Is a continuous map between smoothable manifolds always smoothable? 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.
 A: (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.
