Quotient space of the unit sphere

[hedipaldi]

prove that the quotient space obtained by identifying the points on the southern hemisphere, is homeomorphic to the whole sphere.I am trying to define a homeomorphism between the quotient space and the sphere,and i need help doing it.
Thank's in advance.

 

[gopher_p]

(1) The equator would necessarily need to be part of the "southern hemisphere" in order for the claim to be true.

(2) Are you able to show that the northern hemisphere - minus the equator - is homeomorphic to the sphere less a single point? Don't worry so much yet about finding a formula for a function that does this; just get a basic idea/picture for how the homeomorphism might work.

 

[mathwonk]

nice answer.

 

[hedipaldi]

(1) The equator would necessarily need to be part of the "southern hemisphere" in order for the claim to be true.

(2) Are you able to show that the northern hemisphere - minus the equator - is homeomorphic to the sphere less a single point? Don't worry so much yet about finding a formula for a function that does this; just get a basic idea/picture for how the homeomorphism might work.

Thank's

 

[blue_raver22]

I am happy to help you with this problem. Let's start by defining the quotient space and the unit sphere. The quotient space is obtained by identifying the points on the southern hemisphere of the unit sphere, which means that we are collapsing all the points on the southern hemisphere into a single point. This results in a space that is topologically equivalent to the sphere, but with one point identified.

To prove that this quotient space is homeomorphic to the whole sphere, we need to find a continuous map between the two spaces that is bijective and has a continuous inverse. This map will serve as our homeomorphism.

First, let's define the unit sphere as S^2 = {(x,y,z) | x^2 + y^2 + z^2 = 1}. Now, let's define our map f: S^2 -> S^2 as follows:

f(x,y,z) = (x,y,-z)

This map takes any point on the sphere and reflects it across the xy-plane, essentially flipping it to the opposite hemisphere. This map is continuous because it is a composition of continuous functions (multiplication and addition) and it is also bijective because every point on the sphere has a unique reflection.

Next, let's define our equivalence relation ~ on S^2 such that (x,y,z) ~ (x,y,-z). This means that any point on the southern hemisphere is equivalent to its reflection on the northern hemisphere. This is exactly what we did when we collapsed the points on the southern hemisphere to a single point in the quotient space.

Now, we can define a map g: S^2/~ -> S^2 as follows:

g([x,y,z]) = f(x,y,z)

This map takes a point on the quotient space, which is represented by an equivalence class [x,y,z], and maps it to its reflection on the sphere using our map f. This map is well-defined because it does not matter which point in the equivalence class we choose, the result will always be the same reflection. Furthermore, g is bijective because every point on the sphere has a unique reflection and every point in the quotient space is represented by an equivalence class.

Finally, we need to show that g and g^-1 are continuous. This can be done by showing that the preimage of open sets in S^2 is open in S^2/~ and vice versa. Since g is a bijection, g