Clarifying a point in Guillemin and Pollack on a certain submersion

Question

In Guillemin and Pollack, given $f:X \rightarrow \mathbb{R}^{M}$ we are given a function $F:X \times S \rightarrow \mathbb{R}^{M}$ (where $X$ is a manifold and $S$ is an open ball in $\mathbb{R}^{M}$) defined by $F(x, s) = f(x) + s$. The book's explanation for this map being a submersion is that, for fixed $x$, $F$ is simply a translation of the open ball, then proceeds to say that this obviously implies $F$ is a submersion. I can kind of see this intuitively, but how does this translate to a proof from the definition of submersion?

Answer 1

It's fairly easy. A smooth map $F\ :\ A \to B$ is a submersion if $dF|_a\ : \ T_aA \to T_{F(a)}B$ is surjective for all $a \in A$.

Apply this to $F = f\circ\pi_1 + \pi_2$, where $\pi_1$ and $\pi_2$ are the projection $\pi_1\ :\ X\times S \to X\ :\ x \times s \mapsto x$ and projection-inclusion $\pi_2\ :\ X\times S\to \Bbb R^M\ :\ x \times s \mapsto s$ respectively. $dF = df\circ d\pi_1 + d\pi_2$.

Under the standard identification of $T_sS = T_s\Bbb R^M$ with $\Bbb R^M,\ d\pi_2|_{x \times s}$ is just the projection map $T_xX \times \Bbb R^M \ni (v, w) \mapsto w$. So it is surjective. Now for any such $(v,w)$ if $df(v) = w'$, then $dF(v,w - w') = df(v) + (w - w') = w$, so $dF$ is surjective as well.