The Transversality Condition is Generic

Question

Let $M$ and $N$ be submanifolds of $\mathbb{R}^n$. While I understand that the transversality condition $M \pitchfork N$ is stable (assuming $M$ is compact), I want to show the following property (without assuming $M$ is compact):

If there exists a point $p\in M\cap N$ such that $T_pM+T_pN=\mathbb{R}^n$, there exists a neighborhood $U$ (possibly in $M\cap N$) containing $p$ such that $T_qM+T_qN=\mathbb{R}^n$ for all $q\in U$.

While I intuitively see/believe this is true, is there a way to formalize a proof using the stability of transversality? I first attempted to consider linear independence in $T_pM$, but I'm not sure how to continuously relate this linear independence to "nearby" tangent spaces. Thank you for any help.

Answer 1

Since $N$ is embedded, we can use slice coordinates to (locally) write $N$ as the zero set of independent functions $x^{k+1},\dots ,x^{n}$. We have $\iota^{-1}(N)=M\cap N$, so it follows that $M\cap N$ is locally the vanishing set of $x^{k+1}\circ \iota,\dots,x^n\circ \iota$.

Write $g=(x^{k+1},\dots ,x^{n})$. To apply the preimage theorem, it can be shown that zero is a regular value of $g\circ \iota$ if and only if

$$T_q M+T_q N=T_q\mathbb{R}^n\cong\mathbb{R}^n$$

for all $q\in M\cap N$. Suppose that this equation holds at some $p\in M\cap N$, so that $d(g\circ \iota)_p$ is surjective. It is clear this matter is local in both the domain and codomain, so we reduce to the case where $M$ is an open set in $\mathbb{R}^k$, and the codomain is an open set in $\mathbb{R}^m$. Here, surjectivity gives $k\geq m$, and equivalently, that $d(g\circ \iota)_p$ must have an invertible $m\times m$ submatrix. Call the column indices of this submatrix $i_1,\dots,i_m$. For example, if $d(g\circ \iota)_p=\begin{pmatrix}1 & 1 & 0\\ 0 & 1 & 1\end{pmatrix}$, then we are selecting $i_1,i_3$.

Taking the determinant of this submatrix, continuity yields a neighborhood $U$ about $p\in M$ for which the determinant of the submatrix with column indices $i_1,\dots,i_m$ from the matrix $d(g\circ \iota)_q$ is non-zero. Thus, $d(g\circ \iota)_q$ is surjective for all $q\in U$, and so the equation $$T_q M+T_q N=\mathbb{R}^n$$ holds for all such points.