FUNCTION INVERSE THEOREM IMPLIES IMPLICIT FUNCTION THEOREM
EUGENE LERMAN
We show that the implicit function theorem can be deduced from the inverse function theorem. We consider only the case of f : R2 R. The proof in higher dimensions is analogous. The only real dierence is a more cumbersome notation. We assume that there is a point (a, b) R2 so that f (a, b) = 0. We want to show y that there is a neighborhood U of a in R, V of b in R and a function g : U V so that f (x, g(x)) = c for x U . Here, c = f (a, b). Proof. Consider the map H(x, y) := (x, f (x, y)). It is a map from R to R . We will argue that it is invertible near (a, b). Indeed, its dierential DH(a, b) is 1 0 f f (a, b) y (a, b) x
2 2
Hence det DH(a, b) = 1 f (a, b), is which not zero by assumption. Hence DH(a, b) is y invertible. Hence, by the inverse function theorem, the map H is invertible near (a, b). Denote the inverse by G. It is of the form G(u, v) = (G1 (u, v), G2 (u, v)) for some real-valued functions G2 , G1 dened on a neighborhood of H(a, b) = (a, f (a, b)) = (a, c) in R2 . Since G and H are inverses of each other, (u, v) = H(G(u, v)) = H(G1 (u, v), G2 (u, v)) = (G1 (u, v), f (G1 (u, v), G2 (u, ...