After googling this fact (which is used in Bott and Tu to show the existence of good covers of manifolds) I've gotten the impression this is somewhat difficult to prove. But I also came across this homework problem (problem 0) with a hint: http://www.math.columbia.edu/~thaddeus/geometry/hw10.pdf

which gives me the impression that it is maybe not so difficult as I thought. If anyone has any thoughts about how to prove this result I'd be very appreciative. Thanks for your time.

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    $\begingroup$ Related thread (but without a self-contained proof): mathoverflow.net/questions/4468/… $\endgroup$
    – user31373
    Commented Aug 20, 2012 at 1:39
  • $\begingroup$ Thanks, LVK. Yeah that was something that led me to believe that the proof was long and complicated which makes me puzzled by the fact that it was assigned as a homework problem (albeit for extra credit and at Columbia) $\endgroup$ Commented Aug 20, 2012 at 1:45
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    $\begingroup$ I suppose that the suitable $f$ is $f(q)=\mathrm{dist}(q,\partial U)$ (or a smooth version of it, i.e., a $C^\infty$ positive function comparable to the distance. Such a function is constucted, e.g., in Stein's 1970 book Singular Integrals). This gives you the flow of $fX$ which stays within $U$, and the flow of $X$ on $\mathbb R^n$. It remains to think of a neat way to map one onto the other. $\endgroup$
    – user31373
    Commented Aug 20, 2012 at 1:49
  • $\begingroup$ How about this: pick a small sphere $S\subset U$ centered at $0$. For each point $x\in S$ let $n_x$ be the outward unit vector. Let $\gamma_x$ be the solution of $\dot\gamma=fX(\gamma)$ with $\gamma(0)=x$, $\dot \gamma(0)=n_x$. Also, let $\Gamma_x$ be the solution of $\dot\Gamma=X(\Gamma)$ with $\Gamma(0)=x$, $\dot \Gamma(0)=n_x$. Now map $\gamma_x(t)$ to $\Gamma_x(t)$. $\endgroup$
    – user31373
    Commented Aug 20, 2012 at 1:59
  • $\begingroup$ Thanks again. It will take me a bit to digest this. $\endgroup$ Commented Aug 20, 2012 at 2:16

2 Answers 2


I gave an answar here as well.

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{||x||}$ to the ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ Hence by Mean value theorem and the fact that $\phi$ is $C^1$ $$\phi\left(r\frac{x}{||x||}\right)\leqslant M(A(x)-r)$$ for some constant $M$ and every $r.$ As a result $$\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}$$ diverges. Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$


Let $U$ be an open subset of $\mathbb R^n$ that is star-shaped with respect to the origin $0\in U$. Let $f:U\mapsto (0,\infty)$ be a regularized distance function, i.e., a $C^\infty$ function such that $f(x)/\operatorname{dist}(x,\partial U)$ is pinched between two positive constants.

Aside: how to construct such $U$? Take all maximal dyadic cubes $Q_k\subset U$ such that $\operatorname{dist} (Q_k, \partial U)\ge \operatorname{diam} Q_k $ (aka Whitney decomposition of $U$). Let $\varphi_k$ be a smooth partition of unity associated to the cover of $U$ by larger cubes $\frac32Q_k$. Then $f(x)=\sum \operatorname{dist} (Q_k, \partial U)\,\varphi_k(x)$.

Considering the construction above, it's clear that we can make $f$ constant in a neighborhood of $0$, say $f=K$ there. Pick $r>0$ small enough so that the sphere $r\,\mathbb S^{n-1}$ is contained in this neighborhood. For each unit vector $\xi\in \mathbb S^{n-1}$, let $\gamma_\xi:\mathbb R\to U$ be the solution of the ODE $\dot \gamma_\xi=f(\gamma)\,\gamma$ with initial value $\gamma_\xi(0)=r\xi$. Observe that $\gamma_\xi(t)=e^{Kt}r \xi$ for $t\le 0$. For $t>0$, the integral curve still goes in the direction $\xi$, but slows down approaching $\partial U$ and never leaves $U$. It should be clear that the integral curves sweep out $U$.

Define a map from $\mathbb R^n$ onto $U$ by $$F(\rho\, \xi)= \gamma_\xi( K^{-1}\log \rho ),\quad \rho> 0, \ \xi\in \mathbb S^{n-1} \tag1$$ On the unit ball $F$ is linear. It is $C^\infty$ smooth everywhere. It is a bijection onto $U$. The invertibility of its derivative follows from ODE theorems on the dependence of solutions on initial values, see Hartman.

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    $\begingroup$ Referencing Hartman is a proof by intimidation. $\endgroup$
    – 40 votes
    Commented Jul 16, 2013 at 23:24

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