Does smoothness on the interior and on the boundary (separately) imply smoothness? Let $M,N$ be smooth manifolds with boundary.
Suppose we have a map $\phi:M \to N$ which satisfies the following properties:
$$ (1) \, \, \phi(\operatorname{int}M)=\operatorname{int}N,\phi(\partial M)=\partial N $$
$$ (2) \, \, \phi|_{\operatorname{int}M}:{\operatorname{int}M} \to {\operatorname{int}N} , \phi|_{\partial M}:{\partial M} \to {\partial N} \, \, \text{are both smooth maps}$$
Is it true that $\phi$ is smooth as a map $M \to N$?
I imagine that something can go wrong when we "approach the boundary from the interior".
 A: Consider the case when $M=N=D^2$, a two-dimensional disk. Let $\phi$ be the identity on the interior, and a nontrivial rotation on the boundary. The total map $\phi$ is not even continuous.
A: @lisyarus gave a nice example, where $\phi$ does not need to be continuous.
Here is a continuous two dimensional example (similar in spirit to Jordan Payette's comment).
First we look at the one-dimensional example:
$$ f : ([0, \infty), \{0\}) \to ([0, \infty), \{0\}) : x \mapsto \sqrt{x} $$
We want to construct a similar example where $M=N=D^2$, the unit disk. The idea is to "transfer" the singularity of the one-dimensional example in the origin, to the boundary of the disk. Define
$$ f:D^2 \to D^2,  f(x,y)=(1-\sqrt{1-\|(x,y)\|)}(x,y)$$
Note that $f|_{\partial D^2}=Id_{\partial D^2}$. Also $f(\operatorname{Int} D^2)=\operatorname{Int} D^2$.
$f$ is continuous, and clearly smooth on the interior and the boundary  separately, however it's not smooth as a map $D^2 \to D^2$ (all the boundary is singular in this regard).
