Extension of proper map I am dealing with some problems, which make me get stuck. It seems to me that I need to find an extension, given a proper map on a subset. Here are my questions:

*

*Let $M$ be a smooth manifold, $A \subseteq M$ is a closed subset. Here is a proper map $f:A \rightarrow \mathbb{R}$. Is it right or not: $f$ can be extended continuously to proper map $F: M \rightarrow 
 \mathbb{R}$. I tried to puncture a euclidean space to construct a counterexample but failed.

*Let $M$ be a smooth manifold with boundary, $U \subseteq M$ is an open neighborhood containing $\partial M$. Here is a smooth embedding $h: U\rightarrow \mathbb{H}^{2n+1}$. Can we extend it to a proper map $H: M\rightarrow \mathbb{H}^{2n+1}?$ If such a $H$ exists, can I request that there is a strict positive lower bound $\mu$ such that $H: IntM \rightarrow \mathbb{R}^{2n}\times[\ \mu ,+\infty\ ]$?

In fact, question 2 appears when I try to deal with this problem. By collar neighborhood, the smooth embedding $h$ exists. If we find $H$ above, we can correct proper map to global smooth embedding step by step. Sincere appreciation for any help!
The definition of proper map here requests continuity. $\mathbb{H}^{2n+1}$ denotes closed 2n+1-dimensional upper half-space of $\mathbb{R}^{2n+1}$.

Edit: For question (2), let us assume that $h$ is some smooth embedding obtained from the collar neighborhood theorem. Can the required extension exist?
 A: 

*

*Let $M$ be a smooth manifold, $A \subseteq M$ is a closed subset. Here is a proper map $f:A \rightarrow \mathbb{R}$. Is it right or not: $f$ can be extended continuously to $F: M \rightarrow 
 \mathbb{R}$.


By Tietze extension theorem every such map has a continuous extension. But I assume you want the extension to be proper as well. Such extension does not have to exist. Here's an example: consider $M=\mathbb{R}$, $A_n=[2\cdot n, 2\cdot n+1]$ and $A=\bigcup_{n=0}^\infty A_n$. Clearly $A$ is closed. Now define
$$f:A\to\mathbb{R}$$
$$f(x)=\begin{cases}
x &\text{if }x\in A_n\text{ and }n\text{ is odd}\\
-x &\text{if }x\in A_n\text{ and }n\text{ is even}\\
\end{cases}$$
The map is continuous and proper (because $f^{-1}(X)\subseteq X\cup -X$). I will show that no continuous extension of $f$ is proper.
Assume that $F:\mathbb{R}\to\mathbb{R}$ is its continuous extension. Then by the intermediate value theorem $F(x)=0$ for infinitely many $x$, at least one in between every two consecutive $[2\cdot n, 2\cdot n+1]$ intervals. More precisely for any $n$ there is $x\in [2\cdot n+1, 2\cdot n+2]$ such that $F(x)=0$. Hence the preimage $F^{-1}(0)$ is not bounded and thus not compact.



*Let $M$ be a smooth manifold with boundary, $U \subseteq M$ is an open neighborhood containing $\partial M$. Here is a smooth embedding $h: U\rightarrow \mathbb{H}^{2n+1}$. Can we extend it to a proper map $H: M\rightarrow \mathbb{H}^{2n+1}?$

Not necessarily. Consider $M=\mathbb{H}^1=[0,\infty)$. Then $\partial M=\{0\}$. Now consider $U=[0,1)$ and let $h:U\to\mathbb{H}^1$ be given by $h(x)=tan(\frac{\pi}{2}x)$. This is a smooth embedding, in fact a diffeomorphism (actually any diffeomorphism will work). But $\lim_{x\to 1}h(x)=\infty$ and so $h$ cannot be extended to $1$ in a continuous way.
