Existence of continuous map from $\mathbb{R}^n$ to $\mathbb{R}^{n-1}$ that "respects norms"

Question

Is there a continuous map from $\pi: \mathbb{R}^n \rightarrow \mathbb{R}^{n-1}$ that is $(\sim_n, \sim_{n-1})$-invariant where $\sim_n$ is the equivalence relation of equality up to orthogonal transformation, i.e. for each $a,b \in \mathbb{R}^n$, $\|a\| = \|b\|$ implies that $\|\pi(a)\| = \|\pi(b)\|$?

I don't have a lot of background in topology or geometry so maybe the answer can be found by applying some basic result.

I am working on a dimension reduction idea and was considering the map $\pi: \mathbb{R}^n \setminus A_n \rightarrow \mathbb{R}^2$, defined everywhere outside the $n$-th coordinate axis $A_n$, given by $$\pi(a) = \|a\|\pi_{-1}(a) / \|\pi_{-1}(a)\|$$ where $\pi_{-1}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n-1}$ is just the projection onto the first $n-1$ coordinates $\pi_{-1}(a_1, \ldots, a_n) = (a_1, \ldots, a_{n-1})$.

I don't think it can be continuously extended either since for example, when $n = 2$, the map takes spheres of radius $r$ (except north/south poles) to their equator but cannot be continuously extended.

I might still be able to make this work for my purposes, but just wanted to check if there were any continuous everywhere defined maps or if it is not possible.

Answer 1

As mentioned in the comments, there are a bunch of possible maps. For example all constant maps have the property, but those are typically not very useful.

Here are some more interesting examples, all with the slightly stronger property $\Vert \pi(x) \Vert = \Vert x \Vert$.

First the map $\mathbb{R}^n \rightarrow \mathbb{R}^{n-1}, x \mapsto \Vert x \Vert e_1.$

Secondly, a map that is similar in spirit, but preserves a bit more structure $\mathbb{R}^n\rightarrow \mathbb{R}^{n-1}, (x_1, \dots, x_n) \mapsto (x_1, \dots, x_{n-2}, \Vert (x_{n-1}, x_n)\Vert)$.

Answer 2

If $f : S^{n-1} \to S^{n - 2}$ is any continuous map, then you can take $\pi(x) = f(x/||x||) ||x||$.

For $n > 2$ there are many such maps. For example, the composition of $S^{n - 1} \to D^{n-1}$ projecting to the first $(n-1)$ coordinates, $D^{n-1} \to D^{n-2}$ projecting to the first $(n-2)$ coordinates and $D^{n-2} \to S^{n-2}$ collapsing the boundary to a point produces a surjective map $S^{n-1} \to S^{n-2}$. For $n = 3$, let $r = ||(x, y, z)||$ and $t = x/r$ for $r \ne 0$. Then the above construction provides the explicit map $(x, y, z) \mapsto (r \cos(\pi t), r \sin(\pi t))$ (and $0 \mapsto 0$) where $\pi$ is the circumference of the unit circle, not the map.

Note that these examples are all norm preserving, ie $||\pi(x)|| = ||x||$. Restricting to such examples, $g(x) = \pi(x)/||\pi(x)|| = \pi(x)/||x||$ is a continuous map $\mathbb{R}^n - 0 \to S^{n-2}$, so for $n = 2$ has only $2$ choices (since $\mathbb{R}^n - 0$ is connected and $S^0 = \{\pm 1\}$ is $2$ discrete points). Since $\pi(x) = ||x|| g(x)$, it follows that for $n = 2$, $\pi$ has only two choices $x \mapsto ||x||$ and $x \mapsto -||x||$.

In fact, norm preserving maps $\mathbb{R}^n \to \mathbb{R}^{n-1}$ exactly correspond to continuous maps $\mathbb{R}_{> 0} \times S^{n-1} \to S^{n-2}$ by the correspondence $\pi \leftrightarrow g$, where we have switched to polar coordinates, ie identified $\mathbb{R}^n - 0$ with $\mathbb{R}_{> 0} \times S^{n-1}$.

This also shows that there cannot be a $\pi$ that is a retract, ie one that fixes $\mathbb{R}^{n-1}$ pointwise. Since for such a $\pi$, which would automatically be norm preserving, $g$ restricted to a hemisphere of $1 \times S^{n-1}$ would give a retract $D^{n-1} \to S^{n-2}$ of a disk to its boundary, which is well-known to not exist.