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In Lee Smooth Manifolds, this problem is given:

if $S \subset M$ is smoothly embedded and every $f \in \mathcal{C}^{\infty}(S)$ extends to a smooth function on $\textit{all}$ of $M$, then $S$ is properly embedded.

Properly embedded means the inclusion $i: S \to M$ is proper. For this we have to show $S$ is closed or it's a level set of a continuous function.

I think it should be just a few lines, but I am totally stuck for two hour! Please to help, thanks =)

What I tried: Since $S$ is embedded, it can be covered with slice charts $U_p$. If $V = \cup U_p$ then with bump functions we can make $S$ closed in $V$ by making it a level set. But $V$ is open, so it does not mean $S$ is still closed in the whole manifold. I could not alter this into proof.

I also tried to make an extension of the identity map of $S$, so that then $S$ would be a retract and closed.

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I would try to prove the contrapositive. Here's why.

The utility of smooth functions comes from the existence of bump functions. To extend any $f\in C^\infty(S)$ to a smooth function on $M$, all we need is an $\epsilon$ of wiggle room and we'll be able to extend $f$ to a function that is supported on a small neighborhood of $S$.

By contrast, if $f$ is embedded but not properly embedded, then that means exactly that there's a pathology somewhere: a compact set in $M$, pulling back to a non-compact set in $S$. This should intuitively correspond to a place where $S$ has no "wiggle room" in $M$.

So I would take that compact set $K$ in $M$, look at its pullback $\iota^{-1}(K)\subset S$, and then try to construct a function in $C^\infty(S)$ that is supported on $K$ and has no smooth extension to $M$. You should be able to use the pathology in $K\cap S$ to construct $f\in C^\infty(S)$ with not even a continuous extension to $M$.

Spoiler alert!

Let $K$ in $M$ be compact such that $i^{-1}(K)$ is noncompact in $S$. Because $\iota^{-1}(K)$ is noncompact, there is a sequence $p_1,p_2,\ldots$ contained in $\iota^{-1}(K)$ which has no accumulation points. Let $\phi_j$ be a sequence of smooth compactly supported functions on $S$ such that $\phi_j(p_j) = j$ and for all $j\neq k$, the supports of $\phi_j$ and $\phi_k$ are disjoint. Now let $\phi = \sum_j \phi_j$. Suppose $f$ is an extension of $\phi$ to $M$. Notice that $f$ cannot be continuous, for the sequence $\iota(p_j)$ is contained in $K$, but $f(\iota(p_j))\to\infty$, contradicting that the continuous image of a compact set is compact.

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  • $\begingroup$ I also try this technique for a long time, no success. Is the proof direct? $\endgroup$ – XJD Feb 22 '16 at 4:43
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    $\begingroup$ Yes, it's constructive. I'll put it in a hidden spoiler box. $\endgroup$ – Neal Feb 22 '16 at 4:44
  • $\begingroup$ Ok thanks so much!!! $\endgroup$ – XJD Feb 22 '16 at 4:53
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    $\begingroup$ XJD, I see this is your first question. If you found this answer helpful you can "accept" the answer, which lets other people know you found it helpful and rewards the person who answered it. $\endgroup$ – Gabriel Islambouli Feb 22 '16 at 4:57
  • $\begingroup$ Why is it necessary for the $\{\phi_j\}$ to be compactly supported? I thought assuming that they have disjoint supports for different indices is enough. $\endgroup$ – GFR May 16 '17 at 16:25
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I tried this way: consider $K\subset S$ compact and $M$ and $L=\iota^{-1}(K)$. Let $\{U_a\}$ be an open cover for $L$. Consider a smooth partition of the unity $\phi_a$ associated to $\{U_a\}\cup \{U_0\}$ where $U_0 = S\setminus L$ is open. For each a, $\phi_a$ admits a smoooth extension to all $M$ that agrees on $S$, namely $\tilde{\phi_a}$, so that $\sum_a \tilde{\phi_a}(x)|_S = \sum_a \phi_a(x) = 1$. Now consider the sets $V_a = supp(\tilde{\phi_a}) = \tilde{\phi_a}^{-1}((0,\infty))$. Clearly the family $\{V_a\}$ is an open cover for $K$ which admits a partition of the unity ($\tilde{\phi_a}$) so we can conclude that it is locally finite. So for all $x\in K$ there exists a neighborhood $V_x$ such that only finitely many $\{V_a\}$ intersect $V_x$. Using compactness of $K$ cover it by finitely many such sets, i.e. $$ V_{x_i},i=1,\ldots,n,\ K\subset \cup_{i=1}^n V_{x_i} $$ Now for $i=1,\ldots,n$ pick $\tilde{\phi_a}_k,k=1,\ldots,n_i$ with respect to their support as above. For what we have said it holds $\sum_{k=1}^{n_i} \tilde{\phi_a}_k(x) = 1$ for all $x\in S$. But then if we come back to S we see that the family $\{supp(\phi_a)_k\}_{k=1\ldots,n_i,\ i=1,\ldots,n} \equiv \{ {U_a}_j \}_{j=1,\ldots,m}$ is such that $\sum_{j=1}^m {\phi_a}_j(x) = 1$ for all $x\in S$. We conclude that $\{ {U_a}_j \}_{j=1\ldots,m}$ is a finite subcover and therefore $L$ is compact. This means that $\iota:S\hookrightarrow M$ is proper. The fact that it is an embedding comes from the preceding point in the problem of the book.

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