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Suppose two intersecting smooth manifolds which are both subset of $\mathbb{R}^2$, and their tangent spaces on points of the intersecting parts doesn't coincident. Then is this intersecting part a 1-manifold? If yes, why?

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Is it two dimensional manifolds or are they subsets of $\mathbb{R}^{2}$? In the two dimensional manifold case, the answer is yes. Look at the dimension of the tangent space of the resulting manifold. – Uday Apr 24 '12 at 16:47

Let $X$ and $Y$ be two (nice) submanifolds of $Z,$ intersecting transversely, then the resulting intersection is a manifold of $\text{codim} (X \cap Y)= \text{codim}(X) + \text{codim}(Y).$

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The answer is Implicit function theorem, but I did not get what was the question.

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Well, I think so too. If you are saying you didn't know exactly how the question is stated, I can rephrase it as follows: Let M and N be 2-manifold in $\mathbb{R}^3$. If the tangent spaces to M and N do not coincide at any point of $M\bigcap N$, then show that $M\bigcap N$ is a 1-manifold. – user23153 Apr 25 '12 at 4:44
Locally, one can present $M$ and $N$ as solutions for $f=0$ and $h=0$ for smooth $f$ and $h$; you may assume that $\nabla f$ and $\nabla h$ are linerarly independent at each point in the nbhd. This give a map $\mathbb R^3\to \mathbb R^2$; $(x,y,z)\mapsto (f(x,y,z),h(x,y,z))$. This is a good time to apply implicit function theorem. – Anton Petrunin Apr 25 '12 at 23:17

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