I'm proving that a $C^r$ function $f:\mathbb{R^2}\rightarrow \mathbb{R}$ is not injective. I've seen various solutions but I want to prove it using techniques related to the Inverse function theorem; because it's a problem from Spivaks "Callculus on Manifolds" text in the chapter on the Inverse function theorem.
I've found a function $g:\mathbb{R^2}\rightarrow \mathbb{R^2}$ defined as $g(x,y)=\bigg(f(x,y),y\bigg)$, and proven that it's $C^r$. To prove my main problem, I'm assuming $f$ is injective, and that allows me to apply the inverse function theorem to $g$, in such a way that gives me an open map from an open set in the domain of $g$ to the open image of $g$, such that $g$ has a differentiable inverse $g^{-1}:g(A)\rightarrow A$.
Now I want to prove a contradiction in the injectivity of $f$, by showing that $g^{-1}$ is not defined at certain points, specifically a line, and this contradicts the differentiability of $g^{-1}$. Consider $f(x,y)=b$ then $g(x,y)=(b,y)$, Now I should be able to find that the line that $g^{-1}$ isn't defined on $\forall (b,z)$, $z\neq y$, since that would imply that $f(x,z)=b$.
My Question I don't quite see how $g^{-1}$ being defined on $(b,z)$ implies that $f(x,z)=b$? $\phantom{}$
Here is the original problem. I've seen other solutions suggest to make $g$ satisfy the conditions of $f$ from $2$-$36$, which is why I left it in.
Thanks!
What does it mean for $$ g^{-1}(b, z) = (u, v)? $$ It means that $$ g(u, v) = (b, z). $$ On the other hand, we know that $$ g(u, v) = (f(u, v), v) $$ so setting these two things equal, we see that $$ f(u, v) = b\\ v = z $$ from which we see that $$ f(u, z) = b. $$ I'm not sure if this was what was confusing you, but it seemed to be.