# Is n-euclidean space a quotient of m-euclidean space?

Let $$\mathbb{R}^n$$ have the usual Euclidean topology. Let $$m, n \in \mathbb{N}$$, and $$m < n$$.

Does there exist a quotient map $$f : \mathbb{R}^m \to \mathbb{R}^n$$?

### Notes

• I'm expecting a negative answer. The question is how to prove it.
• The case $$m = 0$$ does not hold because $$f$$ is not surjective.
• Projection provides a quotient map $$g : \mathbb{R}^n \to \mathbb{R}^m$$.
• By the Hahn–Mazurkiewicz theorem, there exists a quotient map $$f : [0, 1]^m \to [0, 1]^n$$ for any $$n, m \in \mathbb{N}^{> 0}$$.
• This question and answer seems to show that a metric, connected, locally connected, locally compact space (i.e. a generalized Peano continuum) is a continuous image of $$\mathbb{R}$$, although I haven't checked it. However, I'm not sure whether it helps to find a quotient even if it is true.
• Hahn-Mazurkiewicz says that we get a continuous surjective map, but I don't think it's obviously a quotient map. Commented Jul 18 at 4:04
• FWIW, you can use Hahn-Mazurkiewicz to get a surjective continuous map $\mathbb{R} \to \mathbb{R}^n$ for all $n$. But as Noah mentioned, there's not much reason why it should be a quotient map. Commented Jul 18 at 4:18
• If $n>m\ge 3$, there is an open continuous map $R^m\to R^n$, hence, a quotient map. See my answer here: math.stackexchange.com/questions/3130389/… Commented Jul 18 at 4:18
• Now that's surprising!
– kaba
Commented Jul 18 at 4:22
• @MoisheKohan For this question I'm already more than content with the result for $m, n \geq 3$, and would accept it as an answer. I came for counterexamples, but left with proofs! :)
– kaba
Commented Jul 18 at 4:43

Does there exist a quotient map $$f : \mathbb{R}^m \to \mathbb{R}^n$$?

Yes. Even a closed map. Even when $$m>n$$ (note that while projections are quotient maps, they are not closed).

Consider $$k\in\mathbb{N}$$ and let $$T_{n,k}=\{v\in\mathbb{R}^n\ |\ k\leq \lVert v\rVert\leq k+1\}$$ So it is an annulus except for $$k=0$$ which gives us a disc. Moreover $$\mathbb{R}^n=\bigcup_{k=0}^\infty T_{n,k}$$. Furthermore let

$$L_{n,k}=\{v\in\mathbb{R}^n\ |\ k=\lVert v\rVert\}$$ $$U_{n,k}=\{v\in\mathbb{R}^n\ |\ \lVert v\rVert= k+1\}$$

which together form the boundary of $$T_{n,k}$$ when $$k>0$$. Of course $$T_{n,k}\cap T_{n,k+1}=U_{n,k}=L_{n,k+1}$$.

The general idea is that we construct surjective continuous maps $$T_{m,k}\to T_{n,k}$$ in such a way that they coincide on boundaries, and thus are gluable. This will automatically guarantee that the map is closed as we will see.

So first we construct a surjective continuous map $$f_0:T_{m,0}\to T_{n,0}$$. We do that by mapping entire boundary $$U_{m,0}$$ to a point, say $$(1,0,\ldots,0)$$. We then map some small closed line inside $$int(T_{m,0})$$ onto $$T_{n,0}$$ (we can do this through a space filling curve, i.e. Hahn-Mazurkiewicz) and then extend this to entire $$T_{m,0}$$ through Tietze for example.

Similarly we define each $$f_k:T_{m,k}\to T_{n,k}$$ map for $$k>0$$. We do this recursively: we map $$L_{m,k}$$ to $$f_{k-1}(U_{m, k-1})$$ (which by recursion is always a point in $$U_{n,k-1}$$) and we map $$U_{m,k}$$ to some point in $$U_{n,k}$$. Well, we don't need recursion, we can predefine all those boundary points, e.g. $$(k,0,0,\ldots,0)$$. We then take a small closed line inside $$int(T_{m,k})$$ and map it onto $$T_{n,k}$$ (Hahn-Mazurkiewicz) and we extend all those pieces to full $$T_{m,k}$$. We can do this in various ways, one way is to realize that $$T_{n,k}\simeq S^{n-1}\times [0,1]$$ and so then we can find an extension on "$$[0,1]$$" coordinate by Tietze. What about "$$S^{n-1}$$" coordinate? If $$m (the general case is considered at the EDIT section) then the (large) inductive dimension of $$T_{m,k}$$ is equal to $$m$$, which is at most $$n-1$$. This implies that the map is extendable (R. Engelking, "Dimension Theory", Theorem 3.2.10).

The way we constructed those $$f_k$$ means that they are compatible on intersections $$T_{m,k}\cap T_{m,k+1}$$ and so we can glue them all together into a single continuous surjective map $$f:\mathbb{R}^m\to\mathbb{R}^n$$. This map has one additional very important property: $$f^{-1}(T_{n,k})=T_{m,k}$$. Which I will call the $$(*)$$ property. And which implies that it is closed.

Indeed, consider $$F\subseteq \mathbb{R}^m$$ closed. Now assume $$(x_k)$$ is a sequence in $$f(F)$$ convergent to some element of $$\mathbb{R}^n$$. Of course $$(x_k)$$ must eventually belong to $$T_{n,s}\cup T_{n,s+1}$$ for some $$s$$ (most of the time it belongs to $$T_{n,s}$$, but we may require additional $$T_{n,s+1}$$ if it converges to the shared boundary and jumps between sides). By the $$(*)$$ property $$(x_k)$$ arises as $$f(y_k)$$ for some $$(y_k)\subseteq F\cap (T_{m,s}\cup T_{m,s+1})$$. Of course $$(y_k)$$ has a subsequence convergent in $$F\cap (T_{m,s}\cup T_{m,s+1})$$ (because it is compact) and therefore $$\lim x_k\in f(F)$$ by the uniqueness of limits.

EDIT: We can do slightly better and get rid of the "$$m" assumption. Yes, I know that there's an obvious quotient map $$\mathbb{R}^m\to\mathbb{R}^n$$ when $$m>n$$, namely one of the projections, but these are not closed unfortunately. I'm not exactly sure whether there is some obvious closed $$\mathbb{R}^m\to\mathbb{R}^n$$ map when $$m>n$$, I don't see it at the moment. But the above construction does work for this case as well, but needs a different argument.

When $$m>n$$ then the "inductive dimension" argument fails. So we need another way to extend our partial $$f_k$$ to whole $$T_{m,k}$$. Luckily we do have the Borsuk homotopy extension theorem (which can be found in R. Engelking, "Dimension Theory", lemma 1.9.7.):

Assume that $$X$$ is a metric space, $$A\subseteq X$$ closed and $$f,g:A\to S^n$$ are homotopic. Then $$f$$ is continuously extendable to whole $$X$$ if and only if $$g$$ is.

(the original lemma is slightly more general and has stronger conclusion, I've simplified it for our needs)

In particular a nullhomotopic map is always extendable. And this applies to our partial $$f_k$$, because it is nullhomotopic. Because its domain is a topological disjoint union of three spaces, and it is nullhomotopic on each of them. On boundaries it already is constant. And the curve has contractible domain so it is nullhomotopic.

• Yes, the proof generalizes to get you a proper surjective map $R^n\to R^m$: Use the fact that $\pi_{n-1}(T_k)=0$ for all $n<m$. Commented Jul 18 at 8:43
• @MoisheKohan I've completely redesigned my answer and made it general from the beginning. I'm still not sure how the homotopy group can be applied here. But it gave me an idea to apply inductive dimension. Commented Jul 18 at 12:31
• That's pretty clever! A reference for future readers on the sphere-extension thing. Engelking's Theory of Dimensions, page 188, Theorem 3.2.10: "A normal space satisfies the inequality $\dim X \leq n \geq 0$ if and only if for every closed subspace $A$ of the space $X$ and each continuous mapping $f: A \to S^n$ there exists a continuous extension $F : X \to S^n$ of $f$ over $X$." Further large (and small) inductive dimension equals covering dimension on separable metrizable spaces.
– kaba
Commented Jul 18 at 13:42
• The argument I had in mind is based on the following observation: Suppose that $f: T_{m,k}\to T_{n,k}$ is a (continuous) map, $m< n$. Then the restriction of $f$ to the boundary is homotopic (as a map to $T_{n,k}$) to a map $\partial T_{n,k}\to \partial T_{n,k}$, with the prescribed values on the boundary spheres. Combining $f$ with this homotopy, if $f$ was surjective, you can replace $f$ with a surjective map $f': T_{m,k}\to T_{n,k}$ that has prescribed boundary values. Commented Jul 18 at 16:15
• @kaba I went through the Engelking's book and I found a theorem that helps us even when $m>n$. Which wasn't obvious for me. Thank you for the inspiration, I've updated the answer. Commented Jul 18 at 18:47

Assume that we have a quotient map from $$\mathbb{R}$$ to $$\mathbb{R}^2$$. Then its square is a quotient map from $$\mathbb{R}^2$$ to $$\mathbb{R}^4$$ ( product of quotient maps OK for locally compact), so inductively, a quotient map $$\mathbb{R}^{2^n} \to \mathbb{R}^{2^{n+1}}$$. Now compose the maps and get a quotient map from $$\mathbb{R}$$ to $$\mathbb{R}^{2^n}$$. Now compose with a projection $$\mathbb{R}^{2^n}$$ to $$\mathbb{R}^{N}$$ ($$N\le 2^n$$) and get a quotient map from $$\mathbb{R}$$ to $$\mathbb{R}^N$$. One can multiply this by an identity map of $$\mathbb{R^m}$$.

Now, with the little we know, can we get a quotient map from $$\mathbb{R}$$ to $$\mathbb{R}^2$$? We know of that famous Peano map from $$[0,1]$$ to $$[0,1]^2$$ ( we can even prescribe where the ends go). Continuous, so closed. Consider a spiral of lattice squares covering the plane ( one could do two spirals, but why complicate). Put together Peano maps from $$[n,n+1]$$ to $$Q_n$$ and get a surjective proper ( hence closed) map from $$[0, \infty)$$ to $$\mathbb{R}^2$$. Now compose with $$\mathbb{R}\to [0, \infty)$$, $$t\to t^2$$.

$$\bf{Added:}$$ If one could wrap one's head around spirals, we could get directly a continuous proper map $$\mathbb{R}\to \mathbb{R}^N$$, and then product it with an identity for a proper map.