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Question: Suppose $M$ is a smooth manifold with boundary, $N$ is a smooth manifold, and $F:M\rightarrow N$ is a smooth map. Let $S=F^{-1}(c)$, where $c\in N$ is a regular value of both $F$ and $F\left|_{\partial M}\right. $. Prove that $S$ is a smooth submanifold with boundary in $M$, with $\partial S=S\cap \partial M$.

Work:

OK so let $n=\dim(N)$ and $m=\dim(M)$ and let $c$ be a regular value of both $F$ and $F\left|_{\partial M}\right.$. Then there are charts $(U,\phi)$ and $(V,\psi)$ centered at $c$ and $F(c)$ such that $c\in U$ and $F(U)\subset V$. Note that $(U,\phi)$ is a boundary chart for $M$.

We have that the function $\widetilde{F}=\psi\circ F\circ\phi^{-1} $ is smooth so there is an open neighborhood $U'$ of $\phi(c)$ such that there is an smooth extension of $G$ of $\widetilde{F}$ on $U'$.

Notice that we have that $G^{-1}(c)\cap H^{m}=\widetilde{F}^{-1}(c)\cap U'$. Now this is where I'm stuck, not sure where to go from here any tips?

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Following the proof given in Milnor's Topology From A Differentiable Viewpoint:

$\require{AMScd}$ $\begin{CD} x \in U \subseteq M @>F>> c \in V \subseteq N \\ @V\phi VV @VV\psi V \\ \phi\left(U\right) \subseteq H^{m} @>>\psi F \phi^{-1}> \psi\left(c\right) \in \psi\left(V\right) \end{CD}$

Because $c \in N$ is a regular value of $F$, for every $x \in F^{-1}\left(c\right) \subseteq M$, there are charts $\left(U,\phi\right)$ at $x$ in $M$ and $\left(V,\psi\right)$ at $c$ in $N$ such that $\psi F \phi^{-1}: \phi\left(U\right) \subseteq H^{m} \to \psi\left(V\right) \subseteq \mathbb{R}^{n}$ is smooth, and has a regular value at $\psi\left(c\right)$.

$\begin{CD} \phi\left(U\right) \subset W \subseteq \mathbb{R}^{m} @>G>> \psi\left(c\right) \in \mathbb{R}^{n} \end{CD}$

Let $W$ be an open subset of $\mathbb{R}^{m}$ such that $W \cap H^{m} = \phi\left(U\right)$; and let $G:W\to\mathbb{R}^{n}$ be the smooth extension of $\psi F \phi^{-1}$ over $W$. Now, we can always choose $W$ small enough so that $G^{-1}\left(\psi\left(c\right)\right)$ does not contain any critical points (Sard's lemma; $\mathbb{R}^{m}$ is regular). Thus, $\psi\left(c\right)$ is a regular value of $G$; and by preimage theorem (for smooth manifolds), $Z := G^{-1}\left(\psi\left(c\right)\right)$ is a (m-n)-dimensional submanifold of $\mathbb{R}^{m}$.

Furthermore, since $G$ is constant over $Z$, $T_{a}Z \subseteq ker\left\{DG\left(a\right)\right\}$ for every $a \in Z$. But $DG\left(a\right):\mathbb{R}^{m}\to\mathbb{R}^{n}$ is surjective, and the rank-nullity theorem implies $dim \left(ker\left\{DG\left(a\right)\right\}\right) =$ (m-n). Thus, $T_{a}Z = ker\left\{DG\left(a\right)\right\}$.

Now, define $\pi: Z \subseteq \mathbb{R}^{m} \to \mathbb{R}$ as $\left(x_{1},\ldots,x_{m}\right) \mapsto x_{m}$.


To show that $0 \in \mathbb{R}$ is a regular value of $\pi$:

Suppose otherwise. That is, suppose $\exists$ $a \in Z \cap \partial H^{m}$ such that $d\pi_{a}:T_{a}Z \to \mathbb{R}$ is not surjective. Then, $ker \left\{d\pi_{a}\right\}$ $=$ $T_{a}Z$ $=$ $ker \left\{DG(a)\right\}$. But, we know that $ker \left\{d\pi_{a}\right\} \subseteq \mathbb{R}^{m-1} \times \left\{0\right\} = \partial H^{m}$. Thus, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ if $0$ is not a regular value for $\pi$.

Now, since $c \in N$ is a regular value of $F|_{\partial M}$ as well, arguing as before, we can show that $\bar{G} := G|_{W\cap\partial H^{m}}$ has a regular value at $\psi\left(c\right)$. That is, for every $a \in Z\cap\partial H^{m}$, $D\bar{G}\left(a\right): \mathbb{R}^{m-1} \to \mathbb{R}^{n}$ is surjective; and by rank-nullity theorem, $dim \left(ker\left\{D\bar{G}\left(a\right)\right\}\right) = $ (m-n-1)

Finally, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ implies $ker \left\{DG(a)\right\} = ker \left\{D\bar{G}(a)\right\}$, which is clearly false (dimension mismatch). Hence, $0$ must be a regular value for $\pi$.


Since $0 \in \mathbb{R}$ is a regular value for $\pi$, $\left\{z \in Z | \pi\left(z\right) \geq 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right)\right)$ is a manifold with boundary $\left\{z \in Z | \pi\left(z\right) = 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right) \cap \partial M \right)$.

$\phi$ being a diffeomorphism, $U \cap F^{-1}\left(c\right)$ is a manifold with boundary $U \cap F^{-1}\left(c\right) \cap \partial M$. Observing that this is true for every $x \in F^{-}\left(c\right)$ completes the proof.

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  • $\begingroup$ Why should $ker \left\{DG(a)\right\} = ker \left\{D\bar{G}(a)\right\}$ be the same? $\endgroup$ – serge Apr 13 '16 at 19:07
  • $\begingroup$ Since $\bar{G}$ is just the restriction of $G$ to $\mathbb{R}^{m−1} \times \{0\}$, the directional derivative of $\bar{G}$ along any direction in this $(m-1)$-dimensional subspace will be equal to the directional derivative of $G$ along the same direction. Now, the kernel of $DG\left(a\right)$ is contained in $\mathbb{R}^{m-1} \times \{0\}$ - what that means is that all the directions along which the derivative of $G$ is zero belong to $\mathbb{R}^{m-1} \times \{0\}$. Hope this helps! (removed my earlier comment which I'd typed on the phone, and had minor syntax mistakes) $\endgroup$ – udit.m Apr 14 '16 at 4:10

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