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The weak Whitney embedding theorem states that any continuous function from an $n$-dimensional manifold to an $m$-dimensional manifold may be approximated by a smooth embedding provided $m > 2n$.

Some differential topology textbooks state that this approximate version of Whitney embedding theorem cannot be improved. However, they only gave simple counterexamples from 1 to 2 dimensions. ('if $S^1$ is mapped into $\mathbb{R^2}$ so that the image curve crosses itself like a figure 8, no sufficiently close approximation can be injective [Hirsch1976, Differential Topology].')

My question is, can we construct such a counterexample to show that $m=2n+1$ is the lower bound? In other words, is it possible to construct a continuous mapping $f: [0,1]^n →R^m$, for any $\epsilon>0$, there is no smooth embedding $g: [0,1]^n →R^m$ such that $|f(x)-g(x)|<\epsilon,\forall x \in [0,1]^n, n < m <=2n$?

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  • $\begingroup$ What do you mean by the "weak Whitney theorem"? What does mean that a continuous function cannot be embedded? If you mean that the manifold cannot be embedded, you did not ay into what. "Hirsh [...] said [...] that the approximation version cannot be improved". Approximation version of what? Your question should be self-contained and if possible (which is possible here) should not require the readers to read external sources to understand the question. $\endgroup$
    – Piotr Hajlasz
    Commented Sep 25, 2023 at 11:33
  • $\begingroup$ I was in a hurry to write and didn't pay attention to these issues. Thank you for your advice. I will update my questions and elaborate in detail later $\endgroup$
    – li ang Duan
    Commented Sep 25, 2023 at 12:38
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    $\begingroup$ Any continuous map where you have two discs which map linearly to a neighborhood of 0 in $0 \times \Bbb R^n$ and $\Bbb R^n \times 0$, respectively, cannot be perturbed to be injective by a small perturbation. (Consider the linking number of two small spheres around 0; this will be 1, but if a small perturbation made this map injective, the linking number would be 0.) It suffices to take $T^n \to \Bbb R^{2n}$ by taking products of your map with itself. $\endgroup$
    – mme
    Commented Sep 25, 2023 at 14:14

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