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consider the surfaces $S_1$ and $S_2$ in $\mathbb{R}^3$ defined by

$$x^2 -xy +yz^2 = 1$$ $$ xz^2 +y^2 -2yz = 0$$ We need to show that there is a smooth map from an interval aboutt $0 \in \mathbb{R}$ to $S_1 \cap S_2$ taking $0$ to $(1,1,1)$

My attempt

I just know that $F(x,y,z) = (x^2 -xy +yz^2 - 1, xz^2 +y^2 -2yz )$ then $F(1,1,1) = 0$ and I have by taking jacobian $\frac{\partial F}{\partial x \partial z}\Big|_{(1,1,1)}$ $$ \left|\begin{matrix} 1 & 2 \\ 1 & 0 \notag \end{matrix}\right| = -2 \neq 0 $$ Then by Implicit function theorem $x$, $z$ can be expressed in terms of $y$ and we have $C^1$ functions '$x$,$z$' with $x(1)= 1$, $z(1) = 1$ with $F(x(y),y,z(y)) = (0,0,0)$

My question is if this approach is right how to proceed from here to get smooth map about $0$?

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  • $\begingroup$ I think the approach is correct and the implicit function theorem includes that your new function solved in terms of the other variables is smooth. So in [en.wikipedia.org/wiki/…, your smooth map is $g$. $\endgroup$ – IAmNoOne Jan 10 '17 at 7:38
  • $\begingroup$ I should mention I didn't check your calculations in depth however. I am just pointing out that I think the procedure is correct. $\endgroup$ – IAmNoOne Jan 10 '17 at 7:39

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