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I think I need a hint in solving the following exercise:

Let $S_1$, $S_2 \subseteq \mathbb R^3$ be surfaces (i.e. 2-dimensional submanifolds), with nonempty intersection $\Gamma := S_1 \cap S_2.$ Show that, if $V :=T_pS_1 \cap T_pS_2$ is a line for each $p \in \Gamma$, then $\Gamma$ is a curve.

Now this is intuitively clear, but the problem is how to prove it formally.
Thoughts:
Fix $p\in S_1$. Since $S_1$ is a surface we can find an open neighbourhood of $p$, $U\subseteq \mathbb R^3 $ and a differentiable map $f: U \to \mathbb R$ with $S_1\cap U = f^{-1}(\{0\})$ and $\nabla f(q) \neq 0$ for each $q\in S_1 \cap U$.

Now $T_pS_1$ = $\ker (D_p f)$ and therefore $D_p f(V) = \{0\}$. Is this going in the right direction? Another idea was to modify $f$ to get a map $\tilde f:W \to \mathbb R$ for some open subset $W \subseteq \mathbb R^2$ which then describes the curve, i.e. $\Gamma = f^{-1}(\{0\})$. Any hints appreciated!

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Let $p \in S_1\cap S_2$. Show that $T_pS_1 \cap T_pS_2$ $($and here I'm guessing the tangent planes are thought of as embedded in $\mathbb R^3)$ is a line if and only if $S_1$ and $S_2$ cross (rather than touch) at $p$. Can you conclude?

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  • $\begingroup$ I thought about that, and I'll try and prove it. But I am not sure how to conclude. Obviously, it's not a curve if they touch at $p$, but is it trivial that $\Gamma$ must be a curve if they cross? Also: how to "formalize" the two surfaces crossing? $\endgroup$ – Staki42 May 30 '18 at 18:00
  • $\begingroup$ Well, what is your definition of curve? And of surface? I guess some well behavedness is necessary somewhere in the definitions, because one can come up with pathological examples if things are left too wide open. Think of a topologist's sine which one extends along another dimension, and a plane intersecting it, The intersection would basically be various lines that accumulate. $\endgroup$ – Fimpellizieri May 30 '18 at 18:08
  • $\begingroup$ Both my surface and curve definitions are equivalent to the one I wrote in the question with differentiable $f$ and non-vanishing gradient (except for a curve it is a open set of $\mathbb R^2$ rather than $\mathbb R^3$). $\endgroup$ – Staki42 May 30 '18 at 18:12
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If $\Gamma=S_1\cap S_2$, then for $p\in \Gamma$, $$ T_p\Gamma =T_pS_1\cap T_pS_2 $$ is a line so that tangent space $T_p\Gamma$ is well-defined. So $\Gamma$ is 1-dimensional so that it is a curve.

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