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Inverse and Implicit Function Theorems

Two cornerstone theorems. If the Jacobian is invertible, the map is locally invertible; and a smooth equation can be solved for some variables in terms of the rest whenever a sub-Jacobian is invertible.

Invertible derivative, invertible map

Linearization governs local behavior. If a C^1 map f: R^n -> R^n has an invertible derivative at a — that is, the Jacobian determinant is nonzero — then near a the map looks like its invertible linear part, and so it should itself be invertible nearby. This is the inverse function theorem.

Inverse function theorem.

Let f: R^n -> R^n be C^1 near a, and suppose Df(a) is invertible,
i.e.  det Df(a) != 0. Then there are open sets U about a and V about
f(a) such that f : U -> V is a bijection, the inverse g : V -> U is
C^1, and by the chain rule applied to g(f(x)) = x:

   Dg( f(a) ) = ( Df(a) )^{-1}.

Proof idea: solving f(x) = y is finding a fixed point of
   T(x) = x + Df(a)^{-1} ( y - f(x) ),
which is a [[contraction-mapping|contraction]] near a; the
[[banach-fixed-point-theorem|Banach fixed-point theorem]] gives a
unique solution, and it varies smoothly with y.
Local invertibility from a nonzero Jacobian determinant, via a contraction.

Solving equations implicitly

A level set F(x, y) = 0 usually defines y as a function of x even when you cannot solve it by formula. The implicit function theorem makes this precise: as long as the partial derivative with respect to the variables you want to solve for is invertible, a smooth local solution exists, and you can compute its derivative by differentiating the equation.

Implicit function theorem (scalar case).

Let F(x, y) be C^1 with F(a, b) = 0 and  dF/dy (a, b) != 0.
Then near a there is a unique C^1 function y = h(x) with h(a) = b
and F(x, h(x)) = 0. Differentiate that identity in x:

   dF/dx + (dF/dy) * h'(x) = 0   =>   h'(x) = - (dF/dx) / (dF/dy).

Example: the unit circle F(x, y) = x^2 + y^2 - 1 = 0 at (a, b)=(0, 1).
   dF/dy = 2y = 2 != 0, so y = sqrt(1 - x^2) exists locally, and
   h'(x) = - (2x) / (2y) = - x / y.   At (0,1): slope 0, as expected.

At (1, 0) instead dF/dy = 0 — and indeed near (1,0) the circle is
NOT a graph y = h(x); the hypothesis fails exactly where it must.
Solving F = 0 for y, and seeing the hypothesis fail at a vertical tangent.

Two sides of one coin

These theorems are equivalent in the sense that each follows quickly from the other. To get the implicit function theorem from the inverse function theorem, apply the latter to the map (x, y) -> (x, F(x, y)); its Jacobian is invertible exactly when dF/dy is, and the inverse hands you the solution y = h(x). Both rest on the same engine: a total derivative that is an invertible linear map controls the nonlinear map nearby.