# (Proof verification) Extension Lemma for Functions on Submanifolds

There is a similar question about this lemma in Lee's book, but I wasn't really satisfied with the detail in the proposed solution. In particular, the OP of the linked question doesn't describe how to get the smooth maps which I am calling $$f_p$$ below. I am hoping someone can give critical feedback on this proof and point out where, if anywhere, I have erred.

Lemma 5.34 (Extension Lemma for Functions on Submanifolds). Suppose $$M$$ is a smooth manifold, $$S\subset M$$ is a smooth submanifold, and $$f\in C^\infty(S)$$.

1. If $$S$$ is embedded, then there exist a neighborhood $$U$$ of $$S$$ in $$M$$ and a smooth function $$\widetilde f\in C^\infty(U)$$ such that $$\widetilde f|_S = f$$.

2. If $$S$$ is properly embedded, then the neighborhood $$U$$ in part (a) can be taken to be all of $$M$$.

Proof of 1:

Let $$n = \dim S$$ and $$m = \dim M$$. Since $$S$$ is embedded, each $$p\in S$$ is in the domain of a slice chart $$(U_p,\varphi_p)$$ in $$M$$ such that $$S\cap U_p$$ is a single slice in $$U$$. Since $$f$$ is smooth, at each $$p\in S$$ we can find a slice chart so that $$\mathbf{f}_p:= f\circ\varphi_p^{-1}\colon \varphi_p(S\cap U_p)\to\Bbb R$$ is smooth. Since $$\varphi_p(S\cap U_p) = \{(x^1,\dots,x^n,0,\dots,0)\in \varphi_p(U_p)\}$$, and $$\mathbf{f}_p$$ is smooth on this set, there is a smooth extension $$\widehat f_p\colon V_p\subset\Bbb R^m\to \Bbb R$$ of $$\mathbf{f}_p$$, where $$V_p$$ is an open set in $$\Bbb R^m$$ containing $$\varphi(p)$$.

(Here is where I see a potential "bug" in this proof. By the definition of a smooth map on $$S$$, the map $$\mathbf{f}_p$$ is "smooth." However, since $$\mathbf{f}_p$$ is defined on a set that is not an open subset of $$\Bbb R^n$$, I believe, and please do address this point, "smooth" means that there is the extension I am calling $$\widehat f_p$$.)

The restriction of $$\widehat f_p$$ to the open set $$W_p := \varphi_p(U_p)\cap V_p$$ is still smooth and gives us a map $$f_p\colon \varphi_p^{-1}(W_p)\to\Bbb R$$ defined by $$f_p(x) = \big(\widehat f_p|_{W_p}\big)\circ\varphi_p(x)$$.

The collection of open sets $$\varphi_p^{-1}(W_p)$$ is an open cover of $$S$$, so let $$(\psi_p)_{p\in S}$$ be a partition of unity subordinate to this cover. For any $$x\in U:=\bigcup_{p\in S}\varphi^{-1}(W_p)$$, define $$\widetilde f(x) = \sum_{p\in S}\psi_p(x)f_p(x)$$. If $$x\in S$$, then $$\widetilde f(x) = \sum_{p\in S}\psi_p(x) f(x) = f(x).$$ Hence $$\widetilde f$$ is an extension of $$f$$. Since the collection of supports of the $$\psi_p$$ is locally finite, each point in $$U$$ has a neighborhood in which the sum is finite, so this defines a smooth map since every point has a neighborhood such that $$\widetilde f$$ restricted to that neighborhood is smooth. Hence $$\widetilde f\in C^\infty(U)$$ is the desired extension. $$\square$$

Proof of 2:

In this case, the only thing that changes is that $$S$$ is actually closed in $$M$$ by a lemma in Lee's book. By appending $$M\smallsetminus S$$ to the collection $$\{\varphi_p^{-1}(W_p)\}$$ and taking a partition of unity $$\{\psi_p: p\in S\}\cup\{\psi_0\}$$ subordinate to this open cover of $$M$$ with $$\operatorname{supp}\psi_0\subset M\smallsetminus S$$, define $$\widetilde f(x) = \psi_0(x) + \sum_{p\in S}\psi_p(x)f_p(x)$$, where $$f_p$$ is defined in the first part. $$\square$$

I heartily welcome feedback, but I believe the proof in my question works. The first section where I thought the proof was lacking can indeed be improved. I got the idea for the improvement from the hint to Exercise 2.3 in Lee's Riemannian Manifolds, restated here for interest:

Exercise 2.3. Suppose $M\subset \widetilde M$ is an embedded submanifold.

(a) If $f$ is any smooth function on $M$, show that $f$ can be extended to a smooth function on $\widetilde M$ whose restriction to $M$ is $f$. [Hint: Extend $f$ locally in slice coordinates by letting it be independent of $(x^{n+1},\dots,x^m)$, and patch that together using a partition of unity.]

Here is the improved proof of 1:

Let $n = \dim S$. Since $S$ is embedded, each $p\in S$ is in the domain of a slice chart $(U_p,\varphi_p)$ in $M$ such that $f|_{S\cap U_p} = f(x^1,\dots,x^n)$ and the $x^i$ are slice coordinates. Extend $f|_{S\cap U_p}$ to $U_p$ by setting $f_p(x^1,\dots,x^n,x^{n+1},\dots,x^m) = f|_{S\cap U_p}(x^1,\dots,x^n)$. The map $f_p\colon U_p\to\Bbb R$ is smooth because it is independent of its last $m-n$ coordinates and it is smooth with respect to its first $n$ coordinates.

The collection of open sets $U_p$ is an open cover of $S$, so let $(\psi_p)_{p\in S}$ be a partition of unity subordinate to this cover, and define $\widetilde f(x) = \sum_{p\in S} \psi_p(x)f_p(x)$. If $x\in S$, then $$\widetilde f(x) = \sum_{p\in S}\psi_p(x) f(x) = f(x).$$ Hence $\widetilde f$ is an extension of $f$.

Since the collection of supports of the $\psi_p$ is locally finite, each point in $U$ has a neighborhood in which the sum is finite, so this defines a smooth map since every point has a neighborhood such that $\widetilde f$ restricted to that neighborhood is smooth. Hence $\widetilde f\in C^\infty(U)$ is the desired extension. $\square$

The proof of 2 is modified by replacing $\varphi_p^{-1}(W_p)$ with $U_p$ defined in the modified proof of 1 above. $\square$

The reason that the original proof still works is that the extensions I describe in the first proof of 1 can be taken to be the $f_p$ explicitly described in the proof in this answer. The approach taken here simplifies some of the notation and construction of the extension $f_p$ by blurring the distinction between points $p$ in the manifold and their local coordinate representations $(x^1,x^2,x^3,\dots)$. Smoothness of the coordinate representation of $f$ is equivalent to smoothness of the actual map defined on the manifold.