I am having a little trouble with this question.

Prove that there does not exist a bijective map from $\mathbb{R}^2 \to \mathbb{R}^3$ where $f$ and $f^{-1}$ are both differentiable.

Thanks for any help.

  • $\begingroup$ I guess this is a variation of the proof that no continuous space filling curve can be everywhere differentiable. I believe the usual proof uses Sard's Theorem but don't quote me on that. Another method would be to use the fact that the Hausdorff dimension of the smooth image of $n$-space is at most $n$-dimensional. of course this proof works for arbitrary surjective maps, so there may be a more elementary proof for the particular case of bijective maps. $\endgroup$ – Dan Rust Jul 12 '13 at 15:01
  • $\begingroup$ Is an elementary proof possible if we add that f^(-1) is also differentiable? $\endgroup$ – Ester Jul 12 '13 at 16:33
  • $\begingroup$ In this case you have a diffeomorphism between $\mathbb{R}^2$ and $\mathbb{R}^3$ which contradicts invariance of domain en.wikipedia.org/wiki/Invariance_of_domain#Consequences. In fact, this holds for continuous $f$ and $f^{-1}$, not just differentiable/smooth. $\endgroup$ – Dan Rust Jul 12 '13 at 16:35

I'd be pretty surprised if this question wasn't already answered somewhere on this site... but here's a sketch.

Suppose $f : \mathbb{R}^2 \to \mathbb{R}^3$ is bijective with both $f$ and $f^{-1}$ differentiable. In particular, $f$ and $f^{-1}$ are continuous i.e. $f$ is a homeomorphism. So the question is: "why is $\mathbb{R}^2$ not homeomorphic to $\mathbb{R}^3$?" The simplest approach is probably to note that $\mathbb{R}^2$ minus a point is not simply connected, but $\mathbb{R}^3$ minus a point is simply connected. Since the property

there exists a point $x \in X$ such that $X \setminus \{x\}$ is not-simply connected

is invariant under homeomorphism, we are done.


Be $f:\mathbb R^2\to\mathbb R^3$ such a function.

Now define on $\mathbb R^3$ the coordinate functions $$\begin{align} \omega_1\colon &(x,y,z)\mapsto x\\ \omega_2\colon &(x,y,z)\mapsto y\\ \omega_3\colon &(x,y,z)\mapsto z \end{align}$$ Those are clearly differentiable, and $\partial_i\omega_j=\delta_{ij}$ everywhere.

Now define the functions $$\alpha_i = \omega_i\circ f$$ By the chain rule, the $\alpha_i$ are differentiable functions from $\mathbb R^2$ to $\mathbb R$. I'm going to write the points of $\mathbb R^2$ as $(u,v)$.

Now consider the gradients at a point $o$, $g_i=\operatorname{grad_{\mathbb R^2}}\alpha_i(o) = (\partial_u\alpha_i(o),\partial_v\alpha_i(o))$. Since those are two-dimensional vectors, they must be linearly dependent, that is, there exist numbers $a, b, c$ of which at least one is not $0$, so that $a g_1+b g_2+c g_3=0$.

Now by the assumptions, $\omega_i=\alpha_i\circ f^{-1}$. Be $p=f(o)$ (and thus $o=f^{-1}(p)$). Then we can calculate $\partial_i\omega_j(p)$ using the chain rule (this is allowed because $f^{-1}$ is differentiable by assumption): $$\partial_i\omega_j(p) = \partial_u\alpha_j(f^{-1}(p))(\partial_i(f^{-1}))_u(p) + \partial_v\alpha_j(f^{-1}(p))(\partial_i(f^{-1}))_v(p) = g_j\cdot(\partial_i(f^{-1}))(p)$$ Especially, we get $a\partial_i\omega_1(p)+b\partial_i\omega_2(p)+b\partial_i\omega_3(p) = (ag_1 + bg_2 + c g_3)\cdot(\partial_i(f^{-1})(p)) = 0$ for all $i$.

But we find from direct calculation that $$a\partial_i\omega_1(p)+ b\partial_i\omega_2(p)+c\partial_i\omega_3(p) = a\delta_{i1}+b\delta_{i2}+c\delta_{i3} = \begin{cases}a & i=1\\b & i=2\\c & i=3\end{cases}$$ and at least one of $a$, $b$, $c$ is $\ne0$.

Therefore we get a contradiction, and thus no such $f$ can exist.


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