Let $M,N$ be smooth manifolds with boundary. Let $A \subseteq N$ be closed, and let $f:A \to M$ be a smooth map.
Suppose $f$ has a continuous extension to $N$. Does it have a smooth extension to $N$?
In case the target manifold $M$ has no boundary, this is known to be true (See corollary 6.27 in Lee's book introduction to smooth manifolds).
I am asking about the case where $M$ has non-empty boundary.
Not necessarily. Here's a counterexample. Let $M=[0,\infty)$, $N=\mathbb R$, $A=[0,\infty)$, and define $f\colon A\to M$ by $f(x)=x$. Then $f$ has a continuous extension to $N$, for example $$ f(x) = \begin{cases} x, & x\ge 0,\\ 0, & x<0. \end{cases} $$ But it has no smooth extension. If $F\colon \mathbb R\to [0,\infty)$ were a smooth extension of $f$, then considered as a real-valued function $F$ would take a global minimum at $0$ so its derivative would be zero there. But by continuity of the derivative, $$ F'(0) = \lim_{x\searrow 0} F'(x) = \lim_{x\searrow 0} f'(x) = 1. $$