# 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

## 1 Answer

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.

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Sorry, I must be having a moment here, but why must the question be able to be localised? Even if locally the graphs are not open, what's to prevent the union over all neighborhoods of the graphs to be an open subset? I feel like there is a really elementary fact that I am overlooking. – Willie Wong♦ Aug 23 '11 at 20:33
oh... wait, I guess Baire category? – Willie Wong♦ Aug 23 '11 at 20:35
I think a little more careful local argument will tell you that not only is the interior of this set empty, but these sets have the form that countable unions of them also have empty interiors. – Ryan Budney Aug 23 '11 at 21:27
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
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