# Tangent space of preimage is the preimage of the tangent space

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.

• I could swear this is an exercise in Pollack's book Commented Dec 12, 2018 at 12:44

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 $$\ker d(\psi f) = \ker d\psi \, df = (df)^{-1} \big(\ker d\psi \big) = (df)^{-1}\big( T_{f(p)} S \big).$$

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.