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I have the following question, more general that Continuous injective map $f:\mathbb{R}^3 \to \mathbb{R}$?

Let $U\subset\mathbb{R}^n$ be a domain. If $f:U\longrightarrow\mathbb{R}^m$ ($m< n$) is a continuous function, how to prove that $f$ is not injective.

Any hint would be appreciated.

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felipeuni
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1 Answers1

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This is really an application of a classical fact known as Invariance of Domain. Namely, take $g:U\to \mathbb{R}^{m}\times \mathbb{R}^{n-m}$, by $g(x)=(f(x), 0)$. Then $g$ is an injective continous function from $U\to \mathbb{R}^n$, so by invariance of domain has an open image. But $Im(g)=Im(f)\times \{0\}$, which cannot possibly be open in $\mathbb{R}^n$ while $n>m$. Thus we have a contradiction.

Pax
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  • A continuous bijection is not necessarily a homeomorphism. – T.J. Gaffney Nov 13 '16 at 03:49
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    @Gaffney It is when your domain is an open subset of $\mathbb{R}^n$ and the range is $\mathbb{R}^m$. This is the content of the qouted theorem, "Invariance of Domain" – Pax Nov 13 '16 at 03:57
  • It may be a consequence, but as stated, it requires $n=m$. – T.J. Gaffney Nov 13 '16 at 04:00
  • The image on the linked wikipedia page seems to show a counter-example to what you're claiming. – T.J. Gaffney Nov 13 '16 at 04:01
  • @Gaffney As $m<n$ one can embed $\mathbb{R}^m$ in $\mathbb{R}^n$ naturally. – Sayantan Nov 13 '16 at 04:03
  • @Sayantan $\mathbb R^m$ is not open in the topology of $\mathbb R^n$ – T.J. Gaffney Nov 13 '16 at 04:04
  • Look, the question is about continuous injective maps from $\mathbb R^m\to\mathbb R^n$. If $m<n$, such maps exists by space-filling curves. Whatever argument you make needs to not apply to this case. But this argument does. – T.J. Gaffney Nov 13 '16 at 04:09
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    @Gaffney The problem is for $f : U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ when $m < n$. One can view $f : U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ as a map $f : U \subseteq \mathbb{R}^n \to \mathbb{R}^n$. Then $f(U)$ is open in $\mathbb{R}^n$ and $f$ is homeomorphism onto $f(U)$. – Sayantan Nov 13 '16 at 04:11
  • @Sayantan What if $f(U)=\mathbb R^m$ ? Then $f(U)$ is not open in $\mathbb R^n$. Just like a line is not open in a plane. – T.J. Gaffney Nov 13 '16 at 04:14
  • @Gaffney That's the point. Such an $f$ cannot exist. – Sayantan Nov 13 '16 at 04:16
  • @Sayantan Wait, you're assuming that such an $f$ cannot exists? – T.J. Gaffney Nov 13 '16 at 04:20
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    @Gaffney You are indeed right, I didn't read the wikipedia page, I learned this theorem from another source. The statement is if you have $U\subseteq \mathbb{R}^n$ and and injection $f:U\to \mathbb{R}^m$ for $m\le n$, then $f$ is a homeomorphism. Sayantan is right though that this is clearly implied by the quoted theorem, just take the map $g:U\to \mathbb{R}^m\times \mathbb{R}^{n-m}$ by $g(x)=(f(x), 0)$. Then your map is still an injection, so the theorem on wikipedia applies. This actually immediately implies the second part, so this actually makes the argument simpler, which I have done. – Pax Nov 13 '16 at 05:56
  • Makes sense; thanks. – T.J. Gaffney Nov 13 '16 at 06:03