\(f: \Omega \to \mathbb{R}^n\), continuously differentiable (\(C^1\)) on a neighborhood of \(a \in \Omega\) & \(\det(f'(a)) \neq 0\) (Jacobian).
Proof Sketch (Case 1: \(f'(a) = I_{n \times n}\)):
By restricting suitably, assume \(f\) is \(C^1\) on \(\Omega\). \(\det: M(n,\mathbb{R}) \to \mathbb{R}\) is a continuous map. \(f'(x) \in M(n,\mathbb{R})\) is \((D_j f_i(x))_{n \times n}\).
\(\{x \in \Omega \mid \det(f'(x)) \neq 0\}\) is open in \(\Omega\). Thus, if \(R\) is a sufficiently small open rectangle centered at \(a\), then on \(R\), \(\det(f'(x)) \neq 0\).
Claim: \(f'(a) = I \implies f(x) = f(a) + (x-a) + \Phi(x-a)\).
\(\lim_{x \to a} \frac{|\Phi(x-a)|}{|x-a|} = 0 \implies \lim_{x \to a} \frac{|f(x) - f(a) - I(x-a)|}{|x-a|} = 0\).
Injectivity: Consider the map \(F: \Omega \to \mathbb{R}^n\) given by \(x \mapsto f(x) - x\). \(F\) is \(C^1\).
\(D_j F_i(x) = D_j f_i(x) - \delta_{ij} \, \forall x \in \Omega\).
Hence, \(\exists\) an open rectangle \(R\) centered at \(a\) such that on \(\bar{R}\), \(|D_j F_i(x)| \le M\).
Then, we get \(|F(x) - F(y)| \le n M |x-y| \, \forall x,y \in \bar{R}\).
i.e., \(|f(x) - x - (f(y) - y)| \le n M |x-y|\).
\(\implies -|f(x) - f(y)| + |x-y| \le n M |x-y| \implies |f(x) - f(y)| \ge (1 - n M) |x-y| \, \forall x,y \in \bar{R}\).
If \(1-nM > 0\), then \(f(x) = f(y) \implies x = y\). So \(f\) is injective on \(R\). The image of \(\bar{R}\) is compact, and \(f\) acts as a homeomorphism onto its image. We can show that \(f^{-1}\) is also differentiable by bounding the error terms similarly.
Case 2: \(Df(a) = \lambda \neq I\):
Define \(\lambda^{-1} \circ f: \Omega \to \mathbb{R}^n\). Its derivative at \(a\) is \(\lambda^{-1} \circ \lambda = I\). By Case 1, it's locally invertible. Composing with \(\lambda\) gives the result for \(f\). ∎
Let \(f: \Omega \subset \mathbb{R}^n \to \mathbb{R}^n\) be open. \(f\) is called a diffeomorphism if \(f\) is differentiable with a differentiable inverse.
- \(f(x,y) = (\sqrt{x^2+y^2}, -\tan^{-1}(y/x))\) for \(\Omega = \{(x,y) \in \mathbb{R}^2 \mid x>0\}\). \(f\) is \(C^1\) & \(\det(f'(x,y)) \neq 0\). Verify that \(f(\Omega)\) is a diffeomorphism (Polar Coordinates).
- \(f: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}^2\), \(f(x,y) = (x^2-y^2, 2xy)\). \[ f'(x,y) = \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}, \quad \det(f'(x,y)) = 4(x^2+y^2) \neq 0 \] So \(f\) is \(C^1\). It corresponds to \(z \mapsto z^2\) in \(\mathbb{C}\). It is not globally injective (e.g., \(f(1,1) = f(-1,-1)\)), but if we restrict to \(\Omega_1 = \{(x,y) \in \mathbb{R}^2 \mid y > 0\}\), \(f\) is injective and forms a diffeomorphism.
Consider \(f: \mathbb{R}^2 \to \mathbb{R}\) (\(\Omega\) connected), \(f\) is \(C^1\). Suppose \(\frac{\partial f}{\partial y}(x_0,y_0) \neq 0\).
Define \(F: \mathbb{R}^2 \to \mathbb{R}^2\) by \(F(x,y) = (x, f(x,y))\). By IFT, \(F\) is a local diffeo near \((x_0, y_0)\). Then \(F(x_0, y_0) = (x_0, 0)\).
There exists an interval \(I\) such that \(F^{-1}(x,0) = (x, K(x,0))\). This gives us \(g(x) = K(x,0)\) such that \(f(x, g(x)) = 0\).
Let \(f: \Omega \subset \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m\) be \(C^1\). Let \((a,b) \in \Omega\), \(f(a,b) = 0\) & \(M = (D_{n+j} f_i(a,b))_{m \times m}\).
Suppose \(\det M \neq 0\), then \(\exists\) an open set \(A \subset \mathbb{R}^n\) containing \(a\), \(B \subset \mathbb{R}^m\) containing \(b\) & a differentiable function \(g: A \to B\) such that for \(x \in A\), \(g(x)\) is a unique element in \(B\) that satisfies \(f(x, g(x)) = 0\). \(g\) is said to be defined implicitly by \(f\).
Proof: We represent an element in \(\mathbb{R}^n \times \mathbb{R}^m\) as \((x,y)\).
Define \(F: \Omega \subset \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^n \times \mathbb{R}^m\) by \(F(x,y) = (x, f(x,y))\).
\(F(a,b) = (a, f(a,b)) = (a,0)\).
Then, \(\det(F'(a,b)) = \det(I)\det(M) = \det(M) \neq 0\).
Hence by Inverse Function Theorem, \(\exists\) a neighborhood \(W\) containing \((a,0)\) and a neighborhood \(A_1 \times B\) of \((a,b)\) such that \(F|_{A_1 \times B}: A_1 \times B \to W\) is a diffeomorphism.
Suppose \((x,y) \in A_1 \times B\) & \(f(x,y) = 0\). \(\implies F(x,y) = (x,0) \in W\).
Conversely, if \((x,0) \in W\), its preimage must be of the form \((x, y)\) with \(f(x,y)=0\).
Thus \(F^{-1}(x,0) = (x, K(x,0))\). Define \(g(x) = K(x,0)\) on a suitable open set \(A \subset A_1\). Then \(f(x, g(x)) = 0\). ∎
\(f(x,y) = x^2 - y^2\), \(f(0,0) = 0\).
Can one determine \(y\) in terms of \(x\) implicitly in the neighborhood of \(0\)?
\(\frac{\partial f}{\partial y}|_{(0,0)} = -2y|_{(0,0)} = 0\).
\(x^2 - y^2 = 0 \implies y^2 = x^2 \implies y = \pm x\).
Let \(g_1(x) = x, g_2(x) = -x\). Then \(f(x, g_1(x)) = 0\) but \(g_1(x) \neq g_2(x)\). A unique implicit function does not exist at \((0,0)\), which matches the theorem since \(\det M = 0\).