Lee - Introduction to Topological Manifolds - How to solve exercises

Question

I am reading the book by Lee - Introduction to topological Manifolds and I like it a lot how it explains the things. I was reading the book by Isidori (Nonlinear Control Systems) and here there is more focus on the explanation of what is a manifold, Riemannian manifold etc. The books are totally different. For an introduction on topological manifolds this (as the title suggests) is better.

Anyway, what I find really hard in this book is to follow the examples sometime and to solve the exercises. I have no idea where to start from and the book does not help much.

An example of an Example that I don't understand (Example 2.25): Let $\mathbb{B}^n \subseteq \mathbb{R}^n$ be the unit ball, and define a map $F:\mathbb{B}^n\rightarrow\mathbb{R}^n$ by:

$F(x)=\frac{x}{1-|x|}$.

Direct computation shows that the map $G:\mathbb{R}^n\rightarrow\mathbb{B}^n$ defined by

$G(y)=\frac{y}{1+|y|}$

is an inverse for $F$. Thus F is bijective, and since $F$ and $F^{-1}=G$ are bot h continuous, $F$ is an omeomorphism.

My questions:

• How can you arrive to the expression of $G $?
• How do you formally prove that $F$ is an homeomorphism?

Next problem I have no clue how to solv, for example, exercises 2.27 and 2.28.

I don't want the solutions but an hint which will point me in the right direction.

Thanks

Answer 1

In order to find the inverse of the function $f$ we can solve $x = \frac{y}{1 - |y|}$ in order to get an expression $y = G(x)$. To prove that $f$ is a homeomorphism, you have to prove that it is a bijection (Ik, since you found an inverse), that $f$ is continuous (ok, since it is the quotient of continuous functions and $1- |x|$ is not zero in the open unit ball, si ce $\|x\| < 1$. As a last part you have to prove that $f^{-1}$ is continuous, but this is just $g$ and the same reasoning shows it is continuous.

Answer 2

Let $X = \mathbb{B}^n$ and $Y = \mathbb{R}^n$. By definition, $\alpha(x):= 1 - \| x \| > 0$ whenever $x \in X$. Because the norm is a continuous real function on $X$ and the difference of continuous real functions on $X$ is again a continuous real function on $X$, the map $\alpha:X \to \mathbb{R}$ must be continuous. Moreover, because $\alpha$ is never zero, the map $\beta:X \to \mathbb{R} \, ; \, x \mapsto 1/\alpha(x)$ is also continuous, since it is a ratio of continuous real functions in which the denominator function is never zero. This implies that $F:X \to Y : x \mapsto \beta(x) \cdot x$ is continuous (since scaling a continuous vector valued function by a continuous scalar valued function results in a continuous vector valued function).

Similar considerations show that $G:Y \to X$ is continuous.

Thus, in order to show that $F$ is a homeomorphism, we need only check that $G \circ F$ and $F \circ G$ are identity functions, and this is a matter of straight plugging in:

$G \circ F(x) = \frac{\frac{x}{1-\|x\|}}{1+\left\lVert\frac{x}{1-\|x\|}\right\rVert} = \frac{\frac{x}{1-\| x\|}}{\frac{1-\|x\|}{1 - \|x\|} + \frac{\|x\|}{1-\|x\|}} = x$

This shows that $G \circ F$ is the identity function on $X$. Analogous computation shows that $F \circ G$ is the identity function on $Y$.