Extending an embedding to the boundary Let $f:int(M^k) \rightarrow N^n$, ($k\leq n$) be an embedding of the interior of a $k$-manifold $M$ into a closed (i.e. compact and without boundary) $n$-manifold $N$.
Can I always extend this map to an embedding $g:M \rightarrow N$? (i.e. $g$ is also an embedding with $g|_{int(M)}=f$)
By manifold I mean a smooth manifold, and by embedding a smooth embedding.
Here are my thoughts: I can extend $f$ the following way: for each point $x\in int(M)\setminus M$, I can choose a sequence $(x_k)\subset int(M)$ converging to $x$. Now since $f$ is in particular continuous, $f(x_k)$ must also converge to some $y\in N$, since $N$ is closed. Now just define $f(x):=y$. Then $f$ is continuous since it is sequentially continuous. But how can I show that the extended $f$ is smooth? Is this even always the case?
Furthermore if I have an isotopy of embeddings $H: int(M)\times [0,1] \rightarrow N$, is the map $\overline{H}:M \times [0,1] \rightarrow N$ obtained by extending each $H_t$ to $M$ as described still a smooth isotopy? I have trouble finding such results, maybe I am using the wrong terminology to find them. Does anybody have a reference or an idea where I can read more about such things?
Thank you for your time.
 A: This is not always possible.
Here is a counter-example.
Let $\bar{M} \subset \Bbb C$ be the closed unit disk.
Let $\bar{\Omega} \subset \Bbb C \subset N=\hat{\Bbb C}$ be the closure a non-empty, simply connected, open domain  different from $\Bbb C$ (we denote by $\hat{\Bbb C}$ the Riemann sphere, which is a closed manifold).
By Poincaré's Theorem, there exists a biholomorphism $f\colon M\to \Omega$, which thus yields an embedding $f\colon M \to N$.
However, there is no reason for $f$ to extend as a homeomorphism up to the boundary $\partial M$, since $\partial M = \Bbb S^1$ and $\partial \Omega$ might not even be homeomorphic to a circle. In fact, $\partial \Omega$ might not even be a manifold.
For instance, take $\Omega = \{z \in \Bbb C \mid |z| < 1 ~ \text{and} ~ z \notin \Bbb R_-\}$, whose boundary $\partial\Omega$ is the union of the unit circle and the segment $[-1,0]$.

Here is another counter-example, more constructive.
Let $\bar{M}$ be the closed unit disk in $\Bbb C$ and $N = \Bbb C \cup \{\infty\}$ be the Riemann sphere.
The map
$$
\begin{array}{r|ccc}
f \colon  & M & \longrightarrow & N \\
& z & \longmapsto & \dfrac{z}{\sqrt{1-|z|^2}}
\end{array}
$$
is a diffeomorphism between $M$ and $\Bbb{C}$.
It uniquely extends as a continuous map $\bar{f}\colon \bar{M}\to N$, with $\bar{f}(z) = \infty$ whenever $|z|=1$,which is thus not injective.
It follows that $f$ cannot be extended to $\bar{M}$ as a homeomorphism.

A third counter-example.
Let $\bar{M} = [0,2\pi]$ and $N$ the unit circle.
The map $f\colon (0,2\pi) \to N$ defined by $f(x) = e^{ix}$ is a homeomorphism, but does not extend as a homeomorphism up to the boundary.
