3

Let $T :\mathbb{R}^m \rightarrow \mathbb{R}^n $ be a linear map, where $m > n$. Then, it is possible to prove that there is no $T$ injective in this case. In fact, if such a map exists the Rank-Nullity Theorem gives us $$ n \geq \dim Im (T) = m > n, $$ so, we obtain a contradiction. A natural question arises here:

It is possible to obtain a injective continuous (nonlinear) map $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ when $m > n$ ?

I try to construct such an example, but it seems so delicate. There exist some way to prove that a such map does not exist ?

Marcelo Ng
  • 374

1 Answers1

1

Suppose there was such an $f$, so $f$ is an embedding $\mathbb{R}^m \to \mathbb{R}^n$. Then since $m>n$, there is a copy of $S^n$ inside $\mathbb{R}^m$, and $f$ must restrict to an embedding $S^n \to \mathbb{R}^n$.

But such maps don't exist because $S^n$ can't be embedded in $\mathbb{R}^n$.

kamills
  • 2,107
  • 11
  • 15
  • Dear @kamills,the part "..such maps don't exist because $S^n$ can't be embedded in $R^n$ caused me confusion. Can you give me a textbook reference? – Marcelo Ng Nov 22 '19 at 19:26
  • I think that the following idea can be related to your answer: Consider the restriction $f \mid_{S^m} \rightarrow f(S^m) \subset R^n$. Then it is a homeomorphism? – Marcelo Ng Nov 22 '19 at 19:27