Why is the differential being onto equivalent to it not being zero?

[yifli]

I have difficulty understanding the following Theorem

If U is open in [itex]ℝ^2[/itex], [itex]F: U \rightarrow ℝ[/itex] is a differentiable function with Lipschitz derivative, and [itex]X_c=\{x\in U|F(x)=c\}[/itex], then [itex]X_c[/itex] is a smooth curve if [itex][\operatorname{D}F(\textbf{a})][/itex] is onto for [itex]\textbf{a}\in X_c[/itex]; i.e., if [tex]\big[ \operatorname{D}F\bigl( \begin{smallmatrix}a \\ b\end{smallmatrix}\bigr)\big]≠0 \mbox{ for all } \textbf{a}=\bigl( \begin{smallmatrix}a \\ b \end{smallmatrix}\bigr)\in X_c [/tex]

I don't understand why the differential of F at a being onto is equivalent to saying the differential is not zero. Can someone explain? Thanks

 

[HallsofIvy]

Think about what happens for simple functions like [itex]f(x)= x^2[/itex] where f'(x)= 0.

 

[yifli]

Think about what happens for simple functions like [itex]f(x)= x^2[/itex] where f'(x)= 0.

the differential of [itex]f(x)=x^2[/itex] is [itex]2x[/itex], so [itex]f'(x)=0[/itex] means x=0.
Now what?