The implicit function theorem stated in Spivak's Calculus of Manifolds is as follows:
My question is:
if the above $f$ is continuously differentiable on an open set of
$(a,b)$ and $f(a,b)=0$and moreover $\exists$ open set $A$ containing $a$ and open set $B$
containing $b$ and a continuously differentiable function $g$ such
that for each $x\in A$ there is a unique point $g(x) \in B$ such that
$f(x,g(x))=0$is it necessarily true that
$M$ is invertible?
I ask this because I was talking with my professor about this (and maybe I misheard) but he said that the converse to implicit function theorem is easy to see.
However, I do not know where to begin with this.
Best Answer
If $M$ fails to be invertible, then you certainly are in trouble with the theorem. Consider the simplest example: $f(x,y) = x^2+y^2-1$ with $(a,b) = (1,0)$. In no neighborhood of that point is $f=0$ the graph of a function $y=g(x)$.
Now, for your question. If you take $f(x,y) = y^2 = 0$ with $(a,b)=(a,0)$ for any $a$, then we have $g(x)=0$ for all $x$ (which is nicely continuously differentiable) and yet $\partial f/\partial y = 0$ at every point of the subset. So $M$ can fail to be invertible. (This is a standard game to play. Instead of taking the "right" equation $y=0$, you take the "wrong" equation $y^2=0$. This phenomenon can be generalized as much as you want working over the real numbers.)