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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.

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    $\begingroup$ Just a reminder: If you can ask a question on this site, then you can accept one of the answers. $\endgroup$
    – Lee Mosher
    Jan 15 at 16:34

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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, $$\text{im}(df_x)+ T_{f(x)}Z = T_{f(x)}Y.$$

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  • $\begingroup$ Thank you very much Ted! I appreciate your comment. $\endgroup$ Jan 14 at 18:58
  • $\begingroup$ By the way, if you haven't done so already, you might want to download the errata list I have compiled for the book. It's available on my webpage, linked in my profile. Also, don't forget to accept an answer when you're satisfied. $\endgroup$ Jan 14 at 18:59
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$\DeclareMathOperator{\Im}{Im}$ $F$ 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.

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  • $\begingroup$ Ok sorry, I should have said that I also don't understand why $F$ is a submersion. I get that fixed $x\in X$, $F(x,\_)$ is a submersion, but I don't get why $F\colon X\times X\rightarrow\mathbb{R}^m$ is also a submersion. Thank you for your answer $\endgroup$ Jan 14 at 10:31
  • $\begingroup$ @CarlosCabezas $d_{(x,s)}F(u,v) = d_xf(u) + v$ is obviously surjective since $v$ can be any vector of $\Bbb R^m$. See my edit $\endgroup$
    – Didier
    Jan 14 at 10:41
  • $\begingroup$ Ok, all clear now, thank you!! $\endgroup$ Jan 14 at 11:04

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