# Surjection that increases dimensions

This question is somewhat inspired by a question on MathOverflow, but it is not necessary to read that question to understand what I am about to ask.

It is well known that one can establish a surjection between sets of different Hausdorff dimensions: in the regime of just set theory the cardinality of the unit interval and the unit square are the same, and in fact we get a bijection. If you add a bit of topology, one can in addition request that this surjection be given by a continuous map, but the map cannot be a bijection, else it'd be a homeomorphism.

What if, instead of continuity, we require a different condition?

Question Fix $N$ a positive integer. Let $B$ be the open unit ball in $\mathbb{R}^N$. Can we find an embedded smooth (or $C^1$) hypersurface $A\subset \mathbb{R}^N$ and a surjection $\phi:A\to B$ such that the vector $a - \phi(a)$ is orthogonal to $A$? Can it be made continuous? Can it be made a bijection?

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I can see the real analysis, and I can see the differential geometry (I think!), but I have no idea where the elementary set theory comes into the question :-) – Asaf Karagila Aug 23 '11 at 20:21
@Asaf: I was wondering if there is a way of getting an answer based on cardinality arguments (something like: if $\gamma$ is a curve that intersects a hypersurface $A$ transversely, then $\gamma\cap A$ has countably many points etc.) – Willie Wong Aug 23 '11 at 20:26
Correct me, but isn't there always a bijection between a hypersurface and the open unit ball, just by cardinality games? – Asaf Karagila Aug 23 '11 at 21:47
@Asaf: yes, which is why there is that funny condition with normality. – Willie Wong Aug 23 '11 at 23:03

If $A$ is a hypersurface (co-dimension one and smooth) what you're describing is the graph of a function on $A$ -- well, locally that's what it is. But the problem boils-down to a local problem. You're asking for functions $f : D^{n-1} \to \mathbb R$ whose graph is an open subset of $\mathbb R^n$. This isn't possible, even if $f$ is discontinuous.
I don't understand what fails in the following argument: Take a plane-filling curve $\gamma$, and write a differential equation for a curve in the plane, $\dot\psi=n\times(\psi-\gamma)$, where $n$ is a vector orthogonal to the plane, whose magnitude can vary along the curve. The solution should be continuous, and the freedom in $n$ should allow us to prevent it from diverging or self-intersecting. Wouldn't $\phi=\gamma\circ\psi^{-1}$ then have the desired properties? – joriki Aug 24 '11 at 8:12