Proof

Without loss of generality we may suppose the origin is an interior point of $D$. We claim that the map $\phi: v \mapsto v/{\|v\|}$ maps the boundary $\partial D$ homeomorphically onto $S^{n-1}$. By convexity, $D$ is homeomorphic to the cone on $\partial D$, and therefore to the cone on $S^{n-1}$ which is $D^n$.

The claim reduces to the following three steps.

1. The restricted map $\phi: \partial D \to S^{n-1}$ is continuous.

2. It’s surjective: $D$ contains a ball $B = B_{\varepsilon}(0)$ in its interior, and for each $x \in B$, the positive ray through $x$ intersects $D$ in a bounded half-open line segment. For the extreme point $v$ on this line segment, $\phi(v) = \phi(x)$. Thus every unit vector $u \in S^{n-1}$ is of the form $\phi(v)$ for some extreme point $v \in D$, and such extreme points lie in $\partial D$.

3. It’s injective: for this we need to show that if $v, w \in \partial D$ are distinct points, then neither is a positive multiple of the other. Supposing otherwise, we have $w = t v$ for $t \gt 1$, say. Let $B$ be a ball inside $D$ containing $0$; then the convex hull of $\{w\} \cup B$ is contained in $D$ and contains $v$ as an interior point, contradiction.

So the unit vector map, being a continuous bijection $\partial D \to S^{n-1}$ between compact Hausdorff spaces, is a homeomorphism.

Proof

$D$ has nonempty interior relative to its affine span which is some $k$-plane, and then $D$ is homeomorphic to $D^k$ by the theorem.

Proof

For instance, the smooth map

has smooth inverse

Proof

Suppose $T$ is a star-shaped open subset of ${\mathbb {R}}^n$ centered at the origin. Theorem 2.29 in Lee 2009 proves that there is a function $f$ on ${\mathbb{R}}^n$ such that $f\gt 0$ on $T$ and $f$ vanishes on the complement of $T$. By applying bump functions we can assume that $f\le 1$ everywhere and $f=1$ in an open $\epsilon$-neighborhood of the origin; by rescaling the ambient space we can assume $\epsilon=2$.

The smooth vector field $V\colon x\mapsto f(x)\cdot x/{\|x\|}$ is defined on the complement of the origin in $T$. Multiply $V$ by a smooth bump function $0\le b\le 1$ such that $b=1$ for ${\|x\|} \gt 1/2$ and $b=0$ in a neighborhood of 0. The new vector field $V$ extends smoothly to the origin and defines a smooth global flow $F\colon \mathbb{R} \times T\to T$. (The parameter of the flow is all of $\mathbb{R}$ and not just some interval $(-\infty,A)$ because the norm of $V$ is bounded by 1.) Observe that for $1/2\lt {\|x\|} \lt 2$ the vector field $V$ equals $x\mapsto x/{\|x\|}$. Also, all flow lines of $V$ are radial rays.

Now define the flow map $p\colon{\mathbb{R}}^n_{\gt 1/2}\to T_{\gt 1/2}$ as $x\mapsto F({\|x\|}-1, \frac{x}{{\|x\|}})$ for ${\|x\|} \gt 1/2$. (The subscript $\gt 1/2$ removes the closed ball of radius $1/2$.) The flow map is the composition of two diffeomorphisms,

hence itself is a diffeomorphism. (Note particularly that the latter map is surjective. In detail: a flow line is a smooth map of the form $L: (A,B) \to T$, where $A$ and $B$ can be finite or infinite. If $B$ is finite and the limit of $L(t)$ as $t \to B$ exists, then the vector field $V$ vanishes at $B$. In our case $V$ can only vanish at the boundary of $T$, which is precisely what we want for surjectivity.)

Finally, define the desired diffeomorphism $d\colon{\mathbb{R}}^n\to T$ as the gluing of the identity map for ${\|x\|} \lt 2$ and as $p$ for ${\|x\|}\gt 1/2$. The map $g$ is smooth because for $1/2\lt {\|x\|} \lt 2$ both definitions give the same value.

Proof

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

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

Now we define $f:\Omega\rightarrow\mathbb{R}^n$ via the formula:

Clearly $f$ is smooth on $\Omega$.

We set $A(x)=\sup\{t\gt0\mid t\frac{x}{\|x\|}\in\Omega\}$. $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{\|x\|}$ to the ray $\mathbf{R}_+\frac{x}{\|x\|}$. Moreover, $f(0\frac{x}{\|X\|})=0$ and

Indeed, if $A(x)=+\infty$, then it holds for obvious reason. If $A(x)\lt+\infty$, then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{\|x\|})=0$. Hence by the mean value theorem and the fact that $\phi$ is $C^1$ due to

for some constant $M$ and every $r$. As a result,

diverges. Hence we infer that $f([0,A(x))\frac{x}{\|x\|})=\mathbf{R}_+\frac{x}{\|x\|}$ and so $f(\Omega)=\mathbf{R}^n$.

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

Suppose that $d_x f(h)=0$ for some $x\in\Omega$ and $h\neq 0$. From definition of $f$ we get that

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 $\lambda(x)\ge1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x \lambda(x)\gt0$, which gives a contradiction.