Proof of Homeomorphism between D^n/S^{n-1} and S^n

[madness]

Homework Statement

Prove that $$D^n/S^{n-1}$$ is homeomorphic to $$S^n$$. (Hint - try the cases n = 1,2,3 first).

Homework Equations

X/Y is defined as the union of the complement X\Y with one point. I showed in a previous section that the equivalence relation x~x' if x and x' are in Y gives X/~ = X/Y. The identification space topology was then used, and I showed that a subset Z of X with $$Y \cap Z$$ non-empty is such that p(Z) is open in X/Y if and only if $$Y \cup X$$ is open in X.

The Attempt at a Solution

First I define the open sets. A set in $$D^n$$ is open if $$U = D^n \cap V$$ for some V open in $$R^n$$. A subset U in $$D^n /S^{n-1}$$ is open if $$U \cap S^{n-1}$$ is empty and U is open in $$D^n$$ or if the intersection is non-empty as in the relevant equations section.
A set U in $$S^n$$ is open if $$U = V \cap S^n$$ for V open in $$R^{n+1}$$.

I need a function mapping open sets to open sets. For n = 1 this was easy since I knew the basis for $$S^1$$ was open arcs, so f = (cos pi(x+1), sin pi(x+1)) (I think this is a homeomorphism) worked. The only was I could explicitly do this for n = 2 would be to use the spherical polar coordinates that I know from physics and other maths courses, which I wouldn't be able to generalise anyway. I'm not really sure how to proceed here.

 

[mighty2000]

Thank you for your post. I would like to offer a proof for the statement you have presented. Let's consider the cases n = 1, 2, and 3 first.

For n = 1, we have D^1/S^0 = D^1/{-1,1}. Since {-1,1} is just two points, we can define a function f: D^1/S^0 -> S^1 by mapping the point -1 to (1,0) and the point 1 to (-1,0). This function is continuous and bijective, and its inverse is also continuous, thus making it a homeomorphism.

For n = 2, we have D^2/S^1 = D^2/{(x,y) in R^2: x^2 + y^2 = 1}. Here, we can use the polar coordinate system to define a function g: D^2/S^1 -> S^2, where g(r,theta) = (r*cos theta, r*sin theta, sqrt(1-r^2)). Again, this function is continuous and bijective, with a continuous inverse, making it a homeomorphism.

For n = 3, we have D^3/S^2 = D^3/{(x,y,z) in R^3: x^2 + y^2 + z^2 = 1}. Similar to the previous case, we can use the spherical coordinate system to define a function h: D^3/S^2 -> S^3, where h(r,theta,phi) = (r*sin phi*cos theta, r*sin phi*sin theta, r*cos phi, sqrt(1-r^2)). This function is also continuous and bijective, with a continuous inverse, making it a homeomorphism.

Now, let's consider the general case for n > 3. We can use the same approach as above, using the n-dimensional spherical coordinate system to define a function from D^n/S^{n-1} to S^n . This function will be continuous and bijective, with a continuous inverse, thus proving that D^n/S^{n-1} is homeomorphic to S^n .

I hope this helps in understanding and proving the statement. If you have any further questions or concerns, please let me know.[