Let $n>m$. Can there exist a map $\phi:\{0,1\}^n\to\{0,1\}^m$ that approximately preserves Hamming distance? I'm defining Hamming distance slightly nonstandardly by dividing by the dimension, so that the maximum distance between two sequences is 1, whatever the dimension. By "approximately preserves" I mean that for every $x,y\in\{0,1\}^n$, $d(\phi(x),\phi(y))$ should be within $\delta$ of $d(x,y)$, where $\delta$ is some small constant like $1/100$.
It seems to me that the answer ought to be no, for the following reason. The fact that distances are approximately preserved implies a kind of continuity of $\phi$ (because in particular small distances go to small distances). But then the Borsuk-Ulam theorem would suggest that there will probably be two antipodal points in $\{0,1\}^n$ that map very close together in $\{0,1\}^m$, contradicting the approximate distance-preserving property.
My question is, is that thought the basis for a well-known argument? Or does the assertion follow easily from a known result?
Edit. As asked the question is not a great one, since, as Felipe points out below, if $m=n-1$ then one can just ignore the last coordinate. There are also arguments based on the pigeonhole principle if $n$ is a fair amount bigger than $m$: then there must be large sets of points in $\{0,1\}^n$ that have the same image, or, more generally, images in the same not too large Hamming ball. Amongst such sets there must be (if the sets are reasonably large) almost antipodal points (see Eoin's comment).
I'm actually interested in the case $m=n/2$. It's not clear to me (but I need to do some proper calculations) that that can be proved by pigeonhole type arguments. However, what is clear is that a topological proof of the kind I speculated about is unlikely to exist.