I'm wondering if the following proposition (from Lee's Introduction to Smooth Manifolds) also applies to smooth manifolds with boundary:
Definition: a map from an arbitrary subset $A\subseteq \mathbb{R}^n$ to $\mathbb{R}^k$ is said to be smooth if in a neighborhood of each point of $A$ it admits an extension to a smooth map defined on an open subset of $\mathbb{R}^n$.
It seems to me that proposition 1.17 does apply to manifolds with boundary, and the proof is essentially the same (picture below), but using the fact that a composition of smooth maps (in the 'general sense' as defined above) is smooth, which can be proven as follows:
Theorem: a composition of smooth maps is smooth i.e. given the arbitrary (non-empty) subsets $A\subseteq\mathbb{R}^n$, $B\subseteq\mathbb{R}^m$, and the smooth maps $f:A\to\mathbb{R}^n$, $g:B\to\mathbb{R}^p$, then $(g\circ f):A\cap f^{-1}(B)\to\mathbb{R}^p$ is smooth.
Proof: choose $p\in A\cap f^{-1}(B)$. There are (open) neighborhoods $U$ of $p$ and $V$ of $f(p)$ and smooth (in the ordinary sense) maps $F:U\to\mathbb{R}^m$ and $G:V\to\mathbb{R}^p$ that agree with $f$ and $g$ on $A$ and $B$ respectively. Since $F$ is smooth it is also continuous, and thus $F^{-1}(V)$ is open. Therefore $U\cap F^{-1}(V)$ is an open subset containing $p$ where $(G\circ F)$ is smooth (in the ordinary sense) and agrees with $(g\circ f)$ on $A\cap f^{-1}(B)$.
Am I correct that the above argument generalizes naturally to manifolds with boundary?