# Is the $\varepsilon$-neighbourhood theorem used in proving Homotopic transverse extension?

In Guillemin & Pollack page 71

I can't see where "For compact mfld Y,the map $\pi:Y^{\varepsilon}\to Y$ is a submersion" is used to show:

"If for $f:M\to N$, closed subset $C\subset M$, closed submanifold $Z\subset N$ we have $f|_{C}\pitchfork Z$, then $\exists g:M\to N$ s.t. $g=f|_{C}$ and $g\pitchfork Z$"

The authors said this need theorem to show that transversality is generic in more arbitrary manifolds.

From the proof of the latter theorem:

" If $x\in C-f^{-1}(Z)\stackrel{Z~closed}{\Rightarrow }\forall x\in X-f^{-1}(Z)$ we have $f\pitchfork Z$

If $x\in f^{-1}(Z)\Rightarrow \exists (W,\phi)$,where $\phi$ is submersion, of f(x) s.t. $f\pitchfork Z$ for $x'\in f^{-1}(Z\cap W)\Leftrightarrow \phi\circ f$ is regular x' $\Rightarrow f\pitchfork Z \forall x\in U$ nbhd of C since x is regular for $f\circ \phi$."

So my guess is $\phi$ exists because of the $\epsilon$-neighbourhood theorem. Is this correct?

Here are typed notes from Guillemin & Pollack: http://www.math.toronto.edu/mgualt/MAT1300/week7.pdf

The two theorems are in page 25.

Thanks

• You have a detailed textbook there. What precisely is your question? What do you understand and what have you tried? – Ted Shifrin Feb 18 '14 at 4:14
• I simple don't see how they used the tubular-theorem in the proof. They don't mention it's title. – TKM Feb 18 '14 at 4:36
• They make it clear. They alter the map $f\colon M\to N\subset \Bbb R^s$ and then use that tubular neighborhood of $N$ to get maps back to $N$. – Ted Shifrin Feb 18 '14 at 4:44
• I still can't find where they use tubular of N to get maps back to N. – TKM Feb 18 '14 at 4:58
• Is the tubular neighbourhood the set W from above? – TKM Feb 18 '14 at 5:00

## 1 Answer

Your guess that $\phi$ exists because of the $\epsilon$-neighbourhood theorem is not correct. Instead $\phi$ is just a map from the chart around $f(x)$ to $\mathbb{R}^k$ and always exists by definition of a manifold. It is used because at the beginning of the proof, $x$ was chosen to be inside $C$. And it was supposed that $f$ is transversal to $Z$ on $C\subset X$. But in the earlier chapters was explained that if a map is transversal to a submanifold $Z$, then its values in $Z$ must be regular. As $x\in C$, there are only two possibilities, either $x\in f^{-1}(Z)$ or not - if not then $f(x)\notin Z$, so $f$ is automatically transversal to $Z$ at that point, therefore suppose $x\in f^{-1}(Z)$, then $f(x)\in Z$, and therefore, $f(x)$ is regular because we supposed $x\in C$. As $\phi$ is chosen to be a diffeo, it means that $\phi\circ f(x)$ is also regular. As the derivative at regular points is surjective we know that $d(\phi\circ f)$ can be written as some matrix with non-vanishing determinant. And we know that the determinant of smooth matrices is smooth, so we know that it will not suddenly drop to zero. Therefore we also know that $\phi\circ f$ is regular in a neighbourhood $N$ of $\phi(f(x))$, but then $\phi^{-1}(N)=W$ is a neighbourhood of $f(x)$ in $Y$ that must be regular. As this is true for all $f(x)$, where $x\in C$, it must be true that $f$ is transversal to an open neighbourhood of $C$. That is the logic behind the use of $\phi$.

To explain where the tubular theorem is used, I can just go on to explain the proof that continuous on the next page: One needs the above theorem to make use of $\gamma$ function of the lemma before. He defines $\tau=\gamma^2$ and then shows that $F(x,\tau(x)s)$ is transversal to $Z$ and equal to $f$ on $C$ where $f$ is transversal anyways. This is the point where the $\epsilon$-nbh theorem is used because in order to construct that $F$ (look at the proof on p. 70), we needed the submersion of $\pi:Y^\epsilon\rightarrow Y$. And in order to show that this submersion really exists, it was necessary to proof the $\epsilon$-nbh theorem, see page 72. (If the version of the book is different, the page number might be slightly different). I hope this answers the question.