I'm studying for preliminary/qualifying exams, and came across the following problem:
Suppose that $S_{1}$ and $S_{2}$ are smooth surfaces in $\mathbb{R}^{3}$ that intersect at a point $p$ and do not have the same tangent plane at that point. Show that $p$ is not an isolated point of $S_{1} \cap S_{2}$.
Since $T_pS_1+T_pS_2=T_p\mathbb{R}^3$, I would like to apply transversality to the intersection $S_1\cap S_2$ to deduce that $S_1\cap S_2 \ni p$ (or potentially a neighborhood in $S_1\cap S_2$ containing $p$) is a 1-dimensional manifold (and hence, without isolated points).
Since the problem statement (a priori) only guarantees $T_pS_1+T_pS_2=T_p\mathbb{R}^3$ at $p\in S_1\cap S_2$, I'm not sure how to realize the transversality condition for all $x\in S_1\cap S_2$. Guillemin-Pollack (pg. 28) indicates that we can reduce the study of the relation $\iota(x)\in S_2$ to the simpler case where $S_2$ is a single point, but I'm having trouble seeing why this is the case (and how it exactly applies to the problem). Thanks!
Let $f_i$ be defined in a neighborhood $U$ around $p$ such that $f_i:U\to\mathbb{R}$ and $U\cap S_i= f_i^{-1}(0)$. Let $g:U\to \mathbb{R}^2, g(x)=(f_1(x),f_2(x))$.
Our aim is to show that $rk(dg_x)=2$ near $p$. This would imply that the codimension of $g^{-1}(0)=U\cap S_1\cap S_2$ is $2$ and hence a $1$ dimensional submanifold as desired.
To this end, we first show that $rk(dg_p)=2$. As $T_pS_1 + T_pS_2=T_p\mathbb{R}^3$, we can find $v_i\in T_pS_i\cap(T_pS_j)^c$, where $i,j$ distinct. Now, given that $(df_i)_p v_j = 0$ iff $i = j$, we have that $dg_p(v_1),dg_p(v_2)$ are linearly independent hence $rk(dg_p)=2$.
This tells us that $dg_p$ has a $2\times 2$ minor with non zero determinant and so by continuity there is some neighborhood of $p$ such that $rk(dg_x)=2$ for all $x$ in said neighborhood as desired.