On his book on differential geometry, Do Carmo has an excercise on section 2-4, essentially excercise 17.
Problem
That states:
Show that the transversal intersection of two regular surfaces $S_1, S_2$ is a regular curve. A transversal intersection is one such that $p \in S_1 \cap S_2 \implies T_p(S1) \neq T_p(S_2)$.
Solution
Slader has one possible solution: But although I am able to follow the math per se. Something isn't clicking. Moreover the author used a theorem that isn't in the prior sections of the book, and it is quite involved (I understand the result but I have never seen the proof and so this won't help me get the intuition right).
My work
The way I tried to do it was, the intersection must be a curve a point or the empty set, otherwise if it is a surface the tangent spaces at that point cannot disagree.
Assume the intersection is a curve $C$. $C$ contains $p$ by construction.
Then since $S_1$ is a regular surface $\forall p \in S_1 \exists f: U\subset R^2 \rightarrow V \subset S_1$ Such that $p \in f(U)$ $f$ is differentiable, a homeomorphism, and the rank of the jacobian of any point in $f(U)$ is 2.
Consider $C_p = C \cap V$ A subsection of $C$ that is contained in $V$ and contains $p$.
$\forall p \in C_p; \exists x \in U$ such that $f(x) = p$, since $f$ is a homeomorphism.
Here I get stuck, I want to justify that either $C_p$ is continuous or that $f^{-1}(C_p)$ is continuous. If I get either one of those conditions, I think I can continue since the homeomorphism condition implies the other must also be continuous and then:
Since $f^{-1}(C_p) = I$ is continuous then since $f$ is differentiable we get that $f'(x_p) = v \in T_p(S_1)$ where $x_p \in I$.
An analogous argument on $S_2$ will give us $w \in T_p(S_2)$.
And then I want to show that it is not possible for both of these vectors to be 0 simultaneously (somehow) which implies the curve is regular since it has a non zero tangent vector everywhere.