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 $\partial S=S\cap \partial M$.


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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.