Can every continuous function between topological manifolds be turned into a differentiable map?

Let $$M$$ and $$N$$ be topological manifolds that admit differential structures and let $$f:M\to N$$ be continuous. Can $$M$$ and $$N$$ always be given differential structures to become differentiable manifolds $$\widetilde M$$ and $$\widetilde N$$ such that $$f:\widetilde M\to\widetilde N$$ is differentiable? What if we impose further restrictions, such as $$\widetilde M$$ and $$\widetilde N$$ being smooth manifolds, or $$f$$ becoming smooth? Are there certain differential structures which we can't do this with (i.e. if we want to make $$f$$ differentiable, we can never make $$N$$ diffeomorphic to $$\overline N$$ where $$\overline N$$ is some differential structure on $$N$$)? What if $$M$$ already has a fixed differential structure?

• Shouldn't the 1-dimensional case yield examples? I mean there are continuous $f : \mathbb{R} \to \mathbb{R}$ which are not differentiable and I somehow think that there shouldn't be more differentiable structures on $\mathbb{R}$ then the natural one?
– user301452
Commented Jan 1, 2021 at 13:33
• There are multiple differential structures on $\mathbb{R}$. They're all diffeomorphic, but they are distinct. For a maximal atlas on $\mathbb{R}$ and a homeomorphism $h:\mathbb{R}\to\mathbb{R}$ we can take a chart $\varphi$ in the atlas and obtain a new chart $\varphi\circ h$. We can define a new atlas consisting of all the charts in the previous one composed with $h$. Clearly the transition maps are differentiable. So if $h$ wasn't itself differentiable, we have a new differential structure, since $h\circ\varphi$ and $\varphi$ clearly aren't compatible. Commented Jan 1, 2021 at 14:07
• Sorry, the last part doesn't work. I confused $\varphi\circ h$ and $h\circ\varphi$. However, if $\varphi$ commutes with $h$, then it's fine, so if, e.g. the identity is a chart (which of course it is in some maximal atlas), then we have a new differential structure. Commented Jan 2, 2021 at 0:38

There is not necessarily a way to make a map smooth. For example, suppose $$M=\mathbb{R}$$, and suppose $$N$$ is any topological manifold of dimension $$2$$ or more. Let $$f:M\rightarrow N$$ be any continuous function whose image contains a non-empty open subset of $$N$$ (e.g., take a space filling curve onto $$\mathbb{R}^n$$ and then think of this $$\mathbb{R}^n$$ as a chart).
Then there are no smooth structures on $$M$$ and $$N$$ which makes $$f$$ smooth. In fact, you cannot even make $$f$$ continuously differentiable. One way to see this is to use Sard's Theorem: if you could make $$f$$ continuously differentiable, then the set of regular values would be open and dense. Because $$M$$ has a lower dimension than $$N$$, regular values are points of $$N$$ which are not in the image of $$f$$. But then the rest that the image of $$f$$ contains an open subset of $$N$$ means the set of regular values of $$f$$ is not dense.