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I would be appreciate some help with the following problem, which has been suggested (but not assigned), by my professor:

Consider the mapping $f : \mathbb R^3 \rightarrow \mathbb R$ defined by $f(x,y,z)=1−(x^2 +y^2 +z^2).$

Show that the set $S = \{(x,y,z) \in \mathbb R^3 : f(x,y,z) = 0\}$ is a surface. Find, for each point $s \in S$, an open set $U \subseteq \mathbb R^2$ and a one-one mapping $g : U \rightarrow S$ onto an open set in $S$ containing $s$.

I have a theorem which states:

Suppose $U \subseteq \mathbb R^k$ is an open set, $f : U \rightarrow \mathbb R^m$ is continuously differentiable, and $S = \{z \in U : f(z) = 0\}$ is non-empty. If $D_f(z)$ is onto for each $z \in S$, then $S$ is a $k−m$ dimensional surface.

Thanks for any help in advance.

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  • $\begingroup$ It's a sphere right? ... $\endgroup$ – user142299 Apr 11 '14 at 14:33
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Taking a bite from DoCarmo's book is the following theorem which is essentially the same as the one which you have written

Theorem: Let $U \subset \mathbb{R}^3$ and $f:U \to \mathbb{R}$ be differentiable with $a$ as a regular value. Then $f^{-1}(a)$ is a regular surface in $\mathbb{R}^3$.

Clearly $0$ is a regular value of your function here since all the partial derivatives $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ do not vanish vanish simultaneously. Hence, the result.

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$S^2$, which is the sphere,is a surface if we can find for each point $ x \in S^2$ a neigbourhood homeomorphic to $\mathbb{R^2}$. Now for $x$, a neighborhood would be $S^2-\{y\}$ where $y \neq x$. By the stereographic projection we know that this is homemorphic to $\mathbb{R^2}$.

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  • $\begingroup$ It is to be noted that $S^2$ is a Hausdorff space. $\endgroup$ – The very fluffy Panda Apr 11 '14 at 15:05

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