6

Let $X$ and $Y$ be topological manifolds and $f:X\to Y$ a continuous map.

Suppose $X$ and $Y$ admit a differentiable structure (at least one).

My question: is it always possible to choose a differentiable structure on $X$ and one on $Y$ in such a way that $f$ turns out to be differentiable?

I know the answer is yes in some cases, e.g. when $f:X\to Y$ is a covering projection. I suspect this is not true in general, but can't find a counterexample.

Thanks in advance.

Lor
  • 5,738
  • Can't you just take a continuous but not differentiable function $f:\mathbb{R} \to \mathbb{R}$ and use the uniqueness of smooth structures on \mathbb{R}? Am I missing something here? – Anthony Carapetis Jun 06 '13 at 08:01
  • 3
    Yes, you're missing that one can put different smooth structures on each instance of $\mathbb{R}$. –  Jun 06 '13 at 08:02

1 Answers1

5

(Note that this answer is more general than can be obtained from any version of Sard's theorem that I know of, since it also rules out "differentiable with discontinuous derivative".)

$\mathbb{R}$ and $\mathbb{R}^2$ are obviously smoothable.
Let $\hspace{.01 in}\: f : \mathbb{R} \to \mathbb{R}^2 \:$ be a continuous extension of a space-filling curve.
Let $\: g : \mathbb{R} \to \mathbb{R} \:$ and $\: h : \mathbb{R}^2 \to \mathbb{R}^2 \:$ be charts for some smooth structure on $\operatorname{Dom}(\:f\hspace{.01 in})$
and $\operatorname{Codom}(\:f\hspace{.01 in})$, respectively. $\;\;$ Let $\: \pi : \mathbb{R}^2 \to \mathbb{R} \:$ be projection to the $x$-coordinate.
$h\circ f\circ g^{-1} \circ \pi \:$ is a function from $\mathbb{R}^2$ to itself, and forward image of the $x$-axis under that function is the unit square. $\:$ Since that $x$-axis has Lebesgue measure zero in $\mathbb{R}^2$ and the unit square does not,
it follows from the answer to my question here that $\: h\circ f\circ g^{-1} \circ \pi \:$ is not differentiable.
Since projection is obviously differentiable, this means $\: h\circ f\circ g^{-1} \:$ is not differentiable,
so $\hspace{.02 in}f$ is not differentiable with respect to the smooth structures on $\operatorname{Dom}(\:f\hspace{.01 in})$ and $\operatorname{Codom}(\:f\hspace{.01 in})$.
Therefore, since that argument works for any charts $g$ and $h$, $\hspace{.05 in}f$ cannot be made differentiable.

Community
  • 1