# Can we extend smooth maps when the target manifold has boundary?

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.

• I don't understand something. Consider the absolute value $|.| : [-1,1] \backslash \{0\} \to \mathbb (-1,2)$. This is a smooth map which admit a continuous extension but no smooth extension. Am I mistaken ? – user171326 Mar 28 '17 at 8:38
• Note that your subset $A$ where $f$ is defined is not closed... Indeed your example shows this is a necessary condition for the theorem to hold – Asaf Shachar Mar 28 '17 at 8:43
• Oh I didn't read it carefully. Sorry for the stupid remark ! – user171326 Mar 28 '17 at 8:45

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, $$F'(0) = \lim_{x\searrow 0} F'(x) = \lim_{x\searrow 0} f'(x) = 1.$$