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".


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.

  • $\begingroup$ Thanks! your example is very nice. However, it made me wonder: What happens if we assume in advance $\phi$ is continuous (as a map $M \to N$)? Can you think of a counter-example in this case also? $\endgroup$ – Asaf Shachar Oct 27 '16 at 16:25
  • 1
    $\begingroup$ @AsafShachar The function $f : ([0, \infty), \{0\}) \to ([0, \infty), \{0\}) : x \mapsto \sqrt{x}$ is a counter-example to your new question. Considering variations of it, you can cook up counter-examples where $M$ is compact or has higher dimension. $\endgroup$ – Jordan Payette Oct 27 '16 at 18:34

@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).


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