Tangent space of preimage is the preimage of the tangent space

Question

Let $M$ and $N$ be smooth manifolds with $S\subseteq N$ a submanifold, and assume a map $f:M\to N$ is smooth and transverse to $S$. Prove that $T_p(f^{-1}(S)) = (df_p)^{-1}(T_{f(p)}S)$ for some $p\in f^{-1}(S)$.

I have found two instances of this question asked here and here, but neither question is given a complete solution (and the OPs seem satisfied with their hints).

I thought that the answer would be a simple unwinding of the definitions of $df_p$ and $T_xM$, but I wasn't able to push it through. It would be most helpful if the definition of $T_xM$ used was the set of derivations, and not the "use a path" definition. This is the definition that was emphasized most in my course, and is the one I am most comfortable with.

This is not homework -- I'm just trying to study for an exam.

Answer 1

I don't currently see a simple "definition-chase" unwinding that doesn't essentially go through the proof that $f^{-1}S$ is a submanifold of $M$, so I'll give a proof of this fact that way, and if I am missing something simpler perhaps someone else can supply it.

In order to show that $f^{-1}S$ is a submanifold, one idea (see this Math:SE question for instance) is to begin with a submersion $\psi: V \to \mathbb{R}^{n-s}$, where $V$ is a neighborhood of $f(p)$ in $N$, such that $S \cap V$ is the preimage $\psi^{-1}(0)$. Such a submersion exists because $S$ is an embedded submanifold.

Note that $T_{f(p)} S = \ker d\psi$.

Using transversiality, you can show that $0$ is a regular value of $\psi f$. (See the argument at the link, for example.)

It follows there's a neighborhood $U \subseteq f^{-1}(V)$ of $p$ such that $\psi f: U \to \mathbb{R}^{n-s}$ is a submersion. From this we see that $(f^{-1}S) \cap U = (\psi f)^{-1}(0)$ is a submanifold, and furthermore $$ T_p (f^{-1}S) = \ker d(\psi f) . $$ But $$ \ker d(\psi f) = \ker d\psi \, df = (df)^{-1} \big(\ker d\psi \big) = (df)^{-1}\big( T_{f(p)} S \big). $$

Answer 2

We will show one inclusion, and conclude with a dimension argument. I assume that we already know that $f^{-1}(S)$ is a submanifold, and I will use the word "path", but try to very emphasize on the derivation meaning.

• The inclusion $T_p(f^{-1}(S))\subset (df_p)^{-1}(T_{f(p)}S)$:

Take any tangent vector $V\in T_p(f^{-1}(S))$. You know that you can always find a smooth path $\gamma:]-\varepsilon,\varepsilon[\to f^{-1}(S)$ such that $\gamma(0)=p$ and $\gamma'(0)=V$. Two little remarks:

1. $\gamma'(0)$ is by definition the derivation $d\gamma_0(\frac{d}{dt}|_0)$, where $\frac{d}{dt}|_0$ is the usual derivation on $\mathcal{C}^{\infty}_0(\mathbb{R})$ given by $\frac{d}{dt}|_0\,f:=f'(0)$ ;

2. Here I will canonically identify the tangent space of a submanifold $V$ as a vectorial subspace of the tangent space of an ambient manifold $W$ with the canonical inclusion $di_p:T_pV\to T_pW$, where $i:V\to W$ is the inclusion.

Then we have that $$df_p(V)=df_p(\gamma'(0))=df_p(d\gamma_0(\frac{d}{dt}))=d(f\circ\gamma)_0(\frac{d}{dt})=\tilde\gamma'(0)=\tilde V,$$ with $\tilde{\gamma}:=f\circ\gamma$ which is a smooth path such that $Im(\tilde\gamma)\subset S$, so $\tilde V\in T_{f(p)}S$, $V\in (df_p)^{-1}(T_{f(p)}S)$, and we get our inclusion.

• The dimension equality:

First, $\dim f^{-1}(S)=\dim M-\mathrm{codim}_NS$, so $\dim T_p(f^{-1}(S))=\dim M-\mathrm{codim}_NS$.

Then, $\dim\,(df_p)^{-1}({T_f(p)}S)\overset{(\star)}{=}\dim \ker (df_p)+\dim df_p(T_pM)\cap T_{f(p)}S$ (classic exercise: if $L:E\to F$ is a linear morphism and $G$ a linear subspace of $F$, apply the rank formula to $L|_{L^{-1}(G)}:L^{-1}(G)\to G$). By transversality, we know that $$df_p(T_pM)+T_{f(p)}S=T_{f(p)}N,$$ so by Grassmann formula: $$\dim(df_p(T_pM)\cap T_{f(p)}S)=\dim df_p(T_pM)+\dim {T_f(p)} S-\dim T_{f(p)}N.$$ In $(\star)$, it gives

\begin{align} \dim\,(df_p)^{-1}({T_f(p)}S)&=\dim \ker (df_p)+\dim df_p(T_pM)+\dim {T_f(p)} S-\dim T_{f(p)}N\\ &=\dim M - \mathrm{codim}_NS. \end{align}

By inclusion and equality of dimensions, we can conclude that these are the same vector spaces.