Skip to main content
MathOverflow

Return to Answer

added 36 characters in body
Source Link
Dmitrii Korshunov
Dmitrii Korshunov
  • 2.9k
  • 20
  • 27
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by JanetWhitney-BurstinGrauert-Morrey embedding theorem) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ again by JanetWhitney-BurstinGrauert-Morrey to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ again by Janet-Burstin to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Whitney-Grauert-Morrey embedding theorem) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ again by Whitney-Grauert-Morrey to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
added 7 characters in body
Source Link
Dmitrii Korshunov
Dmitrii Korshunov
  • 2.9k
  • 20
  • 27
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ again by NashJanet-TognoliBurstin to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ by Nash-Tognoli to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ again by Janet-Burstin to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
added 16 characters in body
Source Link
Dmitrii Korshunov
Dmitrii Korshunov
  • 2.9k
  • 20
  • 27
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ by Nash-Tognoli to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until it$fg$ becomes surjective.
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until it becomes surjective.
  1. Any analytic $n$-manifold admits an analytic surjection $f$ from $\mathbb{R}^n$: pick an analytic Riemannian metric (e.g. by embedding to $\mathbb{R}^N$ by Janet-Burstin) and consider the corresponding exponential mapping. Surjectiveness follows from Hopf-Rinow and analyticity from Cauchy-Kovalevskaya.
  2. Any analytic $m$-manifold with $m\ge n$ admits an analytic map $g$ to $\mathbb{R}^n$ with $g(M)$ having non-empty interior: embed $M$ by Nash-Tognoli to some $\mathbb{R}^L$ and take a generic projection to an $n$-dimensional subspace.
  3. If $fg$ is surjective then we are done. If not, then compose $g$ with dilation of $\mathbb{R}^n$ with the center in the interior of $g(M)$ until $fg$ becomes surjective.
Source Link
Dmitrii Korshunov
Dmitrii Korshunov
  • 2.9k
  • 20
  • 27