According to the Implicit Function Theorem ($2$ dimensional case):
if $F:U\subset \mathbb{R}^2\to \mathbb{R}$ is a $C^1$ function defined on the open set $U$ and $(x_0,y_0)\in U$ such that $F(x_0,y_0)=0$ and $F_y(x_0,y_0)\neq 0$, then for some neighbourhoods $I,\,J$ of $x_0,y_0$, respectively there is a $C^1$ function $f:I\to J$ such that $f(x_0)=y_0$ and $F(x,f(x))=0$ for all $x\in I$.
In othet words, the Implicit Function Theorem gives us conditions for a curve to be a graph of a function (locally).
Let $S=\{(x,y)\in U:F(x,y)=0\}$ be the zero level set of $F$. This level set is non-empty, since $(x_0,y_0)\in S$.
My question is: how are we sure that $S$ is really is a curve? For example, for $F(x,y)=x^2+y^2$, we have $S=\{(0,0)\}.$
Does condition $F_y(x_0,y_0)\neq 0$ imply somehow that $S$ is (indeed) a curve?
Or should we add this on the statement of the theorem? (which I think not...)
Thanks in advance for the help.
At a point $(x_0,y_0) \in F^{-1}(0)$ such that $F_y(x_0,y_0) \neq 0$, the IFT gives us a $C^1$ function $f:I\rightarrow J$ such that $$ F^{-1}(0) \cap (I\times J) = \{(x,f(x)):x \in I \}.$$ Therefore within $I \times J$, $F^{-1}(0)$ may be represented by the $C^1$ curve $$ \gamma: I \rightarrow I\times J, \quad \gamma(x) := (x,f(x)).$$ It follows that around any point $(x_0,y_0)$ such that $DF \neq 0 \iff F_x \neq 0 \text{ or } F_y \neq 0$, there is a neighbourhood $U$ of $(x_0,y_0)$ and a $C^1$ map $\gamma: I \rightarrow \mathbb{R}^2$ such that the image of $\gamma$ is $U \cap F^{-1}(0)$. This is a definition of a one-dimensional $C^1$ submanifold of $\mathbb{R}^2$ (also known as a curve).
By the classification of one-dimensional manifolds, $F^{-1}(0)$ must topologically be a disjoint union of circles and lines.