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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?

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  • $\begingroup$ Welcome to Math.SE. You are a relatively new user here. Once you've understood an answer and are happy with it, you should accept it, so that the question will not stay on the "unanswered" list. If you're not happy with it, you can of course ask for clarification. $\endgroup$ – Ted Shifrin Dec 4 '15 at 21:12
  • $\begingroup$ I apologize for not accepting it sooner. I have been incredibly busy and will get to the answer as soon as possible (it will be this weekend). I just don't want to accept something I have yet to look over and understand thoroughly. $\endgroup$ – user023049 Dec 5 '15 at 3:00
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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.

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