Submanifold of a regular value of a manifold with boundary

Question

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?

Answer 1

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 $\left(x_{1},\ldots,x_{m}\right) \mapsto x_{m}$.

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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 $\pi$.

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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.

Answer 2

Let $ p \in \partial M^m \cap F^{-1}(c)$ and let $ (U, \phi=(x^1, \dots, x^m\geq 0)) $ be a boundary chart about $ p $ and $ \psi $ be a chart about $ c \in N^n $, so $ F $ is locally given by $ \psi\circ F \circ \phi^{-1}: \mathbb{H}^m \to \mathbb{R}^n$ with $ (x^1, \dots, x^m) \mapsto (F^1(\overrightarrow{x}), \dots, F^n(\overrightarrow{x})) $. Let $ G: U' \subset \mathbb{R}^m \to \mathbb{R}^n $ be an extension of $ \psi \circ F \circ \phi^{-1} $ to an open neighborhood $ U' $ of the origin (that's how we defined smoothness over $ \mathbb{H}^m $). Suppose $ U'' \subset U' $ is the open set of points where $ dG $ is surjective, which by definition also includes $ F^{-1}(c) \cap U' $. From this on everywhere that I write $ F $ it will make sense because of the existence of the submersion $ G\big|_{U''} $.

In this setting, $ (x^1,\dots, x^{m-1}) $ is a coordinate system for $ \partial M $ and the fact that $ p $ is a regular point of $ \partial M $ means $n \leq m-1$ and that among the coordinate vectors $ \{ \partial/\partial x^1, \dots, \partial / \partial x^{m-1} \} $, there are $ n $-many whose image under $ dF $ spans $ T_c N $, say $ \partial /\partial x^1, \dots, \partial / \partial x^n $; therefore $$ \theta: (x^1, \dots, x^n,x^{n+1}, \dots, x^m) \mapsto (F^1, \dots, F^n, x^{n+1}, \dots, x^m) $$ is a coordinate change (by considering its Jacobian) and $$ \psi \circ F \circ \phi^{-1} \circ \theta^{-1} : (F^1, \dots, F^n, x^{n+1}, \dots, x^m) \mapsto (F^1, \dots, F^n) $$ is a projection onto the first $ n $ coordinates. The nice point about the transition $\theta$ is that it keeps the $ x^m $ coordinate, so $ \mathbb{H}^m $ is invariant and it takes our boundary chart to another boundary chart. The portion of $ F^{-1}(c)$ lying in this chart is given by $ \{ (0, \dots, 0, x^{n+1}, \dots, x^m) | x^m \geq 0 \} $ which is an $ (m-n) $-dimensional slice and its boundary is exactly given by the points that are also inside $ \partial M $. For the points $ p $ in the interior of $ M $ we apply the usual regular level set theorem to get full charts so we don't see boundary points there.

Answer 3

If $p\in S\cap\text{Int}(M)$, then we can find a neighborhood $U$ of $p$ in $M$ such that $U\cap S$ is embedded in $U$ using the usual proof of the Regular Level Set Theorem. If $p\in S\cap\partial M$, then the idea is to use the following version of the rank theorem for manifolds with boundary (Problem $4$-$3$ in Lee's Introduction to Smooth Manifolds) to obtain such a neighborhood:

Let $M$ be a smooth manifold with boundary and $N$ be a smooth manifold without boundary. Let $F:M\rightarrow N$ be a smooth map with constant rank, and let $p\in\partial M$ such that $\text{ker}\:\text{d}F_{p}\not\subseteq T_{p}\partial M$. Then $F$ has a local coordinate representation of the form $$\widehat{F}(x^{1}, \dots, x^{r}, x^{r+1}, \dots, x^{m}) = (x^{1}, \dots, x^{r}, 0, \dots, 0)$$ in a neighborhood of $p$.

(For a proof, look here.)

Let $W$ be a neighborhood of $p$ such that $F|_{W}$ is a submersion. Since $\text{d}F_{p}$ and $\text{d}(F|_{\partial M})_{p}$ are surjective, we have $\text{dim}(\text{ker}\:\text{d}F_{p}) = m-n$ and $\text{dim}(\text{ker}\:\text{d}(F|_{\partial M})_{p}) = m-n-1$, which implies that $$\text{ker}\:\text{d}(F|_{W})_{p}\not\subseteq T_{p}\partial W.$$ Applying the above theorem, we obtain a neighborhood $U$ of $p$ contained in $W$ such that $U\cap S$ is embedded in $U$ as required.