[proofplan] We reduce the Implicit Function Theorem to the [Inverse Function Theorem](/theorems/51). Define an auxiliary map $\Psi: U \to \mathbb{R}^{n+m}$ by $\Psi(x,y) = (x, F(x,y))$, whose Jacobian at $(x_0, y_0)$ is block-upper-triangular with diagonal blocks $I_n$ and $D_y F_{(x_0, y_0)}$, hence invertible. The Inverse Function Theorem provides a local $C^1$ inverse $\Psi^{-1}$; since $\Psi$ fixes the $x$-coordinates, the second component of $\Psi^{-1}(x, 0)$ yields the desired implicit function $g$. [/proofplan]

[step:Define the auxiliary map $\Psi: U \to \mathbb{R}^{n+m}$ that embeds $F$ into an invertible setting]Define the map \begin{align*} \Psi: U \subseteq \mathbb{R}^{n+m} &\to \mathbb{R}^{n+m} \\ (x, y) &\mapsto (x, F(x, y)). \end{align*} Since $F: U \to \mathbb{R}^m$ is $C^1$ by hypothesis and the projection $(x,y) \mapsto x$ is smooth, the map $\Psi$ is $C^1$.[/step]

[guided]The idea is to reformulate the equation $F(x, y) = 0$ as a fixed-point problem for an invertible map. The equation $F(x, y) = 0$ constrains $m$ of the $n + m$ variables, so we expect the solution set to be locally an $n$-dimensional graph. To make this precise, we need a map between $(n+m)$-dimensional spaces that we can invert. The natural choice is to keep the "free" variables $x$ and replace $y$ with the constraint $F(x,y)$: define \begin{align*} \Psi: U \subseteq \mathbb{R}^{n+m} &\to \mathbb{R}^{n+m} \\ (x, y) &\mapsto (x, F(x, y)). \end{align*} Since $F: U \to \mathbb{R}^m$ is $C^1$ by hypothesis and the coordinate projection $(x, y) \mapsto x$ is smooth, the composite map $\Psi$ is $C^1$. Notice that $\Psi(x_0, y_0) = (x_0, F(x_0, y_0)) = (x_0, 0)$ by the hypothesis $F(x_0, y_0) = 0$. The level set $\{F = 0\}$ near $(x_0, y_0)$ corresponds under $\Psi$ to points of the form $(x, 0)$, which is why this auxiliary map converts the implicit function problem into an inversion problem.[/guided]

[step:Compute $D\Psi_{(x_0, y_0)}$ and verify it is invertible via its block-triangular structure]Write the variables in $\mathbb{R}^{n+m}$ as $(x, y)$ with $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. The [total derivative](/pages/1) of $\Psi$ at $(x_0, y_0)$ is the linear map $D\Psi_{(x_0, y_0)}: \mathbb{R}^{n+m} \to \mathbb{R}^{n+m}$ whose Jacobian matrix has the block form \begin{align*} J\Psi_{(x_0, y_0)} = \begin{pmatrix} I_n & 0 \\ D_x F_{(x_0, y_0)} & D_y F_{(x_0, y_0)} \end{pmatrix} \in \mathbb{R}^{(n+m) \times (n+m)}, \end{align*} where $D_x F_{(x_0, y_0)}: \mathbb{R}^n \to \mathbb{R}^m$ denotes the partial derivative of $F$ with respect to the first $n$ variables and $D_y F_{(x_0, y_0)}: \mathbb{R}^m \to \mathbb{R}^m$ denotes the partial derivative with respect to the last $m$ variables. This matrix is block-lower-triangular. Its determinant factors as \begin{align*} \det J\Psi_{(x_0, y_0)} = \det(I_n) \cdot \det(J(D_y F)_{(x_0, y_0)}) = \det(J(D_y F)_{(x_0, y_0)}) \neq 0, \end{align*} since $D_y F_{(x_0, y_0)}$ is invertible by hypothesis. Therefore $D\Psi_{(x_0, y_0)}$ is invertible.[/step]

[guided]To apply the [Inverse Function Theorem](/theorems/51), we must verify that $D\Psi_{(x_0, y_0)}$ is invertible. The first component of $\Psi$ is the identity on $x$, and the second component is $F(x, y)$. How does this affect the Jacobian? The partial derivative of the first component $(x, y) \mapsto x$ with respect to $x$ is $I_n$, and with respect to $y$ is $0$. The partial derivative of the second component $(x, y) \mapsto F(x, y)$ with respect to $x$ is $D_x F_{(x_0, y_0)}$ and with respect to $y$ is $D_y F_{(x_0, y_0)}$. Assembling these into the Jacobian matrix of $\Psi$ at $(x_0, y_0)$: \begin{align*} J\Psi_{(x_0, y_0)} = \begin{pmatrix} I_n & 0 \\ D_x F_{(x_0, y_0)} & D_y F_{(x_0, y_0)} \end{pmatrix} \in \mathbb{R}^{(n+m) \times (n+m)}. \end{align*} This is a block-lower-triangular matrix. The determinant of a block-triangular matrix is the product of the determinants of its diagonal blocks: \begin{align*} \det J\Psi_{(x_0, y_0)} = \det(I_n) \cdot \det(J(D_y F)_{(x_0, y_0)}) = 1 \cdot \det(J(D_y F)_{(x_0, y_0)}). \end{align*} The hypothesis that $D_y F_{(x_0, y_0)}: \mathbb{R}^m \to \mathbb{R}^m$ is invertible means precisely that the Jacobian matrix $J(D_y F)_{(x_0, y_0)}$ is nonsingular, so $\det(J(D_y F)_{(x_0, y_0)}) \neq 0$. This is the step where the invertibility hypothesis on $D_y F$ is consumed. Without it, $D\Psi_{(x_0, y_0)}$ could be singular and the Inverse Function Theorem would not apply.[/guided]

[step:Apply the Inverse Function Theorem to $\Psi$ to obtain a local $C^1$ inverse]The map $\Psi: U \to \mathbb{R}^{n+m}$ is $C^1$ and $D\Psi_{(x_0, y_0)}$ is invertible. By the [Inverse Function Theorem](/theorems/51), there exist an open neighborhood $A$ of $(x_0, y_0)$ in $\mathbb{R}^{n+m}$ with $A \subseteq U$ and an [open](/pages/1144) neighborhood $B$ of $\Psi(x_0, y_0) = (x_0, 0)$ in $\mathbb{R}^{n+m}$ such that $\Psi|_A: A \to B$ is a $C^1$ diffeomorphism. Denote the inverse by \begin{align*} \Phi := (\Psi|_A)^{-1}: B \to A, \end{align*} which is also $C^1$.[/step]

[guided]We now verify the hypotheses of the [Inverse Function Theorem](/theorems/51). The theorem requires: (i) $\Psi$ is defined on an open set $U \subseteq \mathbb{R}^{n+m}$, which holds by hypothesis; (ii) $\Psi$ is $C^1$, which we established in the first step; (iii) $D\Psi_{(x_0, y_0)}$ is invertible, which we verified in the second step. All three conditions are satisfied. The Inverse Function Theorem therefore provides open sets $A \ni (x_0, y_0)$ and $B \ni \Psi(x_0, y_0) = (x_0, 0)$ with $A \subseteq U$ such that $\Psi|_A: A \to B$ is a bijection, and both $\Psi|_A$ and its inverse \begin{align*} \Phi := (\Psi|_A)^{-1}: B \to A \end{align*} are $C^1$. The key point: within $A$, the map $\Psi$ is injective, so for each $(x, z) \in B$ there is a unique $(x', y) \in A$ with $\Psi(x', y) = (x, z)$. Since the first component of $\Psi$ is the identity on $x$, this forces $x' = x$, meaning $\Phi$ has the form $\Phi(x, z) = (x, \varphi(x, z))$ for some $C^1$ map $\varphi$.[/guided]

[step:Extract the implicit function $g$ from the second component of $\Phi$ restricted to $z = 0$]Write $\Phi(x, z) = (\Phi_1(x, z), \Phi_2(x, z))$ with $\Phi_1: B \to \mathbb{R}^n$ and $\Phi_2: B \to \mathbb{R}^m$. Since $\Psi(x, y) = (x, F(x, y))$, the identity $\Psi(\Phi(x, z)) = (x, z)$ gives $\Phi_1(x, z) = x$ for all $(x, z) \in B$. In particular, $\Phi$ preserves the $x$-coordinate. Since $B$ is open and contains $(x_0, 0)$, there exist open sets $V \subseteq \mathbb{R}^n$ containing $x_0$ and $W_0 \subseteq \mathbb{R}^m$ containing $0$ with $V \times W_0 \subseteq B$. Define \begin{align*} g: V &\to \mathbb{R}^m \\ x &\mapsto \Phi_2(x, 0). \end{align*} The map $g$ is $C^1$ as the composition of $C^1$ maps (the inclusion $x \mapsto (x, 0)$ and $\Phi_2$). Set $W := g(V') \cap \pi_y(A)$ where $\pi_y: \mathbb{R}^{n+m} \to \mathbb{R}^m$ is the projection onto the last $m$ coordinates and $V'$ is chosen small enough so that $W$ is an open neighborhood of $y_0$. After shrinking $V$ if necessary, we may assume $V \times W \subseteq U$.[/step]

[guided]We now extract the implicit function. Write $\Phi = (\Phi_1, \Phi_2)$ with $\Phi_1: B \to \mathbb{R}^n$ and $\Phi_2: B \to \mathbb{R}^m$. The defining relation $\Psi(\Phi(x, z)) = (x, z)$ means \begin{align*} (\Phi_1(x, z),\; F(\Phi_1(x, z), \Phi_2(x, z))) = (x, z). \end{align*} Comparing first components gives $\Phi_1(x, z) = x$. Comparing second components gives $F(x, \Phi_2(x, z)) = z$. Setting $z = 0$: \begin{align*} F(x, \Phi_2(x, 0)) = 0 \quad \text{for all } x \text{ such that } (x, 0) \in B. \end{align*} Since $B$ is open in $\mathbb{R}^{n+m}$ and contains $(x_0, 0)$, we can find open sets $V \subseteq \mathbb{R}^n$ containing $x_0$ and $W_0 \subseteq \mathbb{R}^m$ containing $0$ with $V \times W_0 \subseteq B$. Define $g: V \to \mathbb{R}^m$ by $g(x) = \Phi_2(x, 0)$. This map is $C^1$ because $\Phi_2$ is $C^1$ (as a component of the $C^1$ inverse $\Phi$) and $(x, 0)$ is a $C^1$ function of $x$. After shrinking $V$ if necessary, let $W$ be an open neighborhood of $y_0$ such that $g(V) \subseteq W$ and $V \times W \subseteq U$. Such a $W$ exists because $g(x_0) = \Phi_2(x_0, 0) = y_0$ (since $\Phi(x_0, 0) = \Psi^{-1}(x_0, 0) = (x_0, y_0)$) and $g$ is [continuous](/pages/1147).[/guided]

[step:Verify uniqueness: for each $x \in V$, the point $y = g(x)$ is the only solution to $F(x, y) = 0$ in $W$]Fix $x \in V$ and suppose $y \in W$ satisfies $F(x, y) = 0$. Then $\Psi(x, y) = (x, 0)$. Since $(x, y) \in V \times W \subseteq A$ and $(x, 0) \in V \times W_0 \subseteq B$, and $\Psi|_A: A \to B$ is injective, we conclude $(x, y) = \Phi(x, 0) = (x, g(x))$, so $y = g(x)$.[/step]

[guided]Fix $x \in V$ and suppose $y \in W$ satisfies $F(x, y) = 0$. We must show $y = g(x)$, i.e., that no other solution exists in $W$. By definition of $\Psi$, we have \begin{align*} \Psi(x, y) = (x,\; F(x, y)) = (x,\; 0). \end{align*} For the injectivity of $\Psi|_A$ to apply, we need $(x, y) \in A$. This holds because $(x, y) \in V \times W \subseteq A$ by our choice of $V$ and $W$. We also need the image $(x, 0) \in B$: this holds because $(x, 0) \in V \times W_0 \subseteq B$. Now consider the two points $(x, y)$ and $(x, g(x))$. Both lie in $A$ (since $g(x) \in W$ by construction), and both map under $\Psi$ to the same target: \begin{align*} \Psi(x, y) = (x, 0) \quad \text{and} \quad \Psi(x, g(x)) = (x,\; F(x, g(x))) = (x, 0), \end{align*} where the second equality uses $F(x, g(x)) = 0$ established in the extraction step. Since $\Psi|_A: A \to B$ is a bijection (provided by the Inverse Function Theorem), the preimage of $(x, 0)$ under $\Psi|_A$ is unique. Therefore $(x, y) = (x, g(x))$, which gives $y = g(x)$. This is the step where the local injectivity from the Inverse Function Theorem converts into uniqueness of the implicit function.[/guided]

[step:Confirm all conclusions: $g(x_0) = y_0$, $F(x, g(x)) = 0$, and $g$ is $C^1$] We verify the three conclusions of the theorem: 1. **Initial condition.** $g(x_0) = \Phi_2(x_0, 0) = y_0$, since $\Phi(x_0, 0) = \Psi^{-1}(x_0, 0) = (x_0, y_0)$ (using $\Psi(x_0, y_0) = (x_0, 0)$). 2. **Implicit equation.** For every $x \in V$, we have $F(x, g(x)) = 0$ from the identity $F(x, \Phi_2(x, 0)) = 0$ established in the extraction step. 3. **Regularity.** The map $g: V \to W$ is $C^1$ because it is a component of the $C^1$ diffeomorphism $\Phi$ composed with the smooth inclusion $x \mapsto (x, 0)$. This completes the proof. [/step]