$f:\mathbb R^n \rightarrow \mathbb R^m$ If $n \ne m$ then f is not injective or surjective

Question

The title may be confusing but i want to show that if $f:\mathbb R^n \rightarrow \mathbb R^m$ $f \in C^1$ with $m<n$ then f is not injective and If $m>n$ f is not surjective.

If $f$ was a linear transformations things would be simpler, i tought about recouring to rank theorem but i don't know How, any hints?

$\mathbf{R}$ and $\mathbf{R}^2$ are equipotent, hence there exists a bijection between them. Did you want $f$ to be differentiable with differentiable inverse? — William M., Mar 11 '19 at 20:29
Any conditions on $f$, e.g., continuity? If not, then all Euclidean spaces of dimension $\geq 1$ are pairwise equipotent, so there to exist bijections (which are both injective and surjective), just not continuous ones. — avs, Mar 11 '19 at 20:30

Noah Schweber · Answer 1 · 2019-03-11T20:42:45.057

The statement you want is (without further hypotheses on $f$) false - this was observed by Cantor, and is similar to how there is a bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$. See for example the discussion here.

As a quick explanation of why such a thing might be expected, consider the "interleaving" map $i:[0,1)^2\rightarrow [0,1)$ gotten by interleaving the non-trailing-$9$s decimal expansions of the inputs - e.g. $$i(0.101010..., 0.3333...)=0.13031303...$$ This isn't quite what you want, but it is a bijection. Now how different are $[0,1]$ (which is what Cantor originally phrased his theorem about, if I recall correctly), $[0,1)$, and $\mathbb{R}$ really (that is, in terms of cardinality)?

Now, if you demand that $f$ be continuous, you do indeed get what you want (as a corollary of invariance of domain). But there are in fact bijections between $\mathbb{R}^m$ and $\mathbb{R}^n$ for all integers $m,n>1$.

Any references on this theorem? My analysis book omisses this i think — Daniel Moraes, Mar 11 '19 at 20:32
@DanielMoraes It should appear in any elementary set theory book, and I think it's in "little Rudin" too. I don't think it has a name, unfortunately. This paper of Gouvea may be of interest. (Incidentally, assuming the axiom of choice, every infinite set is in bijection with its square!) — Noah Schweber, Mar 11 '19 at 20:37

avs · Accepted Answer · 2019-03-11T20:34:10.577

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If $f$ is required to be ${\cal C}^{1}$, then you assume $f(0) = 0$ (where each zero denotes the zero vector of the appropriate dimension) and use the Implicit Function Theorem, which says that $f$ at $0$ is locally equivalent to its linear part. This will reduce the problem to your preferred context: linear transformations.

A nice statement and method of proof of the Implicit Function Theorem can be found in V. Arnol'd's Ordinary Differential Equations (use the latest edition).

edited Mar 11 '19 at 20:34

answered Mar 11 '19 at 20:33

avs

6,276

Even continuity is enough, per invariance of domain. – Noah Schweber Mar 11 '19 at 20:34
Thanks! Will look for It – Daniel Moraes Mar 11 '19 at 20:34
@NoahSchweber, true, but invariance of domain may or may not be beyond the asker's level. A likely situation is that they haven't taken general topology yet, and were given the extra smoothness allowance on $f$ to enable using the results that require it. – avs Mar 11 '19 at 21:33
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Actually seen general topology, but can only use results from analysis, this aproach on implicit function is what i was looking for – Daniel Moraes Mar 11 '19 at 22:54

$f:\mathbb R^n \rightarrow \mathbb R^m$ If $n \ne m$ then f is not injective or surjective

2 Answers2