R}^n so cos\frac{\pi}{2}|x| is the (n+1)th component of F(x)]

Now I need to show that {H_{+}}^{n}/~ \cong P^n but I'm not sure how to do this rigorously without getting into a terrible mess.

Anyone has any ideas?

Thanks.

 

[StatusX]

For two spaces to be homeomorphic, 1) there must be a bijection between the underlying sets, and 2) this function and its inverse must be continuous, ie, have the preimage of open sets be open sets. In other words, 2 is asking the following: Do all open sets in the domain map to open sets in the target? Can all open sets in the target be mapped to this way? If so, you have a homeomorphism.

In your case, the bijection shouldn't be too hard to find, so we're left with 2. What are the open sets on your hemisphere space? You can think of them as unions of open disks on the sphere (with a slight subtlety for disks that straddle the equator). Do these map to open sets in projective space? This should be intuitively clear, and you can use an epsilon-delta type argument to prove it (you'll need to go back to the definition of projective space, specifically, what its open sets are). Finally, do the images of these disks form a basis for the topology on projective space? From here, it's just a little set theory to extend from disks to arbitrary open sets.

By the way, at no point am I suggesting using explicit coordinates on these spaces. This is rarely done in topology, where instead clear geometric explanations of a construction are given, which is usually just as rigorous and much easier to follow. Plus sometimes its not possible or feasible to use coordinates, so you should get used to doing things this way.