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The following lemma is my question. (cf GTM218, Introduction to Smooth manifold) I can prove (b) using partion of unity as follows:

$Proof$ for any $p \in S$ choose a slice chart $W_p$ centered at $p$ and extend the restriction $f|W_p \cap S$ to some $f_p \in C^{\infty} (W_p)$. Now ,take a partition of unity $\{\phi_P: p \in S\}\cup\{\chi\}$, which issubordinate to the covering $\{W_P: p \in S\}\cup\{M \smallsetminus S\}$ of $M$( in this book, the condition $S$ is properly embedded is equivalent to the condition that $S$ is closed and embedded). Then ,we set $$ \tilde f = \sum_{p \in S} \phi_p f_p$$ One can check this function is a extension for $f$.

However, this method cannot be applied to show (a), where $S$ is not assumed to be closed.

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    $\begingroup$ How can you extend $f|W_p \cap S$ to $f_p \in C^{\infty}(W_p)$ for part(a)? $\endgroup$ – tattwamasi amrutam Mar 1 '17 at 16:31
  • $\begingroup$ Given $p \in S$, let $\phi_p=(x_p,y_p)$ be a local chart of $M$ around $p$ which is adapted to $S$. Define $f_p: W_p \to \mathbb{R}$ as $f_p(q)=f \circ {\phi_p}^{-1} \circ x_p (q)$. Why is this function a smooth extension of $\left.f\right|_{W_p \cap S}$ to $W_p$? $\endgroup$ – gpr1 Sep 17 '18 at 5:13
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If I haven't missed something, the only thing you cannot do in the case of (a) is adding $M\setminus S$ to the covering. Despite this, $\{W_p\mid p\in S\}$ is still a covering of some neighborhood $$U=\bigcup_{p\in S}W_p$$ of $S$. Therefore, simply taking a partition of unity $(\phi_p)_{p\in S}$ subordinate to this cover allows you to construct the desired function $f\in C^{\infty}(U)$ in exactly the same way as you did for point (b).

Added: In fact, $S$ is properly embedded in $U$. It is embedded, since each point lies in a slice chart, so we only have to see that it is closed. To see this, notice that $W_p\cap S$ is closed in $W_p$ (since $W_p$ is a slice chart). So $W_p\setminus S$ is open in $W_p$, which is open in $M$. Therefore $W_p\setminus S$ is open in $M$ for each $p$, so $$\bigcup_{p\in S}(W_p\setminus S) = U\setminus S$$ is open in $M$. But $U\setminus S\subseteq U$, so $U\setminus S$ is in fact open in $U$. Therefore, $S$ is closed in $U$.

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  • $\begingroup$ Thank you for your answer. Do you mean that "S is embedded" implies "S is closed"? $\endgroup$ – Hang May 27 '15 at 15:51
  • $\begingroup$ @Henry: I mean that $S$ being embedded in $M$ implies that $S$ is properly embedded in $U$ (a neighborhood); i.e. it is closed in $U$, but not necessarily in $M$. (For example, $(-\infty,0)\times\{0\}$ is closed in $(-\infty,0)\times\mathbb R$, but it is not closed in $\mathbb R^2$.) $\endgroup$ – Dejan Govc May 27 '15 at 16:30
  • $\begingroup$ OK, I get it, What's more, could you give a simple example so that this $U$ cannot be the whole $M$? Thanks! $\endgroup$ – Hang May 28 '15 at 11:41
  • $\begingroup$ Yes, take the example from my previous comment, $(-\infty,0)\times\{0\}$ and define $f:(-\infty,0)\times\{0\}\to\mathbb R$ by $f(x,0)=-\frac1x$. This cannot be continuously extended to the point $(0,0)$, since it goes to infinity there, so it also cannot be extended to $\mathbb R^2$. But it can by extended to $(-\infty,0)\times\mathbb R$, take for example $\tilde f(x,y)=-\frac1x$. $\endgroup$ – Dejan Govc May 28 '15 at 12:11

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