Consider the following question:
Suppose that $X$ is a subset of $\mathbb{R}^n$ and $Z$ is a subset of $X$. Show that the restriction to $Z$ of any smooth map on $X$ is a smooth map on $Z$.
(Note: A smooth function is defined to be one that has continuous partial derivatives of all orders)
What exactly is this question asking? I don't see why this does not follow immediately, and so I must be missing something quite crucial. I don't think the question was intended to prompt an answer of the form
Proof:
QED.
Best Answer
For arbitrary sets $X \subset \mathbb{R}^{m}, Y \subset \mathbb{R}^{n}$, a function $f: X \to Y$ is, by definition, smooth, if for any $x \in X$ there exists an open neighborhood $x \in U \subset \mathbb{R}^m$ and a smooth function $F: U \to \mathbb{R}^n$ s.t. $F_{|U \cap X} = f_{|U \cap X}$.
So if $z \in Z \subset X$ and $f$ is smooth, then for any $z \in Z$ there exists an open neighborhood $z \in U \subset \mathbb{R}^m$ and a smooth function $F: U \to \mathbb{R}^n$ s.t. $F_{|U \cap X} = f_{|U \cap X}$. Hence, $f_{|Z}$ is smooth.