Inverse & Implicit Function Theorems

Let $U \subseteq \mathbb{R}^n$ be open. A $C^1$ map $f: U \to \mathbb{R}^n$ is a local diffeomorphism at $a \in U$ if there exist open sets $V \ni a$ and $W \ni f(a)$ such that $f|_V: V \to W$ is:

1. Bijective (one-to-one and onto),
2. $C^1$ (continuously differentiable), and
3. Its inverse $f|_V^{-1}: W \to V$ is also $C^1$.

If $f$ is a local diffeomorphism at every point of $U$, we call it a local diffeomorphism on $U$. If $f: U \to f(U)$ is a bijective local diffeomorphism, it is a (global) diffeomorphism.

The polar coordinate map $f: \mathbb{R}^2 \to \mathbb{R}^2$ defined by

$f(r, \theta) = (r\cos\theta,\; r\sin\theta)$

has Jacobian

$J_f(r, \theta) = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}$

with determinant $\det J_f = r\cos^2\theta + r\sin^2\theta = r$. This is non-zero when $r \neq 0$, so $f$ is a local diffeomorphism everywhere except at the origin. At $r = 0$, the Jacobian is singular — and geometrically, all angles $\theta$ map to the same point $(0, 0)$. The polar coordinate map collapses an entire line ($r = 0$, all $\theta$) to a single point, which is the geometric source of the singularity.

Restricting to $r > 0$ and $\theta \in (0, 2\pi)$, the map becomes a diffeomorphism onto $\mathbb{R}^2 \setminus \{(x, 0) : x \geq 0\}$. The restriction to $\theta \in (0, 2\pi)$ is necessary to prevent $\theta = 0$ and $\theta = 2\pi$ from mapping to the same point.

Define $f: \mathbb{R}^2 \to \mathbb{R}^2$ by

$f(x, y) = (e^x \cos y,\; e^x \sin y).$

The Jacobian is

$J_f(x, y) = \begin{pmatrix} e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y \end{pmatrix}$

with $\det J_f = e^{2x}(\cos^2 y + \sin^2 y) = e^{2x} > 0$ everywhere. The Jacobian is never singular — the IFT guarantees that $f$ is a local diffeomorphism at every point. Yet $f$ is not globally injective: $f(x, y) = f(x, y + 2\pi)$ for all $(x, y)$, because the complex exponential $e^{x+iy}$ is $2\pi i$-periodic. This is a clean example of the gap between “locally invertible everywhere” and “globally invertible.”

The distinction between local and global invertibility is not merely a mathematical subtlety — it drives architectural decisions in machine learning. The IFT only guarantees local invertibility: given a non-singular Jacobian at one point, the function has a smooth inverse in some neighborhood. Normalizing flows need global bijectivity (every input maps to a unique output over the entire domain) to define valid density transformations.

Architectures like Real-NVP and NICE achieve global bijectivity by construction: affine coupling layers $y_{1:d} = x_{1:d}$, $y_{d+1:n} = x_{d+1:n} \odot \exp(s(x_{1:d})) + t(x_{1:d})$ are globally invertible because the triangular structure makes the inverse explicit. The IFT is not needed for these architectures — but it is needed for residual flows $f(x) = x + g(x)$, where global invertibility requires additional conditions (e.g., Lipschitz constant of $g$ less than 1).

Let $U \subseteq \mathbb{R}^n$ be open and let $f: U \to \mathbb{R}^n$ be $C^1$ (continuously differentiable). Let $a \in U$ and $b = f(a)$. Suppose that the Jacobian matrix $J_f(a)$ is invertible (equivalently, $\det J_f(a) \neq 0$).

Then there exist open sets $V \subseteq U$ with $a \in V$ and $W \subseteq \mathbb{R}^n$ with $b \in W$ such that:

1. Bijection: $f: V \to W$ is a bijection.
2. Smooth inverse: The inverse function $g = f^{-1}: W \to V$ is $C^1$.
3. Derivative formula: For every $y \in W$,

$J_g(y) = [J_f(g(y))]^{-1}.$

In words: if the linear approximation to $f$ at $a$ is invertible, then $f$ itself is invertible in a neighborhood of $a$, with a $C^1$ inverse whose Jacobian is the matrix inverse of $J_f$ evaluated at the corresponding point.

We prove the IFT via the Contraction Mapping Theorem (Theorem 2 below). The strategy: reformulate the equation $f(x) = y$ as a fixed-point problem $T_y(x) = x$, show that $T_y$ is a contraction on a closed ball near $a$, and invoke completeness to obtain the unique fixed point $g(y) = f^{-1}(y)$.

Setup. Let $A = J_f(a)$. Since $\det A \neq 0$, the inverse $A^{-1}$ exists. Define the auxiliary map

$T_y(x) = x + A^{-1}(y - f(x)) = x - A^{-1}(f(x) - y).$

Observe that $T_y(x) = x$ if and only if $A^{-1}(y - f(x)) = 0$, which (since $A^{-1}$ is invertible) happens if and only if $f(x) = y$. So finding a fixed point of $T_y$ is equivalent to solving $f(x) = y$.

Step 1: $T_y$ is a contraction near $a$. The Jacobian of $T_y$ with respect to $x$ is

$DT_y(x) = I - A^{-1} J_f(x) = A^{-1}(A - J_f(x)).$

Since $J_f$ is continuous (the $C^1$ hypothesis) and $J_f(a) = A$, for any $\varepsilon > 0$ there exists $\delta > 0$ such that

$\|x - a\| < \delta \implies \|J_f(x) - A\| < \varepsilon.$

Choose $\varepsilon = \frac{1}{2\|A^{-1}\|}$. Then for $\|x - a\| < \delta$:

$\|DT_y(x)\| = \|A^{-1}(A - J_f(x))\| \leq \|A^{-1}\| \cdot \|A - J_f(x)\| < \|A^{-1}\| \cdot \frac{1}{2\|A^{-1}\|} = \frac{1}{2}.$

By the Mean Value Inequality (the multivariable Mean Value Theorem), for any $x_1, x_2 \in B(a, \delta)$:

$\|T_y(x_1) - T_y(x_2)\| \leq \sup_{x \in B(a, \delta)} \|DT_y(x)\| \cdot \|x_1 - x_2\| \leq \frac{1}{2}\|x_1 - x_2\|.$

So $T_y$ is a contraction with Lipschitz constant $\lambda = \frac{1}{2}$ on $B(a, \delta)$.

Step 2: $T_y$ maps $\overline{B}(a, r)$ to itself for $y$ near $b$. Fix $r \leq \delta$ (so the contraction estimate holds on $\overline{B}(a, r)$). For any $x \in \overline{B}(a, r)$:

$\|T_y(x) - a\| \leq \|T_y(x) - T_y(a)\| + \|T_y(a) - a\|.$

The first term is controlled by the contraction:

$\|T_y(x) - T_y(a)\| \leq \frac{1}{2}\|x - a\| \leq \frac{r}{2}.$

For the second term:

$\|T_y(a) - a\| = \|a + A^{-1}(y - f(a)) - a\| = \|A^{-1}(y - b)\| \leq \|A^{-1}\| \cdot \|y - b\|.$

Therefore:

$\|T_y(x) - a\| \leq \frac{r}{2} + \|A^{-1}\| \cdot \|y - b\|.$

This is $\leq r$ provided $\|y - b\| \leq \frac{r}{2\|A^{-1}\|}$. So let $W = B\!\left(b, \frac{r}{2\|A^{-1}\|}\right)$. For every $y \in W$, the map $T_y$ sends $\overline{B}(a, r)$ to itself.

Step 3: Fixed point existence and uniqueness. The closed ball $\overline{B}(a, r) \subset \mathbb{R}^n$ is a complete metric space (closed subsets of complete metric spaces are complete, and $\mathbb{R}^n$ is complete — this is where completeness enters the proof). The map $T_y: \overline{B}(a, r) \to \overline{B}(a, r)$ is a contraction with constant $\frac{1}{2}$.

By the Contraction Mapping Theorem (Theorem 2), $T_y$ has a unique fixed point $g(y) \in \overline{B}(a, r)$. Since $T_y(g(y)) = g(y)$ if and only if $f(g(y)) = y$, we have constructed a function $g: W \to V = B(a, r)$ such that $f(g(y)) = y$ for all $y \in W$.

Injectivity of $f$ on $V$: If $f(x_1) = f(x_2) = y$ for $x_1, x_2 \in V$, then both $x_1$ and $x_2$ are fixed points of $T_y$ on $\overline{B}(a, r)$. By uniqueness of the fixed point, $x_1 = x_2$. So $f|_V$ is injective.

Surjectivity onto $W$: For every $y \in W$, the fixed point $g(y)$ satisfies $f(g(y)) = y$, so $y \in f(V)$. Therefore $f(V) \supseteq W$. (We may shrink $V$ so that $f(V) = W$.)

This establishes part (1): $f: V \to W$ is a bijection with inverse $g$.

Step 4: $g$ is $C^1$. We show that $g$ is continuous, then differentiable, then $C^1$.

Continuity of $g$: For $y_1, y_2 \in W$, let $x_i = g(y_i)$, so $f(x_i) = y_i$. Then:

$\|x_1 - x_2\| = \|T_{y_1}(x_1) - T_{y_2}(x_2)\|$ $= \|(x_1 + A^{-1}(y_1 - f(x_1))) - (x_2 + A^{-1}(y_2 - f(x_2)))\|$ $= \|(x_1 - x_2) + A^{-1}((y_1 - y_2) - (f(x_1) - f(x_2)))\|$ $= \|T_{y_1}(x_1) - T_{y_1}(x_2) + A^{-1}(y_1 - y_2)\|$ $\leq \|T_{y_1}(x_1) - T_{y_1}(x_2)\| + \|A^{-1}\|\|y_1 - y_2\|$ $\leq \frac{1}{2}\|x_1 - x_2\| + \|A^{-1}\|\|y_1 - y_2\|.$

Rearranging: $\|g(y_1) - g(y_2)\| = \|x_1 - x_2\| \leq 2\|A^{-1}\|\|y_1 - y_2\|$. So $g$ is Lipschitz (hence continuous).

Differentiability: Since $f(g(y)) = y$ and $f$ is differentiable with invertible Jacobian $J_f(g(y))$ (invertibility persists in a neighborhood because $\det J_f$ is a continuous function that is nonzero at $a$, hence nonzero near $a$), we differentiate both sides using the chain rule:

$J_f(g(y)) \cdot J_g(y) = I.$

Solving: $J_g(y) = [J_f(g(y))]^{-1}$.

$C^1$ regularity: The map $y \mapsto J_g(y)$ is the composition

$y \xrightarrow{g} g(y) \xrightarrow{J_f} J_f(g(y)) \xrightarrow{\text{inv}} [J_f(g(y))]^{-1}.$

Each arrow is continuous: $g$ is continuous (shown above), $J_f$ is continuous (by the $C^1$ hypothesis on $f$), and matrix inversion $M \mapsto M^{-1}$ is continuous on $GL(n)$ (the set of invertible matrices). Therefore $J_g$ is continuous, which means $g$ is $C^1$.

$\square$

Let $(X, d)$ be a complete metric space and let $T: X \to X$ be a contraction: there exists $\lambda \in [0, 1)$ such that

$d(T(x), T(y)) \leq \lambda \, d(x, y) \quad \text{for all } x, y \in X.$

Then $T$ has a unique fixed point $x^* \in X$ (i.e., $T(x^*) = x^*$). Moreover, for any starting point $x_0 \in X$, the iterates $x_{k+1} = T(x_k)$ converge to $x^*$, with the convergence rate bound

$d(x_k, x^*) \leq \frac{\lambda^k}{1 - \lambda} \, d(x_0, x_1).$

Existence. Pick any $x_0 \in X$ and define $x_{k+1} = T(x_k)$. We show $(x_k)$ is Cauchy. By the contraction property:

$d(x_{k+1}, x_k) = d(T(x_k), T(x_{k-1})) \leq \lambda \, d(x_k, x_{k-1}) \leq \cdots \leq \lambda^k d(x_1, x_0).$

For $m > n$, the triangle inequality gives:

$d(x_n, x_m) \leq \sum_{k=n}^{m-1} d(x_k, x_{k+1}) \leq \sum_{k=n}^{m-1} \lambda^k d(x_0, x_1) \leq \frac{\lambda^n}{1 - \lambda}\, d(x_0, x_1).$

Since $0 \leq \lambda < 1$, we have $\lambda^n \to 0$ as $n \to \infty$. For any $\varepsilon > 0$, choose $N$ such that $\frac{\lambda^N}{1-\lambda} d(x_0, x_1) < \varepsilon$. Then for all $m > n \geq N$, $d(x_n, x_m) < \varepsilon$. So $(x_k)$ is Cauchy.

Since $X$ is complete, the Cauchy sequence $(x_k)$ converges to some $x^* \in X$.

$x^*$ is a fixed point. Since $T$ is a contraction, it is Lipschitz, hence continuous. Therefore:

$T(x^*) = T\!\left(\lim_{k \to \infty} x_k\right) = \lim_{k \to \infty} T(x_k) = \lim_{k \to \infty} x_{k+1} = x^*.$

Uniqueness. Suppose $T(x^*) = x^*$ and $T(y^*) = y^*$. Then:

$d(x^*, y^*) = d(T(x^*), T(y^*)) \leq \lambda \, d(x^*, y^*).$

Since $\lambda < 1$, this implies $(1 - \lambda) d(x^*, y^*) \leq 0$, so $d(x^*, y^*) = 0$, hence $x^* = y^*$.

Rate bound. Letting $m \to \infty$ in the estimate $d(x_n, x_m) \leq \frac{\lambda^n}{1-\lambda} d(x_0, x_1)$:

$d(x_n, x^*) = \lim_{m \to \infty} d(x_n, x_m) \leq \frac{\lambda^n}{1 - \lambda}\, d(x_0, x_1).$

$\square$

Define $f: \mathbb{R}^2 \to \mathbb{R}^2$ by $f(x, y) = (x^2 - y^2,\; 2xy)$ — this is the real-variable form of the complex map $z \mapsto z^2$, where $z = x + iy$.

The Jacobian is:

$J_f(x, y) = \begin{pmatrix} 2x & -2y \\ 2y & 2x \end{pmatrix}, \qquad \det J_f = 4x^2 + 4y^2 = 4|z|^2.$

At $a = (1, 1)$: $\det J_f(1, 1) = 8 \neq 0$, so the IFT guarantees a local inverse near $b = f(1, 1) = (0, 2)$.

The inverse Jacobian is:

$[J_f(1,1)]^{-1} = \frac{1}{8}\begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} 1/4 & 1/4 \\ -1/4 & 1/4 \end{pmatrix}.$

This tells us that a small perturbation $\Delta y = (\Delta u, \Delta v)$ near $b = (0, 2)$ produces a perturbation $\Delta x \approx [J_f(1,1)]^{-1} \Delta y$ near $a = (1,1)$.

The Jacobian is singular when $|z| = 0$, i.e., at the origin — exactly where the complex squaring map $z^2$ has a branch point. The IFT fails at the origin, and indeed $z^2$ is two-to-one in every neighborhood of $0$ (except at $0$ itself).

The IFT as stated gives a $C^1$ inverse when $f$ is $C^1$. But the regularity transfers: if $f$ is $C^k$ (all partial derivatives up to order $k$ exist and are continuous), then the local inverse $g$ is also $C^k$. If $f$ is $C^\infty$ (smooth), so is $g$. If $f$ is real-analytic, so is $g$.

The key to the induction is the derivative formula $J_g(y) = [J_f(g(y))]^{-1}$. If $g$ is $C^{k-1}$ and $J_f$ is $C^{k-1}$ (which follows from $f$ being $C^k$), then $J_g$ is a composition of $C^{k-1}$ maps with matrix inversion, hence $C^{k-1}$. But $J_g$ being $C^{k-1}$ means $g$ is $C^k$.

Let $f: V \to W$ be a $C^1$ diffeomorphism with inverse $g = f^{-1}: W \to V$. Then for every $y \in W$:

$J_g(y) = [J_f(g(y))]^{-1}.$

In components: if $f = (f_1, \ldots, f_n)$ and $g = (g_1, \ldots, g_n)$, then

$\frac{\partial g_i}{\partial y_j}(y) = \left([J_f(g(y))]^{-1}\right)_{ij}.$

In the scalar case $n = 1$, this reduces to the familiar formula $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$.

Differentiate $f(g(y)) = y$ with respect to $y$ using the chain rule:

$J_f(g(y)) \cdot J_g(y) = I_n.$

Since $J_f(g(y))$ is invertible (invertibility persists in a neighborhood of $a$ by continuity of $\det J_f$), multiply both sides on the left by $[J_f(g(y))]^{-1}$:

$J_g(y) = [J_f(g(y))]^{-1}.$

$\square$

Polar coordinates: $f(r, \theta) = (r\cos\theta, r\sin\theta)$ with inverse $g(x, y) = \left(\sqrt{x^2 + y^2},\; \arctan(y/x)\right)$ (for $x > 0$).

Via the formula: At the point $(r, \theta) = (2, \pi/6)$, so $(x, y) = (\sqrt{3}, 1)$:

$J_f(2, \pi/6) = \begin{pmatrix} \cos(\pi/6) & -2\sin(\pi/6) \\ \sin(\pi/6) & 2\cos(\pi/6) \end{pmatrix} = \begin{pmatrix} \sqrt{3}/2 & -1 \\ 1/2 & \sqrt{3} \end{pmatrix}.$

The determinant is $r = 2$. So:

$J_g(\sqrt{3}, 1) = [J_f(2, \pi/6)]^{-1} = \frac{1}{2}\begin{pmatrix} \sqrt{3} & 1 \\ -1/2 & \sqrt{3}/2 \end{pmatrix} = \begin{pmatrix} \sqrt{3}/2 & 1/2 \\ -1/4 & \sqrt{3}/4 \end{pmatrix}.$

Direct computation (verification):

$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{\sqrt{3}}{2}, \quad \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} = \frac{1}{2},$

$\frac{\partial \theta}{\partial x} = \frac{-y}{x^2+y^2} = \frac{-1}{4}, \quad \frac{\partial \theta}{\partial y} = \frac{x}{x^2+y^2} = \frac{\sqrt{3}}{4}.$

The results agree. The formula $J_g = [J_f(g(y))]^{-1}$ saves us the trouble of computing the inverse function and differentiating it directly — which in higher dimensions may not be feasible.

An affine coupling layer in Real-NVP splits the input $z = (z_a, z_b) \in \mathbb{R}^d \times \mathbb{R}^{d'}$ and defines:

$f(z_a, z_b) = (z_a,\; z_b \odot \exp(s(z_a)) + t(z_a))$

where $s, t: \mathbb{R}^d \to \mathbb{R}^{d'}$ are neural networks and $\odot$ is elementwise multiplication. The Jacobian is block-triangular:

$J_f = \begin{pmatrix} I_d & 0 \\ \frac{\partial}{\partial z_a}[z_b \odot e^{s(z_a)} + t(z_a)] & \operatorname{diag}(e^{s(z_a)}) \end{pmatrix}.$

The determinant is $\det J_f = \prod_{j=1}^{d'} \exp(s_j(z_a)) = \exp\!\left(\sum_j s_j(z_a)\right)$, which is always positive — so the IFT is satisfied everywhere. The inverse is explicit: $g(y_a, y_b) = (y_a,\; (y_b - t(y_a)) \odot \exp(-s(y_a)))$. The inverse Jacobian is:

$J_g = [J_f]^{-1} = \begin{pmatrix} I_d & 0 \\ * & \operatorname{diag}(e^{-s(y_a)}) \end{pmatrix}$

with $\log|\det J_g| = -\sum_j s_j(y_a)$. The entire normalizing flow density computation reduces to evaluating the $s$ network — this is why coupling layers are popular.

Let $F: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m$ be a $C^1$ map. An implicit equation is the system $F(x, y) = 0$, where $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. The level set (or solution set) is

$S = \{(x, y) \in \mathbb{R}^{n+m} : F(x, y) = 0\}.$

We say $y = g(x)$ is an implicit function defined by $F$ near a point $(a, b)$ if $g$ is a $C^1$ map defined on a neighborhood of $a$ such that $F(x, g(x)) = 0$ for all $x$ in that neighborhood, and $g(a) = b$.

Let $F: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m$ be $C^1$. At a point $(a, b)$, the partial Jacobian with respect to $x$ is the $m \times n$ matrix

$D_x F(a,b) = \left(\frac{\partial F_i}{\partial x_j}(a,b)\right)_{\substack{1 \leq i \leq m \\ 1 \leq j \leq n}}$

and the partial Jacobian with respect to $y$ is the $m \times m$ matrix

$D_y F(a,b) = \left(\frac{\partial F_i}{\partial y_k}(a,b)\right)_{\substack{1 \leq i \leq m \\ 1 \leq k \leq m}}.$

The full Jacobian of $F$ viewed as a map $\mathbb{R}^{n+m} \to \mathbb{R}^m$ is the $m \times (n+m)$ block matrix $J_F(a,b) = \begin{pmatrix} D_x F(a,b) & D_y F(a,b) \end{pmatrix}$.

Let $F: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^m$ be $C^1$ in a neighborhood of $(a, b) \in \mathbb{R}^{n+m}$. Suppose:

• $F(a, b) = 0$, and
• the partial Jacobian $D_y F(a, b)$ is invertible ($\det D_y F(a, b) \neq 0$).

Then there exist open sets $U \ni a$ in $\mathbb{R}^n$ and $V \ni b$ in $\mathbb{R}^m$ such that:

1. Existence and uniqueness: For every $x \in U$, there is a unique $y = g(x) \in V$ with $F(x, g(x)) = 0$.
2. Smoothness: The implicit function $g: U \to V$ is $C^1$.
3. Derivative formula:

$J_g(x) = -[D_y F(x, g(x))]^{-1} \, D_x F(x, g(x)).$

We derive the ImFT from the IFT by a standard augmentation trick.

Step 1: Construct an auxiliary map. Define $\Phi: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^n \times \mathbb{R}^m$ by

$\Phi(x, y) = (x, \, F(x, y)).$

So $\Phi$ keeps $x$ unchanged and applies $F$ to get the second component. Note $\Phi(a, b) = (a, F(a,b)) = (a, 0)$.

Step 2: Compute the Jacobian of $\Phi$. The Jacobian of $\Phi$ at $(a, b)$ is the $(n+m) \times (n+m)$ block matrix:

$J_\Phi(a, b) = \begin{pmatrix} I_n & 0 \\ D_x F(a,b) & D_y F(a,b) \end{pmatrix}.$

The determinant of this block-triangular matrix is:

$\det J_\Phi(a, b) = \det I_n \cdot \det D_y F(a, b) = \det D_y F(a, b) \neq 0$

by hypothesis. So $J_\Phi(a, b)$ is invertible.

Step 3: Apply the IFT to $\Phi$. By the Inverse Function Theorem (Theorem 1), $\Phi$ is locally invertible near $(a, b)$: there exist neighborhoods $\widetilde{V} \ni (a, b)$ and $\widetilde{W} \ni (a, 0)$ and a $C^1$ inverse $\Psi = \Phi^{-1}: \widetilde{W} \to \widetilde{V}$.

Since $\Phi(x, y) = (x, F(x, y))$, the first component of $\Phi$ is the identity on $x$. Therefore the inverse $\Psi$ has the form $\Psi(x, w) = (x, \psi(x, w))$ for some $C^1$ function $\psi$, because the first component must invert the identity.

Step 4: Extract $g$. Set $w = 0$: the equation $\Phi(x, y) = (x, 0)$ means $F(x, y) = 0$. Its unique solution near $(a, b)$ is $(x, y) = \Psi(x, 0) = (x, \psi(x, 0))$. Define $g(x) = \psi(x, 0)$. Then:

• $g(a) = \psi(a, 0) = b$ (since $\Psi(a, 0) = (a, b)$),
• $F(x, g(x)) = 0$ for all $x$ near $a$,
• $g$ is $C^1$ (as a restriction of the $C^1$ map $\psi$).

Step 5: Derive the derivative formula. Differentiate $F(x, g(x)) = 0$ with respect to $x$ using the chain rule:

$D_x F(x, g(x)) + D_y F(x, g(x)) \cdot J_g(x) = 0.$

Solving for $J_g(x)$ (using the invertibility of $D_y F$):

$J_g(x) = -[D_y F(x, g(x))]^{-1}\, D_x F(x, g(x)).$

$\square$

Let $F(x, y) = x^2 + y^2 - 1$. The level set $F = 0$ is the unit circle $S^1$.

We have $F_x = 2x$ and $F_y = 2y$. The ImFT applies at any $(a, b) \in S^1$ where $F_y(a, b) = 2b \neq 0$, i.e., where $b \neq 0$.

Near $(0, 1)$: $F_y(0, 1) = 2 \neq 0$. The ImFT gives $g(x) = \sqrt{1 - x^2}$ (the upper semicircle) with

$g'(x) = -\frac{F_x}{F_y} = -\frac{2x}{2\sqrt{1-x^2}} = -\frac{x}{\sqrt{1-x^2}}.$

Near $(0, -1)$: $F_y(0, -1) = -2 \neq 0$. The ImFT gives $g(x) = -\sqrt{1 - x^2}$ (the lower semicircle).

At $(\pm 1, 0)$: $F_y = 0$. The ImFT does not apply, and indeed the circle has vertical tangent lines there — $y$ cannot be expressed as a function of $x$ near these points. However, $F_x(\pm 1, 0) = \pm 2 \neq 0$, so by applying the ImFT with the roles of $x$ and $y$ reversed, we can solve for $x = h(y)$ near these points.