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