A circle is topologically equivalent to a triangle, because there is a continuous deformation to transform one into the other. But this misses out on a key fact of the triangle, which is that it has 3 corners, where the derivative is not defined, yet the line tangent to the circle is defined at all points on the circle. Similarly, a cube is topologically equivalent to a sphere, but again, a cube has sharp corners where a sphere does not. Is there a generalization of “smooth transformation” that would account for this difference, and see the circle and triangle as different, cube and sphere different? What would the equivalence between the cube and all shapes like it be called?
Another example: Consider $f(x)=x^2$ and $g(x)=|x|$. Again, there is a “continuous transformation” (I’m not exactly sure what that means formally) between $f(x)$ and $g(x)$, so they should be considered topologically equivalent. Yet, there is no continuous transformation between $f’(x)$ and $g’(x)$, since $g’(x)$ has a discontinuity at $x=0$ while $f’(x)$ does not. Therefore, we should say that $f’(x)$ and $g’(x)$ are not topologically equivalent.
Is there a notion like topological equivalence where we require not just $f$ and $g$ to have a smooth transformation between the two, but for all $n$, $f^{(n)}$ and $g^{(n)}$ can be smoothly transformed into each other? Or perhaps for a fixed $m$ and for all $n<m$ the previous condition holds?
This question might seem a little imprecise. In particular, I am not sure how to generalize this notion beyond the one-dimensional case, as I am not sure what the similar notion of “derivative” is there.