differentiable maps between topological spaces without using the idea of manifolds

Question

Is it possible to define differentiable maps between topological spaces without using the idea of manifolds?

Answer 1

On a general topological space: no, not in a way that is consistent with the common meaning. A self-homeomorphism of $\mathbb R^2$ can be nowhere differentiable: e.g., $F(x,y)= (x,y+f(x))$ where $f:\mathbb R^2\to\mathbb R^2$ is the Weierstrass function. If $\Phi:\mathbb R^2\to\mathbb R^n$ is a differentiable map, then $\Phi\circ F$ is nowhere differentiable: composition with $F $ completely destroys differentiability. On the other hand, being a homeomorphism, $F$ preserves all properties that can be defined in terms of topology. Hence, differentiability cannot be defined in terms of topology.

One can hope for better results with metric spaces, because on many spaces (including all $\mathbb R^n$) every metric-preserving bijection of the space is a diffeomorphism. Thus, a map preserving metric structure also preserves differentiability; this gives hope that metric structure can capture differentiability. And indeed, first-order derivatives can be made to work on many metric spaces that are not manifolds: see the survey Nonsmooth Calculus by Heinonen. The emphasis in this field is on the first-order weak derivatives (rather than classical/pointwise derivatives), which are more robust: e.g., they survive a bi-Lipschitz change of the metric. Thus, one can work with differentiable maps on topological spaces with a distinguished equivalence class of metrics, equivalence being bi-Lipschitz. This sort of structure falls in between the metric and topological spaces.