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(My question is inspired by this math.SE question, whose
negative answer I showed by a dimension-increasing map.)

Is it the case that for all smoothable manifolds $M_0$ and $M_1$ with the same dimension,
for all continuous maps $\: \hspace{.04 in}f : M_0 \to M_1 \:$, $\:$ there exists a smooth
structure on $M_0$ and a smooth structure on $M_1$ that make $\hspace{.04 in}f$ smooth?

If no, what if "smoothable" and "smooth" are replaced with "differentiable" and "differential" respectively?
If yes, what if $\: \operatorname{dim}(M_1) < \operatorname{dim}(M_0) \;$?

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    $\begingroup$ I haven't thought this through carefully and I might be missing something, but what if you take a function $f:\mathbb R\to\mathbb R$ that isn't locally monotone anywhere (say the Weierstrass function). That can't happen for a differentiable function, and your coordinate maps have to be monotone. $\endgroup$ Jul 27, 2014 at 0:57
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    $\begingroup$ That would probably do it. $\;$ $\endgroup$
    – user5810
    Jul 27, 2014 at 1:15
  • $\begingroup$ In fact, coordinate maps in one dimension are monotone, so differentiable a.e., and it's enough to start out with any function that's not differentiable on a positive measure set. $\endgroup$ Jul 27, 2014 at 1:19

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