# Diffeomorphism: Unit Ball vs. Euclidean Space

In my differential geometry class we are being asked to prove that the open unit ball

$B^n$ = { $x$ $\in$ $\mathbb{R}$$^n such that |x| < 1} is diffeomorphic to \mathbb{R}$$^n$

I am having a hard time with this as I am brand new not only to differential geometry, but also topology.

I know that I need to construct a smooth, differentiable bijection between the two with a differentiable inverse, but beyond that, I am unsure of where to start. Some guidance in the right direction would be greatly appreciated.

• Try doing it for $n=1$ first.
– user98602
Nov 17, 2014 at 2:32
• For a text, I'd recommend Lee's Introduction to Topological Manifolds starts out with an accessible introduction to the relevant topology with a view toward manifolds, and is a readable, compact introduction to the titular subject. (Full disclosure: I know Lee personally.) Nov 17, 2014 at 3:32
• Ah, thanks! I've heard that his Introduction to Smooth Manifolds book is also great. By any chance would you recommend that I read one before the other, or does it matter? Nov 17, 2014 at 6:12
• My books are written to be read in the order Topological Manifolds -> Smooth Manifolds -> Riemannian Manifolds. You'll get a lot more out of them in that order. Happy reading! Nov 17, 2014 at 16:08
• Wow, I wasn't expecting to hear back from the author himself! Thanks for the tip, it's very much appreciated. Nov 19, 2014 at 5:28

Let $\phi \colon [0,1) \to [0, \infty)$ a diffeomorphism with inverse $\psi$. Some possible choices: $t \mapsto \frac{t}{1-t}$, $t \mapsto \tan (\frac{\pi}{2}\cdot t)$.

The map $$x \mapsto \phi(||x||) \cdot \frac{x}{||x||}$$

is a diffeomorphism from $B^n$ to $\mathbb{R}^n$ with inverse $$y \mapsto \psi(||y||) \cdot \frac{y}{||y||}$$

$\bf{Added:}$ It turns out that the choice of the diffeomorphism from $[0,1)$ to $[0,\infty)$ matters a lot, since $x \mapsto ||x||$ is not smooth at $0$. This was brought to my attention by @Freeze_S and I thank him a lot! One can check that the map obtained for $\phi(t) = \frac{t}{1-t}$ is only $C^1$ at $0$... However, we can use the map so kindly suggested by @Jesus RS: ( big thanks! ) $\phi(t) = \frac{t}{\sqrt{1-t^2}}$ with inverse $\psi(s) = \frac{s}{\sqrt{1+s^2}}$ and it will work just fine. The diffeomorphisms are, as written by @Jesus RS: $$x \mapsto \frac{x}{\sqrt{1-||x||^2}} \\ y \mapsto \frac{y}{\sqrt{1+||y||^2}}$$

In fact, as long as $\phi(t)$ is an odd function of $t$ things will work OK. So, another example is

$$x \mapsto \frac{\tan (\frac{\pi}{2} \cdot ||x|| )}{||x||} \cdot x$$

• One must also take (as you surely have in mind) that $\phi$ is increasing, so that $\lim_{t \to 0^+} \phi(t) = 0$. Also, one must check that the map $x \mapsto \phi(||x||) \cdot \frac{x}{||x||}$ (well, pedantically, its extension by $0$ to continuity) is smooth at $0$. Is this true for the candidates $\phi$ you suggest? Nov 17, 2014 at 3:36
• @Travis: Oh, let's just take the half closed intervals, thanks for the observation! I see the issue with the smoothness at $0$, perhaps harmless since in fact $\frac{\phi(||x||)}{||x||}$ is smooth at $0 \in \mathbb{R}$. Nov 17, 2014 at 3:38
• @orangeskid: No, not harmless! Consider for example: $\phi(|x|)=\frac{1}{1-|x|}$ But the problem is not that it doesn't tend to zero or that it's not increasing the problem is that it is not smooth at zero. That's why for example you have to consider a more sophisticated construction, as is done in Lee's SM. Jan 4, 2015 at 15:35
• @Freeze_S Thank you. Yes, there is an issue with the smoothness at $0$. Let me see if anything can be salvaged. Note that the map from the ball to the full space is $x \mapsto \frac{1}{1-||x||} \cdot x$ Jan 6, 2015 at 0:41
• @orangeskid: Ah yes the notions agree then. Ah and my statement was wrong: The open annulus cannot be even homeomorphic to the euclidean space since it is not simply connected. Jan 7, 2015 at 11:56

Or you can try $f(x)=x/\sqrt{1-|x|^2}$ for $x\in B^n$.

• Thank you, this is the one. And I used it to salvage my answer. Jan 6, 2015 at 0:59

Standard Approach

Consider the identification: $$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_E^2}}$$ Its inverse is given explicitely by: $$\Psi:\mathbb{R}^n\to\mathbb{B}^n:\Psi(y):=\frac{y}{\sqrt{1+\|y\|_E^2}}$$ The argument of the roots never vanish: $$1-\|x\|_E^2\neq0\quad1+\|y\|_E^2\neq0$$ So they're both differentiable.

Alternative Approach

Consider the identification: $$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{1-\|x\|_E^2}$$ By the inverse function theorem: $$\mathcal{N}d\Phi(x)\equiv(0)\implies\mathrm{d}\Psi(y)=\mathrm{d}\Phi(\Psi(y))\in\mathcal{L}(\mathbb{R}^n)$$ But this holds globally as the identification is a bijection.

Problematic Approach

Consider the identification: $$\Phi:\mathbb{B}^n\to\mathbb{R}^n:\Phi(x):=\frac{x}{\sqrt{1-\|x\|_\infty^2}}$$ This is most likely not differentiable.

• I am sorry, I am confused that how can we ensure the norm is smooth?
– user198206
Oct 18, 2015 at 19:02

Not every smooth function induces a smooth map: $$\Phi:\mathbb{B}\to\mathbb{R}:\quad\varphi(|x|):=\frac{1}{1-|x|}$$ Just have a careful look at its diagram:

(Note that it is not even differentiable at zero!)

The problem is that the norm is not smooth in general: $$\|\cdot\|:V\to[0,\infty)$$

One can master this almost only by patching a bump on top: $$\varphi(r):=e^{-\frac{1}{r^2}}\cdot\frac{1}{1-r}$$

For a mere diffeomorphism still the whole story highly depends on the chosen norm!

As to the book recommendations I suggest the following:

(i) A course in Differential Geometry (W. Klingenberg)

(ii) Elementary Differential Geometry (A. Pressley)

(iii) Differential-Geometrie und Minimal-Flächen (J. Jost) (in german but it's a wonderful book).

Some more advanced books dealing with differential geometry in the context of differentiable manifolds I'd suggest:

(v) Differential Geometry: Manifolds, Curves and Surfaces (M. Berger, B. Gostiaux)

(vii) A Comprehensive Introduction to Differential Geometry (M. Spivak) (all the volumes)

I hope it helps.

As to your question about the diffeomorphism between the open unit ball and $\mathbb R^n$ the idea is defining a function which will stretch the open ball in all directions.