Extension of Smooth Functions on Embedded Submanifolds

Question

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.

Answer 1

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.

Answer 2

I know this question is old, but it seems that the proof actually follows simply from a previous result in Lee's book. I've added it for completeness. This approach is similar to what @Neal posted.

Suppose for contradiction, $S$ is an embedded submanifold of $M$ that is not properly embedded but any $f \in C^{\infty}(S)$ smoothly extends to $M$. Then there exists a compact set $K$ in $M$ such that $K \cap S$ is not compact. Prop 2.28 in Lee's book states that any smooth manifold has a "smooth exhaustion function", i.e. $S$ has an $f \in C^{\infty}(S)$ s.t. for any $c \in \mathbb{R}$, $f^{-1}((-\infty, c])$ is compact (makes use of compactly-supported partition of unity). By assumption, $f$ can be smoothly extended to some $F \in C^{\infty}(M)$. Since $F$ is continous on $M$, $F(K) \subset (-\infty, b]$ for some $b \in \mathbb{R}$.

Thus $F^{-1}((-\infty, b]) \cap S = f^{-1}((-\infty, b])$ is compact and contains $K \cap S$. Since $S$ has the subspace topology, $K \cap S$ is closed in $S$'s topology and thus compact, a contradiction. $\square$

Thus if $S$ is not properly embedded then there is a smooth function on $S$ that cannot even be continuously extended to $M$.

Answer 3

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.