5

Suppose $N,M$ are smooth manifolds and $f:N\rightarrow M$ is a smooth map intersecting transversally with a submanifold $S$ of $M$. The question is to prove that $f^{-1}(S)$ is a smooth submanifold of $N$.

There is a proof in Lee’s smooth manifolds book but that seems to be incomplete or I am misunderstanding something.

The idea given there is to somehow show that $f^{-1}(S)$ is a regular level set of some smooth map $N\rightarrow M’$ for some smooth manifold $M’$.

As $S$ is an embedded submanifold, it is locally a regular level set i.e., given $p\in S$ there exists open $U$ in $M$ containing $p$ and a smooth map $\varphi :U\rightarrow \mathbb{R}^k$ such that $U\cap S=\varphi^{-1}(0)$.

We then have $f^{-1}(U\cap S)=f^{-1}\varphi^{-1}(0)=(\varphi\circ f)^{-1}(0)$. By $f$ in $\varphi\circ f$ I mean restriction of $f$ to $\varphi^{-1}(U)$. This would only tell me (after proving that $0$ is a regular value for composition) that $f^{-1}(U\cap S)=f^{-1}(U)\cap f^{-1}(S)$ is a submanifold of $f^{-1}(U)$. I do not see why this would imply $f^{-1}(S)$ is a submanifold of $N$. Is it that straightforward?

Any suggestion is welcome.

  • 6
    Take a look at Proposition 5.16 in my book: It says $S$ is an embedded submanifold if and only if every point of $S$ has a neighborhood $U$ in $M$ such that $U\cap S$ is a level set of a smooth submersion. You might find that useful here. – Jack Lee Oct 31 '17 at 21:29
  • 1
    @JackLee Master, I saw that proposition just now, i understand that. Thank you. I like your Riemannian Manifolds - An Introduction to Curvature book very much. Thanks for writing one such. –  Nov 01 '17 at 02:30
  • @JackLee to use that result i have to prove that the map $f^{-1}(U)\rightarrow U\rightarrow \mathbb{R}^k$ is a submersion second map is a submersion, I do not see why first map is a submersion. –  Nov 01 '17 at 06:29

1 Answers1

3

“$ f $ is transverse to $ S $” means that $ {T_{f(x)}}(M) = {(\mathrm{d} f)_{x}}({T_{x}}(N)) + {T_{x}}(S) $ for any $ x \in N $.

Consider a chart $ \varphi: U \to \Omega $ of $ M $ that is adapted to $ S $ and centered at $ f(x) $. “Adapted” means that $ \varphi[U \cap S] = \Omega \cap (\mathbb{R}^{k} \times \{ O \}) $, where $ k $ denotes the dimension of $ S $.

Observe that $ {T_{f(x)}}(S) = {(\mathrm{d} \varphi^{- 1})_{0}}(\operatorname{Span}(e_{1},\ldots,e_{k})) $.

Let $ \psi = (\varphi^{k + 1},\ldots,\varphi^{n}) $; we will prove that $ O $ is a regular value of $ \psi \circ f $. As $ \varphi $ is a diffeomorphism, \begin{align} \mathbb{R}^{n} & = {(\mathrm{d} \varphi)_{f(x)}}({T_{f(x)}}(M)) \\ & = {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) + {T_{x}}(S) \\ & = {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) + {(\mathrm{d} \varphi)_{f(x)}}({T_{f(x)}}(S)) \\ & = {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) + \mathbb{R}^{k} \times \{ O \}, \end{align} meaning that $ \{ O \} \times \mathbb{R}^{n - k} \subseteq {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) $. In particular, $$ \mathbb{R}^{n - k} \subseteq {(\mathrm{d} \psi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)), $$ so $ O $ is a regular value.

Consequently, we deduce that $ {(\psi \circ f)^{- 1}}[O] = {f^{- 1}}[S \cap U] $ is a sub-manifold of $ N $ of co-dimension $ n - k $. To conclude, simply cover $ S $ with adapted charts $ U_{i} $ and then take the inverse image of the intersection. This will prove that $ {f^{- 1}}[S] $ can be covered by charts $ {f^{- 1}}[S \cap U_{i}] $, so it will be a sub-manifold.

Berrick Caleb Fillmore
  • 5,269
Overflowian
  • 5,771