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Assume that the equation $F(x,y,p)=0$ defines a regular submanifold $M$ of $R^3$. Consider the projection $\pi :M \rightarrow R^2$, given by $\pi (x,y,p)=(x,y)$.

By the implicit function theorem, in a neighborhood of a regular point of this map, $M$ is the graph of a smooth function $p=v(x,y)$. This is the part that I don't understand, since I would apply the inverse function theorem in the following way: let $U$ be an open subset of $R^2 \times R$, consider the smooth function $F :U \rightarrow R$, if there exists a point $(x_0,y_0,p_0)\in U$ where $F(x_0,y_0,p_0)=0$ and $\partial F / \partial p (x_0,y_0,p_0)\neq 0$, then there exists a neighborhood $A \times B$ of $(x_0,y_0,p_0)$ in $U$ and a unique function $v:A \rightarrow B$ such that in $A \times B$,
$F(x,y,p)=0$ if and only if $p=v(x,y)$.

So my problem would be solved if I knew why $\partial F / \partial p (x_0,y_0,p_0)\neq 0$, where $(x_0,y_0,p_0)$ is a regular point of $\pi$. Thanks

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  • $\begingroup$ The Implicit Function Theorem only gives you $p=v(x,y)$ if you know that $\partial F/\partial p(x_0,y_0,p_0)\ne 0$. Are you asking how to prove that this is true by using the Inverse Function Theorem? Your question is confusing. ... Or perhaps you have a concrete application with a particular $F$, etc.? $\endgroup$ – Ted Shifrin Jun 2 '13 at 18:59
  • $\begingroup$ No. What I'm asking is, why if $(x_0,y_0,p_0)$ is a regular point of $F$ and also of $\pi$, then $\partial F / \partial p (x_0,y_0,p_0)\neq 0$. $\endgroup$ – inquisitor Jun 2 '13 at 19:12
  • $\begingroup$ Oh ... If we let $(x_0,y_0,p_0) = a$, you're asking that if we know $D\pi_a\colon T_aM\to\mathbb R^2$ is an isomorphism, then how we conclude $\partial F/\partial p (a)\ne 0$? $\endgroup$ – Ted Shifrin Jun 2 '13 at 19:15
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The geometric answer is this: Let's let $a=(x_0,y_0,p_0)\in M$. The normal vector to $T_aM$ is the gradient vector $\nabla F(a)$. If this tangent plane is to project isomorphically onto the $xy$-plane, then its normal vector cannot be horizontal (i.e., in the $xy$-plane); that is, it must have a nonzero $p$-component, so $\partial F/\partial p(a)\ne 0$. In other words, if we had $\partial F/\partial p(a)=0$, then the vertical vector $(0,0,1)$ would be an element of $T_aM$; but this vector maps to $0$ under $D\pi(a)$.

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By a dimension argument, we have that: $$\pi:M\to\mathbb R^2\textrm{ is regular at }a\ \Longleftrightarrow\ \dim(\ker T_a\pi\cap\ker T_aF)=0$$ But $\ker T_a\pi=\{(u,v,w): u=v=0\}$, while $\ker T_aF=\{(u,v,w):dF(a)\cdot(u,v,w)=0\}$, so $$\ker T_a\pi\cap\ker T_aF=\{(0,0,w):\dfrac{\partial F}{\partial p}(a)w=0)\}.$$ Therefore $\pi$ is regular at $a$ if and only if $\frac{\partial F}{\partial p}(a)\ne 0$.

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