Describing maps $\mathbb{R}^n \to \mathbb{R}^m$ up to homotopy I'm trying to describe continuous maps $f : \mathbb{R}^n \to \mathbb{R}^m$, where $f^{-1}(0) = \{0\}$, up to homotopy. 

Without the condition that $f^{-1}(0) = \{0\}$, it seems that all maps are homotopic to each other; since all the homotopy groups of $\mathbb{R}^n$ and $\mathbb{R}^m$ are trivial, and this is evidence that maps between the two spaces are trivial up to homotopy too (though I can't seem to come up with a proof of this fact?)

With the condition that $f^{-1}(0) = \{0\}$, I'm not sure how I would go about approaching this problem. The main thing that seems to cloud my vision is that I can only really imagine the cases where $n,m \leq 3$; and even then I'm not sure that my intuition holds.
This looks like it may be equivalent to computing $\pi_{n-1}(S^{m-1})$.
Each $f$ restricts to a map from $\Bbb  R^n-\{0\}$ to $\Bbb  R^m-\{0\}$ to
and these spaces are homotopy equivalent to $S^{n-1}$ and $S^{m-1}$.
Conversely each $g:S^{n-1}\to S^{m-1}$ extends radially to an $f$ of your form.

I haven't checked the details on whether these correspondences are preserved by homotopy etc.
Hint: For $f, g: R^n \to R^k$, consider, for each $P \in R^n$, the line segment from $f(P)$ to $g(P)$ as a parameterized arc (with parameter $s$, say, running from $0$ to $1$). Suppose that $f(0) = g(0) = 0$. What will this line segment look like?