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I have a question concerning differential manifolds. I need to prove that $$M=\{z-x=\sqrt{x+y^2},0<x<z\}$$ is a $2$ dimensional manifold. I define the function $F(x,y,z)=z-x-\sqrt{x+y^2}=0$. Obviously it is $C^1$ and its differential has maximum rank so it's a differential manifold. Now I know it's a manifold how should I determine its dimension? I know it's a pretty basic question but I would love to get some help.

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    $\begingroup$ I have not here the exact reference, but I recall that, by the implicit function theorem, since $F:R^3 \to R$, then the manifold $F^{-1}(0)$ has dimension 3-1=2 $\endgroup$
    – Miguel
    Dec 30, 2015 at 12:13
  • $\begingroup$ Further, you have the obvious 2-dimensional coordinates $(x,y)$, with $z=x+\sqrt{x+y^2}$. $\endgroup$
    – Miguel
    Dec 30, 2015 at 12:23
  • $\begingroup$ The implicit function theorem may be viewed as a nonlinear generalization of the rank-nullity theorem in linear algebra. The "kernel of $F$" is $M$, and the "nullity" is the dimension of $M$. $\endgroup$ Dec 30, 2015 at 13:25

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The regular value theorem implies that the zero set of $F : \Bbb R^3 \to \Bbb R$, where $F$ has surjective differential, is either empty or a submanifold of dimension $3-1 = 2$.

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