Trouble understanding transversality I'm reading "Differential Topology" by Guillemin, V and Pollack, A. While reading the chapter about transversality, I got through this theorem :https://i.stack.imgur.com/5wYBo.jpg
(I'm not allowed to put pictures yet).
After the proof he says that this theorem easily implies that transverse maps are general when the target manifold ($Y$) in the euclidean space $\mathbb{R}^m$. If you take $S$ to be an open ball, $F(x,s)=f(x)+s$, and you fix $x\in X$ you get that $F$ is a traslation of the ball, and therefore a submersion. Everything ok until now. But then he says "So, of course, $F$ is a submersion of $X\times S$ and therefore transversal to any submanifold $Z\subset\mathbb{R}^m$". Can you help me with that step? Thank you very much.
 A: It's an easy trick/observation. This sort of argument occurs many times in the text and in the exercises.
Because $S$ is an open ball in $\Bbb R^m$, in fact the $x$ variable is irrelevant. Fixing $x$, the derivative $dF_{(x,s)}$ maps $\{0\}\times T_sS \cong \Bbb R^m$ by the identity map to $\Bbb R^m$. That is, $dF_{(x,s)}(0,v)=v$ for all $v\in\Bbb R^m$. There's nothing more to do.
Any submersion $f\colon X\to Y$, by the way, is transverse to any submanifold $Z\subset Y$: For any $x\in f^{-1}(Z)$, $df_x\colon T_x X\to T_{f(x)}Y$ is surjective, so for any $Z$ whatsoever,
$$\text{im}(df_x)+ T_{f(x)}Z = T_{f(x)}Y.$$
A: $\DeclareMathOperator{\Im}{Im}$
$F$ being transversal to a submanifold $Z$ means that either $\Im(F) \cap Z = \varnothing$ or for any $p=F(x,s) \in \Im(F)\cap Z$, we have $\Im(d_{(x,s)}F) + T_pZ = \Bbb R^m$.
This latter condition is obvious if $F$ if a submersion, because then $\Im(d_{(x,s)}F)$ is already equal to $\Bbb R^m$.
So it all boils down to show that $F$ is a submersion.
Direct computations show that for $u\in T_xX$ and $v \in \Bbb R^m = T_sB(0,1)$, it holds that
$$
d_{(x,s)}F(u,v) = d_xf(u) + v.
$$
For $w\in \Bbb R^m$, and $u\in T_xX$ fixed, chosing $v= w-d_xf(u)$ yields $d_{(x,s)}F(u,v) = w$, so that $F$ is a submersion.
