# Stability of transversality

I'm having some troubles to prove that transverality is a stable property.
First, some definitions.

1. $$F:M\rightarrow{N}$$ and $$A\subset N$$. $$F$$ is transveral to $$A$$ $$\$$ ($$F\pitchfork A$$) $$\$$ if $$\forall x\in f^{-1}(A)$$, it holds $$\begin{equation*} T_xf(T_xM)+T_{f(x)}A=T_{f(x)}N. \end{equation*}$$
2. A property is said to be stable if for any function $$f_0:M\rightarrow N$$ with that property and any homotopy $$F:M\times[0,1]\rightarrow N$$, with $$F|_{M\times \{0\}}=f_0$$, there exists $$\epsilon>0$$, such that each $$f_s$$ still has that property for $$s<\epsilon$$.

Stability theorem

Let $$M$$ and $$N$$ be manifolds and suppose $$M$$ compact. Then, the property of being transversal to a given submanifold $$Z\subset N$$ is a stable property.
Proof. Let $$k:=\text{codim}_NZ$$. Take any $$a\in f_0^{-1}(Z)$$, since $$f_0\pitchfork Z$$, we can choose a chart of $$M$$ centered in $$a$$ and a chart of $$N$$ centered in $$f_0(a)$$ such that the local representation of $$f_0$$ in these charts has the following property: $$$$\label{matrix f0} \det\begin{pmatrix} \dfrac{\partial f_0^1}{\partial x^1}(a)&...&\dfrac{\partial f_0^1}{\partial x^k}(a)\\ \vdots&\ddots&\vdots\\ \dfrac{\partial f_0^k}{\partial x^1}(a)&...&\dfrac{\partial f_0^k}{\partial x^k}(a) \end{pmatrix}\neq0,$$$$ otherwise the transversality condition couldn't be satisfied.
Now, since $$f_0=F(\cdot,0)$$, we have $$T_af_0=T_{(a,0)}F$$ and so $$F$$ satisfies in the same charts $$$$\label{matrix of F} \det\begin{pmatrix} \dfrac{\partial F^1}{\partial x^1}(a,0)&...&\dfrac{\partial F^1}{\partial x^k}(a,0)\\ \vdots&\ddots&\vdots\\ \dfrac{\partial F^k}{\partial x^1}(a,0)&...&\dfrac{\partial F^k}{\partial x^k}(a,0) \end{pmatrix}\neq0.$$$$ Since $$F$$ is smooth, partial derivatives and determinant are continuous functions, thus, by the theorem of sign permanence, there will be an open neighborhood $$U_a$$ of $$(a,0)$$ such that the previous determinant is different from 0 in that neighborhood.
The question is: how can I choose neighborhoods $$U_a$$ such that $$\bigcup_{a\in f_0^{-1}(Z)}(U_a)$$ forms an open covering of $$M\times\{0\}$$ in such a way I can take a finite subcovering of it?

• Do you know the so-called Tube Lemma? Any neighborhood of $M\times \{0\}$ in $M\times [0,1]$ contains a tube $M\times [0,\epsilon)$ for some $\epsilon>0$. Commented Jul 19, 2021 at 19:48
• I didn't know it. However, I don't have any neighborhood of $M\times\{0\}$ to use at the moment. All I have is a collection of neighborhoods of $(a,0)$ with $a\in f_0^{-1}(Z)$. Nothing tells me that $f_0^{-1}(Z)$ is dense in $M$, so I cannot cover the whole $M$. Commented Jul 19, 2021 at 20:18
• But if $a\notin f^{-1}(Z)$, you have transversality automatically (I like to say “by default”), and so there is a neighborhood of $(a,0)$ for such an $a$ as well. Commented Jul 19, 2021 at 20:24
• That's right. Thank you very much! Commented Jul 19, 2021 at 20:28
• You’re welcome. Why don’t you write an answer? :) Commented Jul 19, 2021 at 21:24

Take the neighborhoods as $$U_a=V_a\times [0,\delta_a)$$, with $$V_a$$ open neigborhood of $$a$$ in $$M$$. This holds $$\forall a\in f_0^{-1}(Z)$$, hence $$\bigcup_{a\in f_0^{-1}(Z)}U_a$$ is an open covering of $$f_0^{-1}(Z)\times\{0\}$$.
Now, take $$a\notin f_0^{-1}(Z)$$. Since $$F$$ is continuous, $$\exists\epsilon_a$$ such that $$a\notin f_s^{-1}(Z)$$, $$\forall s<\epsilon_a$$. Then, $$a\notin \bigcup_{s<\epsilon}f_s^{-1}(Z)$$. This means that if we define $$\tilde{V}_a:=\bigcap_{s<\epsilon_a}\left(f_s^{-1}(Z)^c\right),$$ the collection $$\bigcup_{a\in (f_0^{-1}(Z))^c}\tilde{V}_a\times[0,\epsilon_a)$$ is a covering of $$(f_0^{-1}(Z))^c$$, open in $$(f_0^{-1}(Z))^c\times[0,1]$$.
If we combine the two coverings, we find a covering of $$M\times\{0\}$$: $$\bigcup_{a\in f_0^{-1}(Z)} U_a\cup\bigcup_{a\notin f_0^{-1}(Z)} \tilde{U}_a.$$ (But there is a problem that the second part is not a collection of open sets).
Since $$M\times\{0\}$$ is compact, we can find finite subset $$C_1$$ and $$C_2$$ of $$f_0^{-1}(Z)$$ and $$f_0^{-1}(Z)^c$$, respectively such that $$\bigcup_{a\in C_1} U_a\cup\bigcup_{a\in C_2} \tilde{U}_a$$ covers $$M\times\{0\}$$. By the finiteness of $$C_1$$ and $$C_2$$, eixsts $$\epsilon$$ and $$\delta$$ such that $$\begin{equation*} \bigcap_{a\in C_1}[0,\epsilon_a)\supset[0,\epsilon)\qquad \bigcap_{a\in C_2}[0,\delta_a)\supset[0,\delta) \end{equation*}$$ So it is enough to choose $$\tilde{\epsilon}=\min\{\epsilon,\delta\}$$ and so $$f_s\pitchfork Z$$ $$\forall s<\tilde{\epsilon}$$.