Elementary Differential Geometry Questions

[Saketh]

I'm trying to teach myself differential geometry from the internet, and I've hit a snag in proving homeomorphisms.

First, show that \Re^n is homeomorphic to any open ball in \Re^n. (I'm not sure how to write the conventional "R" using Latex.)

I'm trying to prove this statement, but I am having issues proving the continuity of the map and its inverse. It's clear that there exists a bijective map because the cardinality of both the open ball and the space itself are the same, but how should I go about proving continuity?

The next problem is to prove that a sphere with a point removed is homeomorphic to \Re^2. I have absolutely no idea how to do this. If someone could show me how to do this without skipping any steps, I would appreciate it.

 

[George Jones]

Can you think of how to show \mathbb{R} is homeomorphic to an open ball in \mathbb{R}, i.e., an open interval? Hint: visualize a function that does this.

\mathbb{R} is written using \mathbb{R}.

 

[Saketh]

The tangent function on the interval (-\frac{\pi}{2}, \frac{\pi}{2}) seems to be a homeomorphism because it is bijective, continuous, and its inverse (arctan) is continuous. I think I could use this to prove the first thing for \mathbb{R} , right?

But how do I generalize this to \mathbb{R}^n? And how am I supposed to come up with a specific example to prove that the punctured sphere is homeomorphic to \mathbb{R}^2?

Thank you for your help -- I think I am on the verge of having everything click.

 

[Saketh]

I just read that the punctured sphere isn't a solid ball with a point removed, as I had thought, but a shell. This suddenly makes the second problem clear. I still don't know how to prove the first one though.

How can I make this intuitive explanation of the punctured sphere problem rigorous:

From point p, draw curves outward that pierce the sphere and map each point to a point on R^2. Thus, this map is bijective and continuous, whose inverse is continuous (how do I show this rigorously?), therefore it is a homeomorphism.

 

[Hurkyl]

How can I make this intuitive explanation of the punctured sphere problem rigorous:

From point p, draw curves outward that pierce the sphere and map each point to a point on R^2. Thus, this map is bijective and continuous, whose inverse is continuous (how do I show this rigorously?), therefore it is a homeomorphism.

How do you show it rigorously? Directly. Explicitly prove that it's one to one and onto. Explicitly show that the inverse image of an open set is open, both for your construction and the inverse of your construction.

Or, you could compute an explicit formula for the transformation.

 

[LorenzoMath]

The following map f from R^n to B^n gives homeomorphism:

f(x)=x/2 if |x|is less than or equal to 1,
f(x)={1 - (1/2|x|)}(x/|x|) if |x| is greater than 1.

 

[Saketh]

How do you show it rigorously? Directly. Explicitly prove that it's one to one and onto. Explicitly show that the inverse image of an open set is open, both for your construction and the inverse of your construction.

I would really appreciate it if someone could show me how to start such a proof. I am unable to find examples of any proofs of homeomorphism on the internet, which is why I'm struggling.

The following map f from R^n to B^n gives homeomorphism:

f(x)=x/2 if |x|is less than or equal to 1,
f(x)={1 - (1/2|x|)}(x/|x|) if |x| is greater than 1.

I don't understand how this map maps R^n to B^n. Isn't it mapping R to B? I'm sorry for not understanding, but it's not clear to me how this map applies to sets of n-tuples.

 

[Hurkyl]

I would really appreciate it if someone could show me how to start such a proof. I am unable to find examples of any proofs of homeomorphism on the internet, which is why I'm struggling.

Let P be any point on the punctured sphere. {insert stuff here} Therefore, we have constructed a unique point on the plane. Call it f(P).

Let P be any point on the plane. {insert stuff here} Therefore, we have constructed a unique point on the punctured sphere. Call it g(P).

Let P be any point on the plane. {insert stuff here} Therefore, g(f(P)) = P.

Let P be any point on the punctured sphere. {insert stuff here} Therefore, f(g(P)) = P.

Let U be any open set in the plane. Let V be the set of points on the punctured sphere with f(P) lying in U. {insert stuff here} Therefore, V is open in the punctured sphere.

Let U be any open set in the punctured sphere. Let V be the set of points on the plane with f(P) lying in U. {insert stuff here} Therefore, V is open in the plane.

 

[Hurkyl]

I don't understand how this map maps R^n to B^n. Isn't it mapping R to B? I'm sorry for not understanding, but it's not clear to me how this map applies to sets of n-tuples.

He is using x to denote an element of Rⁿ.

 

[Saketh]

I'm going to try to fill in the blanks.

Let p be a point on the punctured sphere. Draw a curve from P to a unique point f(P) on the plane.

Let q be a point on the plane. Draw a curve from q to a unique point g(Q) on the punctured sphere.

Given p and q, f maps p to q, and g maps q to p. Therefore, f(g(q))=q, and g(f(p))=p.

Let U be an open set on the plane. For each point q in U, g(q) is open (is this true? I feel like my logic here is wrong). Thus, from the definition of a topological space, the union of all g(q), termed as V, is also open. Similar logic will prove that the image of an open set on the plane is open as well.

Thus f is continuous and bijective with a continuous inverse (g) , and thus the punctured sphere is homeomorphic to the plane.

(If I made any logical errors, I think I'd learn better at this point if someone simply corrected them.)

 

[LorenzoMath]

Yes, I used x to denote a point in R^n. So, x means (x_1, x_2, ... , x_n).

By the way, it is essentially the same thing, but let me give you the intuition with which I constructed the map proposed.

Saketh, you agree that R and I:=[-1,1] are homeomorphic, right? you said so yourself by introducing tangent function.

Then, by taking cartesian product with product topology, you can say, R^n is homeomorphic to I^n.

Then, it is clear, or at worse with computation, I^n is homeomorphic to B^n.

 

[cipher_42]

Let p be a point on the punctured sphere. Draw a curve from P to a unique point f(P) on the plane.

(If I made any logical errors, I think I'd learn better at this point if someone simply corrected them.)

Saketh,
Your logic here is fine; you definitely have the right mental image in your head, and that's where the proof has to start. But, just like when proving that R is homeomorphic to (-1,1), you have to have an explicit function
to take it to the level of rigor that you want. In proving that R is homeomorphic to (-1,1) you found the explicit mapping in the form of the tangent function, and then you used it's known properties (bijectivity, continuity, etc) to prove that it was a homeomorphism. So now to show that the punctured sphere is homeomorphic to the plane, you need to find another explicit[\i] mapping (what you're calling 'f'). In other words, if I parameterize the punctured sphere with the coordinates (u,v), then you need to find a formula for f(u,v): A -> R^2, where A is the domain of your parameterization of the punctured sphere. Then you can set about showing that f is a homeomorphism by showing that it's a continuous bijection with a continuous inverse. You can use that intuition about drawing lines through points on the sphere to construct f(u,v).

Let U be an open set on the plane. For each point q in U, g(q) is open (is this true? I feel like my logic here is wrong).

Once you have an explicit formula for g (the inverse of f), then you have to prove that it's true (which it will be, but that's where the fun comes in!).

I hope that makes some sense. These kind of proofs take a while to get used to, but keep at it!

- Jason