TITLE: What are the open subsets of $\mathbb{R}^n$ that are diffeomorphic to $\mathbb{R}^n$
QUESTION [64 upvotes]: I would like to know if there is a known necessary and sufficient
property on an open subset of $\mathbb{R}^n$ to be diffeomorphic to $\mathbb{R}^n$ :
For example :

Are all open star-shaped subsets of $\mathbb{R}^n$ diffeomorphic to $\mathbb{R}^n$ ?

Reciprocally, are all open subsets of $\mathbb{R}^n$ which are diffeomorphic to $\mathbb{R}^n$, star-shaped ?


Thank you for your answers and proofs

REPLY [2 votes]: Yet another proof in the case where $\Omega$ is star-shaped, in little details unless anyone is interested.
Assume for simplicity that $\Omega$ is star-shaped around $0$. Define the function $r$ mapping $\theta\in\mathbb S(\mathbb R^n)$ to the only $r(\theta)$ such that $\Omega\cap \mathbb R_+\theta = [0,r(\theta))\theta$ (possibly $r(\theta)=\infty$). Then $r:\mathbb S(\mathbb R^n)\to[0,+\infty]$ is lower semicontinuous and uniformly positive, so there exists a sequence $(r_k)_{k\geq0}$ of smooth uniformly positive functions such that $r=\sum_{k\geq0}r_k$ (e.g. use this post to write $r$ as an increasing limit of Lipschitz functions and reduce to the Lipschitz case, then iteratively fit a smooth function $r_{k+1}$ strictly between $(r-r_k)/2$ and $r-r_k$ by using a partition of unity). Without loss of generality, $r_0$ is actually a (positive) constant.
The idea is that we are looking for a diffeomorphism $\mathbb R^n\to\Omega$ such that the image of the sphere of radius $k$ is the surface $\lbrace |x|=r_k(x/|x|)\rbrace$. The final diffeomorphism will end up being a bit different, but I think it is a good image to keep in mind.
Find a collection of smooth functions $(\rho_k)_{k\geq0}$ from $\mathbb R_+$ to $[0,1]$ such that

for all $k>0$, $\rho_k$ starts at zero, then increases and finally stays constant equal to one, in such a way that $\rho_k'$ has support in $(k-1,k+2)$ and is (uniformly) positive over $[k,k+1]$;
$\rho_0$ starts at zero, then increases and finally stays constant equal to one, in such a way that $\rho_0'$ is one in a neighbourhood of zero, has support in $[0,2)$, and is (uniformly) positive over $[0,1]$.

Then $$\Phi:x=|x|\cdot\frac x{|x|}=r\cdot\theta\mapsto\sum_{k\geq0}\rho_k(r)r(\theta)\theta$$ is smooth at zero (it is a constant multiple of the identity) and everywhere else (it is locally a finite sum of smooth functions), it is injective (because it preserves rays and is strictly increasing on rays) and surjective with image $\Omega$ (because $r=\sum_{k\geq0}r_k$), and its differential is invertible (away from zero because it is strictly increasing with positive derivative along the rays and it preserves the angles). It means that $\Phi$ has an inverse $\Phi^{-1}$ that is locally smooth, so $\Phi$ is actually a diffeomorphism.