Euclidean Spaces: Embedding

Question

Given the real line $\mathbb{R}$ and plane $\mathbb{R}^2$.

Are there maps: $$\eta\in\mathcal{C}(\mathbb{R}^2,\mathbb{R}),\vartheta\in\mathcal{C}(\mathbb{R},\mathbb{R}^2):\quad \vartheta\circ\eta=1_{\mathbb{R}^2}$$

What about measurables?

Answer 1

A continuous map from $\Bbb R^2$ to $\Bbb R$ cannot be injective. This is trivial; proof below. It's higher-dimensional versions, like the fact that a continuous map from $\Bbb R^3$ to $\Bbb R^2$ cannot be injective, that I suspect are non-trivial.

Given two points $a,b\in\Bbb R^2$, let $[a,b]$ denote the line segment from $a$ to $b$. Say $f:\Bbb R^2\to\Bbb R$ is continuous. Suppose $f(0)=0$. Let $p=(1,0)$, $q=(0,1)$, $r=(1,1)$. Consider the three numbers $f(p)$, $f(q)$, and $f(r)$. If one of them is zero then $f$ is not injective. Otherwise, either (i) at least two of them are strictly positive or (ii) at least two of them are strictly negative.

Assume wlog that $f(p)>0$ and $f(q)>0$. If $f(p)=f(q)$ then $f$ is not injective. If not, wlog $0<f(p)<f(q)$. Now the intermediate value theorem shows that there exists $x\in[0,q]$ with $f(x)=f(p)$. Hence $f$ is not injective.

EDIT: So now the OP asks about measurable functions. As measure spaces the euclidean spaces are all the same. There are ways to see this that are perhaps simpler than the following, but the following is more fun:

Let $p:[0,1]\to[0,1]^2$ be that Peano curve. In particular it's a continuous surjection. Of course there are various constructions of such a $p$; the one I have in mind satisfies $$p([0,1/4])=[0,1/2]^2.$$And in fact for any $j$ and $N$ there exist $m$ and $n$ such that $$p\left(\left[j4^{-N},(j+1)4^{-N}\right]\right)=[m2^{-N},(m+1)2^{-N}]\times [n2^{-N},(n+1)2^{-N}].$$This says that $p$ is measure-preserving.

Of course $p$ is not injective; it can't be because the interval and the square are not homeomorphic. But $p$ is injective off a certain null set; modifying $p$ in irrelevant ways on that null set gives a measure-preserving bijection between the unit interval and the unit square.

Better yet, modify $p$ on a null set to give a measure-preserving bijection from $[0,1)$ onto $[0,1)^2$. These half-open things tile the line and the plane nicely, giving a measure-preserving bijection from $\Bbb R$ onto $\Bbb R^2$.

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One could construct maps from $\Bbb R$ to $\Bbb R^n$ similarly, and then compose to get maps from $\Bbb R^n$ to $\Bbb R^m$. Or note this: If $f:\Bbb R^2\to\Bbb R$ then $x\mapsto f(f(x_1,x_2),x_3)$ mape $\Bbb R^3$ to $\Bbb R$...

Answer 2

Suppose there were such continuous maps. Then $\eta$ would be injective. Hence it's restriction to the closed unit disc is a homeomorphism. The image of that disc would be a closed bounded interval. But the disc is not homeomorphic to an interval: removing an interior point from the interval leaves a disconnected set, while removing any point from the disc leaves a connected set.

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Added later: Let me take a crack at the measurable case too. ("Measurable" will mean "Borel measurable".) Recall that each $x\in (0,1)$ has a unique decimal expansion $x = .x_1 x_2\dots $ that doesn't end in a string of $9$'s; all decimal expansions mentioned will be of this sort. We want to define a map $f: (0,1)\times (0,1) \to (0,1)$ as follows:

$$f((.x_1 x_2\dots , .y_1 y_2\dots)) = .x_1 y_1 x_2 y_2 \dots.$$

Then $f$ is injective, which is nice, but there are a few problems with $f$: i) Why is $f$ measurable? ii) $f$ isn't onto (for example, $.0909\overline {09}$ is not in the range). To settle i) consider the function $d_n:(0,1) \to \mathbb R$ defined by $d_n(.x_1 x_2\dots) = x_n.$ Then $d_n$ is measurable on $(0,1).$ (I'm going to leave this to you for now; it's a kind of fun little exercise). Since $f(x,y) = \sum d_n(x)/10^{2n-1} +\sum d_n(y)/10^{2n},$ $f$ is measurable.

To take care of ii), let $E_n = \{x\in (0,1): d_k(x) = 9, k = n,n+2,n+4, \dots \}.$ Then each $E_n$ is measurable, hence so is $E= \cup E_n.$ You can check that $f$ is a bijection between $(0,1)\times (0,1)$ and $(0,1)\setminus E.$ Furthermore, $f^{-1}$ is measurable, since

$$f^{-1}(x)=(\sum d_{2n-1}(x)/10^{n},\sum d_{2n}(x)/10^{n}).$$

So now let $g$ be a diffeomorphism of $\mathbb {R}^2$ onto $(0,1)\times (0,1).$ Also define $h: \mathbb {R}\to [(0,1)\setminus E]$ in the obvious way: $h(x) =x,x\in (0,1)\setminus E, h=0$ elsewhere; clearly $h$ is measurable. Then we can take $\eta = f\circ g,$ $\vartheta = g^{-1}\circ f^{-1}\circ h.$ Recalling that compositions of Borel measurable functions are Borel measurable, we see $\eta, \vartheta$ do the job.