# 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.

• Just a reminder: If you can ask a question on this site, then you can accept one of the answers. Jan 15 at 16:34

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

• Thank you very much Ted! I appreciate your comment. Jan 14 at 18:58
• 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. Jan 14 at 18:59

$$\DeclareMathOperator{\Im}{Im}$$ $$F$$ 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.

• 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 Jan 14 at 10:31
• @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 Jan 14 at 10:41
• Ok, all clear now, thank you!! Jan 14 at 11:04