Let $f:M\to N$ be a smooth map between smooth manifolds, Z is a regular submanifold of N, for $p\in M$ and $f(p) \in Z$, we say $f$ is transversal to $Z$ at $p$ if $f_{*p}(T_{p}(M))+T_{f(p)}(Z)=T_{f(p)}(N)$, now try to prove $T_{p}${$f^{-1}(Z)$}$=f_{*p}^{-1}(T_{f(p)}(Z))$.
I find it's hard for me to form a clear image of what this question is asking about, can someone give me a simple example to help me clarify this question or just tell me how to prove the statement above? thanks!
$\newcommand{\Reals}{\mathbf{R}}$In the spirit that the implicit function theorem is a non-linear generalization of the rank-nullity theorem, it's generally helpful to ask what this type of condition (or assertion) means for linear transformations.
Let $f:V \to W$ be a linear transformation (between finite-dimensional real vector spaces), $Z$ an affine subspace of $W$. For $p$ in $V$ and $f(p)$ in $Z$, we say $f$ is transverse to $Z$ at $p$ if $f(V) + Z = W$.
Prove that if $K = \ker(f)$ and $V_{0}$ is a complementary subspace (i.e., $V = K \oplus V_{0}$), then $$ f^{-1}(Z) = K \oplus (f^{-1}(Z) \cap V_{0}). $$
The preceding conclusion is not a direct translation; in this framework, your question ("the tangent space to the preimage is the preimage of the tangent space") is trivial because affine spaces can be identified with their tangent space at an arbitrary point.
As for examples, let $f:\Reals^{2} \to \Reals^{3}$ be defined by $f(u, v) = (u, 0, 0)$.
$f$ is transverse to the plane $\{x = c\}$ at each point $(c, v)$.
$f$ is not transverse to the $y$-axis at any point.
$f$ is not transverse to the $(x, y)$-plane at any point.
