The tangent space to the preimage of $Z$ is the preimage of the tangent space of $Z$. 
More generally, let $f: X \to Y$ be a map transversal to a submanifold $Z$ in $Y$. Then $W = f^{-1}(Z)$ is a submanifold of $X$. Prove that $T_x(W)$ is the preimage of $T_{f(x)}(Z)$ under the linear map $df_x:T_x(X) \to T_{f(x)}(Y)$.
("The tangent space to the preimage of $Z$ is the preimage of the tangent space of $Z$.") (Why does this imply The tangent space to the intersection is the intersection of the tangent spaces.?) 

I am very lost at this question. My idea is that to show
$$T_x(W) = \{v \in T_x(X)\;|\;df_x(v) \in T_{f(z)}(Z)\}.$$
Not sure if this is the right track, and I have no clue how to proceed.
Any ideas? Thanks.
 A: @1LiterTears: You can try it yourself, you just need some definitions:


*

*$Z$ a submanifold of $Y$ of codimension $k$ if $Z$ is locally the preimage of zero by a submersion $\varphi : Y\to\mathbb{R}^k$, that is any $y\in Z$ admits a neighborhood $V$ (in $Y$) such that $D_y\varphi : T_yV=T_yY\to\mathbb{R}^k$ is surjective and $V\cap Z=\varphi^{-1}(0)$.

*For any $x\in f^{-1}(V\cap Z)=(\varphi\circ f)^{-1}(0)$,
$$f\pitchfork Z\Longrightarrow x\ \text{is a regular point of}\ \varphi\circ f.$$

*The tangent space to a submanifold $Z$ in a point $y$ is
$$T_yZ=\big\{\gamma'(0),\ \gamma:\,(-\varepsilon,\varepsilon)\to Z\ \text{smooth},\ \gamma(0)=y\big\}$$
With this you can prove that $T_yZ=\ker D_y\varphi$, $T_xW=\ker D_x(\varphi\circ f)=\ker D_{f(x)}\varphi\circ D_x f$. Then, $T_xW=(T_xf)^{-1}(T_{f(x)}Z)$.
A: Your idea doesn't make sense as written, since $df_x$ is defined on $T_x(X)$,
not $X$ itself, and so $df_x(x)$ makes no sense (and neither does $df_w(w)$).
To answer this question, you will need to have a much stronger understanding
of basic ideas like derivatives (i.e. the maps $df_x$) and tangent spaces than
you currently seem to.  Perhaps you should try some more basic questions first.
Also, this implies the result about tangent spaces of intersections pretty
immediately: to say that submanifolds $X$ and $Y$ of $Z$ are transverse is the
same as saying the the inclusion $\iota:X \to Z$ is transverse to $Y$.  The preimage
of $Y$ under $\iota$ is precisely $X \cap Y$.  Now just apply the general statement
in this particular case.
