# Is there a topological notion of the derivative?

My topology book says that "A function $f:U \to \mathbb{R}^m$ from an open set $U$ in $\mathbb{R}^n$ into $\mathbb{R}^m$ is smooth provided that $f$ has continuous partial derivatives of all orders. A function $f:A \to \mathbb{R}^m$ from an arbitrary subset $A$ of $\mathbb{R}^n$ to $\mathbb{R}^m$ is smooth provided that for each $x$ in $A$ there is an open set $U$ containing $x$ and a smooth function $F:U \to \mathbb{R}^m$ such that $F$ agrees with $f$ on $U \cap A$."

It really just leaves things at that, assuming knowledge of calculus including partial derivatives (which I do have). What I'm curious about is...

Is there a topological notion of the derivative? If there is not, is there a generalization of the derivative designed to allow the notion to make sense in a purely topological context?

I have never seen any references to such an idea. There is a topological notion of limit (see here), but can this be used to define a topological definition of the derivative?

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 General and algebraic topology, not really to my knowledge. Differential topology though studies diffeomorphisms (smooth maps) - what you define above. As we talk locally euclidian then, we naturally can rely on knowledge of such spaces (such as partial derivatives). This doesn't answer your question, but you probably won't find a 'topological derivative' outside differential topology. – gnometorule Jan 7 at 17:47 This question may be of interest as well – Ilya Jan 7 at 18:33 I agree with gnometorule that outside diff geometry there's likely not enough structure to discuss derivatives. Do you know about: en.wikipedia.org/wiki/Tangent_bundle ? For example differential equations evolving on manifolds are generated by vector fields that are contained in the tangent bundles. – alancalvitti Jan 9 at 4:23

 Do you mean that if we have two spaces, $X$ and $Y$, a function $f:X \to X$, and a homeomorphism $h:X \to Y$, then it's not necessarily true that for $x \in X$ we have $f'(x) = f'(h(x))$? I don't know if that necessarily makes sense (does $f$ 'act the same' on $X$ and $Y$ since they're homeomorphic?), I just need some clarification on what exactly you're saying. I do get the general idea though. – Alex Petzke Jan 8 at 3:10 @AlexPetzke: In the situation you describe one would in fact be interested in the derivative at $h(x)$ of the "conjugate" $g=h\circ f\circ h^{-1}$. If $h$ were a diffeomorphism that derivative would be related to $f'(x)$ by $g'(h(x))=h'(f(x))\circ f'(x)\circ h'(x)^{-1}$, but if $h$ is just a homeomorphism this makes no sense, and there is no clear relation at all between $f'(x)$ and $g'(h(x))$. – Marc van Leeuwen Jan 8 at 7:54 Alright. Thanks for the input. – Alex Petzke Jan 8 at 16:31