I'm currently reading through Guillemin and Pollack and there's a question in it which says the following
The diagonal $\Delta$ in $X \times X$ is the set of points of the form $\langle x, x \rangle$. Show that $\Delta$ is diffeomorphic to $X$, so $\Delta$ is a manifold if $X$ is.
Now the thing is $X$ (at least to my understanding), can be any topological space.
The problem is though, that the definition of a smooth function is only defined as follows
A mapping $f$ of an open set $U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ is called smooth if it has continuous partial derivatives of all orders.
A map $f : X \to \mathbb{R}^m$ defined on an arbitrary subset $X \subseteq \mathbb{R}^n$ is smooth if for each $x \in X$ there exists an open set $U \subseteq \mathbb{R}^n$ and a smooth map $F : U \to \mathbb{R}^m$ such that $F$ equals $f$ on $U \cap X$
Therefore it seems that $X$ in this question must be some subset of a Euclidean space, $\mathbb{R}^n$, am I correct in saying that?
Furthermore is there a more general definition of smoothness that can be applied to functions between arbitrary topological spaces? Say $g : (X, \mathcal{T}) \to (Y, \mathcal{K})$
It should not be too hard to compute the partial derivatives of $f(x) = (x,x)$ between $X$ and $\Delta$. I assume $X$ has locally Euclidean neighbourhoods? Check that the same holds for $\Delta$ as a subspace of $X^2$. The diagonal map is easily seen to be a diffeomorphism for $X = \mathbb{R}^n$, and this extends to the local atlases as well, I believe.