I'm trying to prove that the open ball of radius 1 centered at the origin in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$. I believe the "shrinking map" from $\mathbb{R}^n$ to the ball given by $x \mapsto \dfrac{x}{1 + |x|}$ does the job, but I'm having trouble showing it's a homeomorphism, particularly the "continuous inverse" part. What's a good way to do this?
General Topology – How to Find a Homeomorphism from R^n to the Open Unit Ball?
general-topology
Related Solutions
Let me outline a proof; I shall leave the details of this proof as exercises.
Exercise 1: Prove that the unit circle $S^1$ is homeomorphic to the space obtained from the unit interval $[0,1]$ by identifying the endpoints $0$ and $1$. (Hint: the exponential mapping $\theta\to e^{i\theta}$ can be composed with a linear mapping to give a homeomorphism.)
Let $C$ be the image of a simple closed curve in the plane, that is, let $C$ be the image of a continuous function $f:[0,1]\to \mathbb{R}^2$ such that $f:(0,1)\to\mathbb{R}^2$ is injective and $f(0)=f(1)$.
Exercise 2: Prove that $C$ is homeomorphic to the unit circle $S^1$. (Hint: show that $f:[0,1]\to C$ induces a continuous bijection $\tilde{f}:S^1\to C$ using Exercise 1 and then show that $f$ is an open mapping using the compactness of $S^1$ and the fact that the plane is Hausdorff.)
We now need an elementary lemma of point-set topology:
Exercise 3: Let $X$ be a topological space and let $X=A\cup B$ where $A$ and $B$ are closed subspaces of $X$. Prove that if $g:A\to Y$ and $h:B\to Y$ are continuous functions into a topological space $Y$ such that $g(x)=h(x)$ for all $x\in A\cap B$, then there is a unique continuous function $f:X\to Y$ such that $f(a)=g(a)$ and $f(b)=h(b)$ for all $a\in A$ and $b\in B$.
Finally, we can prove the result of your question:
Exercise 4: Prove that the unit square is the image of a simple closed curve in the plane and conclude that it is homeomorphic to the unit circle. (Hint: you can use Exercise 3 to "glue" continuous mappings together.)
Exercise 5: Give examples of images of simple closed curves in the plane (using Exercise 3) and conclude (using Exercise 2) that they are homeomorphic to the unit circle $S^1$.
I hope this helps!
Injectivity
$F(x)=F(y)$, with $x\ne0$ and $y\ne0$ means
$$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}= \tan\biggl(\frac{\pi \|y\|}{2}\biggr)\frac{y}{\|y\|} $$
So we can assume $x\ne0$ and $y\ne0$. Then, taking norms, $$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)=\tan\biggl(\frac{\pi \|y\|}{2}\biggr) $$ because $x/\|x\|$ has norm $1$ and the same for $y/\|y\|$; moreover $\tan t>0$ if $t>0$. By the injectivity of the tangent you get $$ \frac{\pi \|x\|}{2}=\frac{\pi \|y\|}{2} $$ so $\|x\|=\|y\|$ and, finally $x=y$.
It's clear that $F(x)=0$ if and only if $x=0$, so the proof is complete.
Surjectivity
$0=F(0)$, so we only need to show that, for $z\ne0$, we can find $x$ with $F(x)=z$. We should have $$ z=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|} $$ so $$ \|z\|=\tan\biggl(\frac{\pi \|x\|}{2}\biggr) $$ hence $$ \frac{\pi\|x\|}{2}=\arctan\|z\| $$ that is, $$ \|x\|=\frac{2\arctan\|z\|}{\pi}. $$ Thus the candidate is $$ x=\frac{2\arctan\|z\|}{\pi\|z\|}z $$ Verify it's the right one.
Continuity
The continuity of $F$ is obvious in the points different from $0$, because it's obtained by continuous functions. Since your domain consists of vectors with norm $<1$, there's no problem with the tangent function, because you consider only arguments in $(0,\pi/2)$. The continuity of the inverse is obvious as well outside $0$
Are $F$ and its inverse continuous at $0$?
A well known fact is that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$; now $$ \|F(x)\|=\biggl\|\tan\biggl(\frac{\pi\|x\|}{2}\biggr)\frac{x}{\|x\|}\biggr\| =\tan\biggl(\frac{\pi\|x\|}{2}\biggr) $$ that obviously satisfies the requested property. The same for the continuity at $0$ of $F^{-1}$.
Best Answer
To figure out the inverse, note that if $|x|=\lambda$, then $$\left|\frac{x}{1+|x|}\right| = \frac{\lambda}{1+\lambda}.$$ Thus, given $y\in\mathbb{R}^n$ with $0\leq |y|\lt 1$, you want to find $\lambda$ such that $\lambda = (1+\lambda)|y|$. Letting $|y|=\mu$, we have $(1-\mu)\lambda = \mu$, or $\lambda = \frac{1}{1-\mu}$.
So the map you want for the inverse is $$y\longmapsto \frac{y}{1-|y|}.$$ Note that this is well-defined, since $0\leq |y|\lt 1$, so $0\lt 1-|y|\leq 1$. Also, the compositions are the identity: $$\begin{align*} x &\longmapsto \frac{x}{1+|x|}\\ &\longmapsto \left(\frac{1}{1- \frac{|x|}{1+|x|}}\right)\frac{x}{1+|x|} = \left(\frac{1+|x|}{1+|x|-|x|}\right)\frac{x}{1+|x|}\\ &= \vphantom{\frac{1}{x}}x.\\ y &\longmapsto \frac{y}{1-|y|}\\ &\longmapsto \left(\frac{1}{1 + \frac{|y|}{1-|y|}}\right)\frac{y}{1-|y|} = \left(\frac{1-|y|}{1-|y|+|y|}\right)\frac{y}{1-|y|}\\ &= \vphantom{\frac{1}{y}}y. \end{align*}$$ Now simply verify that both maps are continuous.