The following is John Lee's Introduction to Smooth manifolds Problem 6-10.
Suppose $F:N\to M$ is a smooth map that is transverse to an embedded submanifold $X\subset M$, and let $W = F^{-1}(X)$. For each $p\in W$, show that $T_pW = (dF_p)^{-1}(T_{F(p)}X)$.
My attempt: Let $\dim M = m,\dim X = k$. As $X$ is an embedded submanifold, each point in $X$ has a neighborhood $U\subset M$ such that $X\cap U$ is a regular level set of local defining map $\Phi:U\to\Bbb R^{m-k}$. So for each $p\in F^{-1}(X\cap U)$, $\ker d\Phi_{F(p)} = T_{F(p)}X$. Since $F$ is transverse to $X$, $dF_p(T_pN)+T_{F(p)}X = T_{F(p)}M$. Since $W$ is an embedded submanifold of $N$, $\color{red}{\Phi\circ F:F^{-1}(U)\to\Bbb R^{m-k}\ \text{is a local defining map of}\ W.}$ Now for each $p\in F^{-1}(X\cap U)$, $T_pW = \ker(d\Phi_{F(p)}\circ dF_p)$, $v\in T_pW\iff dF_p(v)\in\ker d\Phi_{F(p)} = T_{F(p)}X\iff v\in (dF_p)^{-1}(T_{F(p)}X)$. Hence, $T_pW = (dF_p)^{-1}(T_{F(p)}X)$.
I'm not sure the highlighted red part is true since $F$ is just a smooth map. Does transversality ensure this?
Best Answer
We may assume $0$ is a regular value of $\Phi$ such that $\Phi^{-1}(0) = X\cap U$. Let $p\in (\Phi\circ F)^{-1}(0)\subset F^{-1}(U)$. Then we get \begin{align*} T_pN\xrightarrow{dF_p}T_{F(p)}M = dF_p(T_pN)+T_{F(p)}X\xrightarrow{d\Phi_{F(p)}}T_{\Phi\circ F(p)}\Bbb R^{m-k}. \end{align*} where the equality follows from the fact that $F$ is transverse to $X$. Since $T_{F(p)}X$ is in the kernel of $d\Phi_{F(p)}$, $d(\Phi\circ F)_p:T_pN\to T_{\Phi\circ F(p)}\Bbb R^{m-k}$ is a surjective. Hence, $0$ is a regular value of $\Phi\circ F$ and \begin{align*}(\Phi\circ F)^{-1}(0) = F^{-1}\circ\Phi^{-1}(0) = F^{-1}(X\cap U) = F^{-1}(X)\cap F^{-1}(U) = W\cap F^{-1}(U).\end{align*} Hence, $\Phi\circ F:F^{-1}(U)\to\Bbb R^{m-k}$ is a local defining function of $W$.