# Is $\mathbb{R}^n$ properly homotopy equivalent to $\mathbb{R}^m$ if $n \neq m$?

$\DeclareMathOperator{\id}{id} \newcommand{\R}{\mathbb{R}}$ If $f,g : X \to Y$ are two maps (all maps considered are continuous here), a homotopy between $f$ and $g$ is a map $H : [0,1] \times X \to Y$ such that $H(0,-) = f$ and $H(1,-) = g$. Two spaces $X$ and $Y$ are said to be homotopy equivalent if there exist $f : X \to Y$ and $g : Y \to X$ such that $g \circ f \sim \id_X$ and $f \circ g \sim \id_Y$. This creates an equivalence relation on spaces.

Let $n, m \ge 0$ be two integers. It is very easy to see that $\R^n$ and $\R^m$ are homotopy equivalent, as they are in fact both contractible, and one can write an explicit deformation retraction of $\R^k$ onto the origin: $H(x,t) = tx$ (then $H(0,-)$ is constant and $H(1,-) = \id_{\R^k}$).

But this homotopy is not proper, though: the preimage $H^{-1}(0) = \R^k \times \{0\} \cup \{0\} \times [0,1]$ is not compact. One can easily adapt all the definitions of the first paragraph by defining a notion of proper homotopy between two proper and proper homotopy equivalence between two spaces by requiring all the maps involved to be proper.

Is $\R^n$ properly homotopy equivalent to $\R^m$ for $n \neq m$?

This question is motivated by this other one: $\R \times [0,1]$ is properly homotopy equivalent to $\R$ (the closed interval properly deformation retracts onto a point), so if one can show that $\R$ and $\R^2$ are not properly homotopy equivalent, then $\R^2$ and $\R \times [0,1]$ cannot possibly be homeomorphic.

• Any proper map $f\colon X \to Y$ induces a continuous map between the end spaces of $X$ and $Y$, $\epsilon(f) \colon\epsilon(X) \to \epsilon(Y)$ and so this is a functor $\epsilon\colon \mathbf{TOP}\to\mathbf{STONE}$. If $f$ is a homotopy equivalence as well then $\epsilon(f)$ will be an isomorphism. This at least rules out a proper homotopy equivalence between $\mathbb{R}^n$ and $\mathbb{R}$ for $n\geq 2$ as $\mathbb{R}$ has two ends and $\mathbb{R}^n$ has only one. – Dan Rust Jun 8 '15 at 14:43
• Sorry I forgot to compose with $\pi_0$. $\epsilon$ is a functor from $\mathbf{TOP}$ to the category $\mathbf{PRO}-\mathbf{TOP}$ of pro-objects in the topological category. – Dan Rust Jun 8 '15 at 15:00
• Thanks @Daniel! It at least solves the question asked as motivation, though (but it wouldn't work to prove that $\R^{n+1}$ and $\R^n \times [0,1]$ aren't homeomorphic for $n>1$). – Najib Idrissi Jun 8 '15 at 15:02
• It looks like some of the tools mentioned on this page provide a robust set of invariants to use, so maybe a relative form of the "Brown–Grossman homotopy groups" would be the right invariant to look at? – Dan Rust Jun 8 '15 at 15:05

The invariants which distinguish the different $\mathbb{R}^n$'s up to proper homotopy are the "homotopy groups at infinity". For example, let's distinguish $\mathbb{R}^2$ from $\mathbb{R}^n$ with $n \ge 3$ by using "simple connectivity at infinity".

For every sequence of nonempty compact sets $K_1 \subset K_2 \subset \cdots \subset \mathbb{R}^2$ whose union equals $\mathbb{R}^2$, there is exactly one component $U_i$ of $\mathbb{R}^2 - K_i$ whose closure is noncompact, we have an inclusion $U_1 \supset U_2 \supset \cdots$, and we have a sequence of injections of fundamental groups of the form $$\pi_1(U_1) \leftarrow \pi_1(U_2) \leftarrow \cdots$$ The inverse limit of this sequence of groups is well-defined up to isoomorphism, and it is isomorphic to $\mathbb{Z}$.

But if you do the same thing with $\mathbb{R}^n$ where $n \ge 2$, the inverse limit of the sequence of groups will be trivial.

Of course one still has to prove that this inverse limit is a proper homotopy invariant, and one has to prove this for higher homotopy groups, and one has to do the appropriate calculations for the $\mathbb{R}^n$. The upshot is that the first nontrivial homotopy group at infinity for $\mathbb{R}^n$ is the $n-1^{\text{st}}$.

Here is a paper of Davis and Meier where you'll find some details about homotopy groups at infinity.

• Could you comment on why only the $n-1$st homotopy group at infinity for $\mathbb{R}^n$ is non-trivial? My intuition is that we can make the $K_i$ just be increasing closed balls and then the $U_i$ all deformation retract onto an $n-1$ sphere, so the inverse limit of homotopy groups for this sequence of $K_i$ would be the same as those of $S^{n-1}$. If, as you say, the limit is independent of choice of $K_i$, then shouldn't this be a valid calculation? Apologies if the answer is buried in the linked paper, I'm feeling lazy. – Dan Rust Jun 8 '15 at 23:26
• @DanielRust: Oops, I miswrote that, I'll correct it. – Lee Mosher Jun 9 '15 at 0:12
• Awesome, order is restored. Great answer by the way. – Dan Rust Jun 9 '15 at 0:13
• I don't know if I like the answer, since it suggests that we need a new theory for solving the problem. But everybody who carefully studied Poincare duality already has a tool to distinguish $\mathbb R^n$'s up to proper homotopy equivalence. That is why I also don't really like the first sentence "The invariants which..." – Daniel Valenzuela Jun 9 '15 at 6:15

In algebraic topology we want to find invariants to distinguish spaces up to certain equivalences. For proper homotopy equivalence one of such is compactly supported cohomology. By Poincaré duality for $0\neq n\neq m$:

$$H_0(\mathbb R^n) \cong H^n_c(\mathbb R^n) \not \cong H^n_c(\mathbb R^m) =0$$

• Note that it is actually also easy to compute the relevant groups by definition. – Daniel Valenzuela Jun 9 '15 at 2:44
• I know proper maps induce pullbacks on $H^*_c$, but is it clear that cohomology with compact support is preserved by proper homotopy equivalence? I checked a few references, none of them mention that... – Najib Idrissi Jun 9 '15 at 7:22
• Okay, I just found a reference that explicitly proved that. Thanks for your answer! – Najib Idrissi Jun 9 '15 at 12:07
• I actually think it is even mentioned in Hatcher's Algebraic topology. Maybe under the chapter for cohomology with local coefficients. – Daniel Valenzuela Jun 10 '15 at 0:37