Can anyone give an example of an injective map from $\mathbb{R}^2$ to $\mathbb{R}$? Clearly, such a map cannot be continuous (for instance by Borsuk-Ulam Theorem).
Thanks in advance.
Real Analysis – Injective Map from R^2 to R
real-analysis
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Any constant map $f:\mathbb{S}^1\to\mathbb{R}$ is continuous and would satisfy $f(x)=f(y)$ for every $x,y\in\mathbb{S}^1$, of which there are certainly uncountably many.
In fact, a continuous map $f$ that is only constant on a non-empty open set of $\mathbb{S}^1$ would have $f(x)=f(y)$ for uncountably many $x,y\in\mathbb{S}^1$, because any non-empty open set of $\mathbb{S}^1$ is uncountable.
Theorem 1. For every $n> m\ge 3$ there exists a continuous open mapping $f: R^m\to R^n$.
Proof. I will give a proof which is a variation on my answer to this question.
The key result is a rather nontrivial theorem due to John Walsh (he proved something stronger, I am stating a special case):
Theorem 2. Fix $n, m\ge 3$. Then for any pair of compact connected triangulated manifolds (possibly with boundary) $M, N$ of dimensions $m, n$ respectively, every continuous map $g: M\to N$ inducing surjective map of fundamental groups $\pi_1(M)\to \pi_1(N)$ is homotopic to a surjective open continuous map $h: M\to N$.
See corollary 3.7.2 of
J. Walsh, Monotone and open mappings on manifolds. I. Trans. Amer. Math. Soc. 209 (1975), 419-432.
This deep theorem is a generalization of earlier results on existence of open continuous dimension-raising maps from $m$-cubes to compact triangulated manifolds due to Keldysh and Wilson.
The next part of the proof uses some basic algebraic topology covered, say, in Hatcher's "Algebraic Topology".
Take $N=T^n$, the $n$-dimensional torus ($n$-fold product of circles). Its fundamental group is ${\mathbb Z}^n$. Let $S$ be a compact connected oriented surface of genus $n$. Its fundamental group admits a surjective map to ${\mathbb Z}^{2n}$ (given by the abelianization) and, hence, to ${\mathbb Z}^{n}$. Consider the manifold $M$ which is the product $S\times T^{m-2}$. Its fundamental group admits an epimorphism to ${\mathbb Z}^{n}$. The universal covering spaces of the manifolds $M$ and $N$ are homeomorphic to ${\mathbb R}^m$ and ${\mathbb R}^n$ respectively.
Since the manifold $N$ is $K( {\mathbb Z}^n, 1)$, Whitehead's theorem implies that the epimorphism $$ \pi_1(M)\to \pi_1(N) $$ is induced by a continuous map $g: M\to N$. Applying Walsh's theorem, we obtain that $g$ is homotopic to an open map $h: M\to N$. Lifting $h$ to the universal covering spaces we obtain a continuous open map $\tilde{h}: {\mathbb R}^m\to {\mathbb R}^n$. I claim that $\tilde{h}$ is a surjective map. Indeed, the map $h$ is surjective (since otherwise the image $h(M)$ is a proper closed and open subset of $N$ contradicting connectivity of $N$). Since the map $\tilde{h}$ is equivariant with respect to the actions of the fundamental groups of $M, N$ on the respective universal covering spaces, the image $\tilde{h}({\mathbb R}^m)$ is invariant under the covering group $\Gamma$ of the universal covering ${\mathbb R}^n\to T^n$. Therefore, surjectivity of $h$ implies surjectivity of $\tilde{h}$.
Theorem 1 follows. qed
Best Answer
How about this construction: Express $(x,y)\in\mathbb{R}^2$ in decimals ($x=\sum a_k 10^k$, $y=\sum b_k 10^k$) and define the image of $(x,y)$ as the real number which you obtain by interlacing the decimals (i.e. take $c_{2k} = a_k$ and $c_{2k+1} = b_k$).