Constructing Onto & Continuous Function from D^{2n} to \mathbb{C}P^n

[MathematicalPhysicist]

I am asked to construct an onto and continuous function from $$D^{2n}$$ onto $$\mathbb{C}P^n$$ such that it's one- to -one on the interior of $$D^{2n}$$.

I was thinking of sending the line that joins two antipodal points on the boundary of this ball, is this right, or should I be looking for something else?

Thanks.

 

[lavinia]

I am asked to construct an onto and continuous function from $$D^{2n}$$ onto $$\mathbb{C}P^n$$ such that it's one- to -one on the interior of $$D^{2n}$$.

I was thinking of sending the line that joins two antipodal points on the boundary of this ball, is this right, or should I be looking for something else?

Thanks.

For the 2 sphere - CP1 - just map the boundary of the disk to the south pole. Generalize this.

 

[MathematicalPhysicist]

I don't understand how does this maps D^2n onto CP^n?

 

[MathematicalPhysicist]

I mean CP^n is homeomorphic to a sphere of S^{2n+1}, so if I map S^2n to a point at the south pole of S^2n, I don't see how does this mapping cover all of S^{2n+1} ~ CP^n?

 

[quasar987]

What's a "sphere of S^{2n+1}"? Anyway, CP^n is definitely not a sphere for n>1. It is obtained from quotienting S^{2n+1} (seen as sitting in C^{n+1}) by the obvious circle action (the circle seen as sitting in C). In particular, there is a natural surjective map pr:S^{2n+1}-->CP^n.

In particular, CP^0 = {pt} and CP^1=S^2. Why? Because there is a homeomorphism between the open set U:={[z0:z1]| z0 not equal to 0} and C obtained by sending [z0:z1] to z1/z0. And what is the complement of U in CP^1? Just 1 point! (the point [0:1]) Thus, CP^1 = C^1 u CP^0. Great, so we see how to build our map D²-->CP^1 from this decomposition: just map int(D²) to C^1 homeomorphically in the obvious way, and map ∂D²=S^1 surjectively onto CP^0 according to the god-given quotient map pr:S^{0n+1}-->CP^0. Now generalize this!