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Let ${\bf x}=(x(u,v),y(u,v),z(u,v))$ be a $C^1$ vector-valued function on an open neighborhood $U$ of $(u_0,v_0)$ such that $({\bf x}_u\times{\bf x}_v)\cdot {\bf k}\neq 0$ at $(u_0,v_0)$. Is it true that $z$ can be locally solved in terms of $x,y$? i.e. there exists a $C^1$ function $f$ defined in a neighborhood $O$ of $(x(u_0,v_0),y(u_0,v_0)$ such that $$z(u,v)=f(x(u,v),y(u,v))$$ for all $(x(u,v),y(u,v))\in O$.

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Excuse me, but could you explain what (xu×xv)⋅k≠0 means? I don't see what product you do between the partial derivatives. And just to be sure, the vector k is the vector (0,0,1), right? –  Fezvez Nov 22 '11 at 6:14

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Your nonvanishing condition is the nonvanishing of the determinant of the Jacobian matrix of the function $(x,y)$. Therefore, by the inverse function theorem, locally, you can make a change of variables from $(u,v)$ to $(x,y)$. Therefore, $z$, which is a function of $(u,v)$, is also a function of $(x,y)$.

Geometrically, your nonvanishing condition means that the tangent plane to the surface at that point is not vertical.

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