I'm studying Analysis on Manifolds by Munkres, and at page 199, it is given that
Let $S$ be a subset of $\mathbb{R}^k$; let $f: S \to \mathbb{R}^n$. We
say that $f$ is of class $C^r$, on $S$ if $f$ may be extended to a
function $g: U \to \mathbb{R}^n$ that is of class $C^r$ on an open set
$U$ of $\mathbb{R}^k$ containing $S$.
It is clear from this definition that, even if we were working on a subspace $M$ of $\mathbb{R}^n$ (or on the set $M$ with different topology other subspace topology), we still consider the opens sets of $M$ as a subset of $\mathbb{R}^n$, and show the differentiability according to that.
However, for example, if we were to show the continuity of a function $f : \mathbb{R}^k \to M$, we would consider open sets of $M$, as open sets of $M$, i.e not $\mathbb{R}^n$. In this sense, the continuity of a function is depends on the topology of the domain & codomain of that function, whereas the differentiability does not, as far as I have seen.
So my question is that, is there any definition of differentiability that is dependent on the underlying topologies of domain & codomain ?
Clarification
For example, normally, for $f: A \to \mathbb{R}^m$, the concept differentiability is defined for $x \in Int(A)$, but the very definition of interior needs the definition of what we mean by an open set, which is dependent on the underlying topology, so say (trivially) $A = \{1,2\}$, then with the discrete topology both $1,2 \in Int(A)$, but can we define differentiability in this space ?
Or let say, $A = (0,1]$ as a subspace of $[0,1]$ with the standard topology (subspace topology inherited from $\mathbb{R}$).Now if we consider our bigger space as $[0,1]$, then we should be able to define differentiability at $x = 1$ because $(0,1]$ is open in $[0,1]$, hence $1\in Int[0,1]$.
Best Answer
Take a look also at No topologies characterize differentiability as continuity by Geroch, Kronheimer and McCarty (1976).