Implicit Function Theorem
Let be a smooth map, and such that the first columns of are linearly independent (i.e. is of full rank). There there is a neighbourhood of in , and neighbourhood of in , a neighbourhood of in , and a smooth map such that for all points .
Proof Step 1: Construction of the Auxiliary Mapping. To establish the local parameterization of , we introduce an auxiliary augmented mapping defined by preserving the coordinates of while embedding the output of into the first component: Since is a smooth map on the open set , is inherently a smooth mapping from an open subset of into . Step 2: Verification of Local Invertibility. Let us evaluate the total derivative matrix (Jacobian matrix) of at the point . Expressing as a block matrix matching the coordinate decomposition yields:
\begin{bmatrix} \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\[1.5ex] \frac{\partial z}{\partial y} & \frac{\partial z}{\partial z} \end{bmatrix} = \begin{bmatrix} D_y f|_{(y_0, z_0)} & D_z f|_{(y_0, z_0)} \\[1.5ex] \mathbf{0}_{k \times n} & \mathbf{Id}_k \end{bmatrix}.$$ This block matrix exhibits an upper-triangular structure. Consequently, its determinant is simply the product of the determinants of its diagonal blocks: $$\det\left( DG|_{(y_0, z_0)} \right) = \det\left( D_y f|_{(y_0, z_0)} \right) \cdot \det(\mathbf{Id}_k) = \det\left( D_y f|_{(y_0, z_0)} \right).$$ By the hypothesis of the theorem, the first $n$ columns of $Df|_{x_0}$ are linearly independent, which explicitly means that the $n \times n$ matrix $D_y f|_{(y_0, z_0)}$ is of full rank and invertible. Hence, $\det\left( D_y f|_{(y_0, z_0)} \right) \neq 0$, implying that the full derivative $DG|_{(y_0, z_0)}$ is an invertible linear operator on $\mathbb{R}^{n+k}$. Step 3: Inverse Function Theorem. Since $G$ is a smooth mapping and its derivative at $x_0 = (y_0, z_0)$ is non-singular, the Inverse Function Theorem implies that $G$ is a local smooth diffeomorphism. Specifically, there exists an open neighborhood $C \subseteq U$ of $x_0 = (y_0, z_0)$ in $\mathbb{R}^{n+k}$ and an open neighborhood $V \subseteq \mathbb{R}^{n+k}$ of $G(y_0, z_0) = (f(y_0, z_0), z_0)$ such that the restricted map $$G\big|_{C}: C \to V$$ is a bijection with a smooth local inverse $G^{-1}: V \to C$. Step 4: Identification of the Smooth Map $F$. Because the second component of $G(y, z)$ is identically the coordinate vector $z$, the inverse mapping $G^{-1}$ must strictly preserve this component when maps back from $V$. Let $(w, z) \in V$ represent the coordinates in the target space, where $w = f(y, z) \in \mathbb{R}^n$. We can structurally express the local smooth inverse as: $$G^{-1}(w, z) = (\Phi(w, z), z),$$ where $\Phi: V \to \mathbb{R}^n$ is a smooth map representing the first $n$ component channels of $G^{-1}$. By shrinking the neighborhoods if necessary, we can choose an open neighborhood $B \subseteq \mathbb{R}^n$ of $f(y_0, z_0)$ and an open neighborhood $A \subseteq \mathbb{R}^k$ of $z_0$ such that their product satisfies $B \times A \subseteq V$. We then define our desired smooth map $F: B \times A \to \mathbb{R}^n$ by setting: $$F(z, w) := \Phi(w, z).$$ Note that the arguments are flipped here to strictly match the formulation $F: B \times A \to \mathbb{R}^n$ given in the theorem statement. Since $\Phi$ is smooth, $F$ is naturally a smooth map. Step 5: Final Algebraic Closure. For any point $(y, z) \in C$, its image under $G$ is given by definition as $G(y, z) = (f(y, z), z) \in V$. Applying the left-inverse identity $G^{-1}(G(y, z)) = (y, z)$, we compute: $$(y, z) = G^{-1}(f(y, z), z) = (\Phi(f(y, z), z), z).$$ By equating the first components of the vectors on both sides, we immediately obtain: $$y = \Phi(f(y, z), z).$$ Substituting our defined function $F(z, w) = \Phi(w, z)$ with $w = f(y, z)$ into the relation above yields: $$y = F(z, f(y, z)).$$ This identity holds strictly for all points $(y, z) \in C$, completing the proof. >[!theorem] The Rank Theorem > >Let $f : U \subset \mathbb{R}^m \to \mathbb{R}^n$ be a smooth map, such that the rank $k$ of the derivative $Df|_x$ is constant on $U$. Then for any $x \in U$ there exist neighbourhoods $A$ of $x$ in $\mathbb{R}^m$ and $B$ of $f(x)$ in $\mathbb{R}^n$, and diffeomorphisms $\varphi : A \to C \subset \mathbb{R}^m$ and $\eta : B \to D \subset \mathbb{R}^n$ such that >$$\eta \circ f \circ \varphi^{-1}(x^1, \dots, x^m) = (x^1, \dots, x^k, 0, \dots, 0)$$ >for all $(x^1, \dots, x^m) \in C$. *Proof* >[!theorem] Constant Rank Level Set Theorem > >Let $M$ and $N$ be smooth manifolds, and let $F: M \to N$ be a smooth map with constant rank $k$. Each level set of $F$ is a closed embedded submanifold of codimension $k$ in $M$.