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: \begin{equation} \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, \end{equation} 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 \begin{equation} \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. \end{equation} 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?

  • $\begingroup$ 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 $\epsilon>0$. $\endgroup$ Jul 19, 2021 at 19:48
  • $\begingroup$ 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 $M$. $\endgroup$ Jul 19, 2021 at 20:18
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 19, 2021 at 20:24
  • $\begingroup$ That's right. Thank you very much! $\endgroup$ Jul 19, 2021 at 20:28
  • $\begingroup$ You’re welcome. Why don’t you write an answer? :) $\endgroup$ Jul 19, 2021 at 21:24

1 Answer 1


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 $f_s\pitchfork Z$ $\forall s<\tilde{\epsilon}$.


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