Step 1: Construct auxiliary map. Let $A = Jf(\mathbf{a})$, which is invertible. For fixed $\mathbf{y} \in \mathbb{R}^n$, define

$T_{\mathbf{y}}(\mathbf{x}) = \mathbf{x} + A^{-1}(\mathbf{y} - f(\mathbf{x})).$

Note: $T_{\mathbf{y}}(\mathbf{x}) = \mathbf{x}$ iff $f(\mathbf{x}) = \mathbf{y}$. So solving $f(\mathbf{x}) = \mathbf{y}$ is equivalent to finding a fixed point of $T_{\mathbf{y}}$.

Step 2: $T_{\mathbf{y}}$ is a contraction on a ball around $\mathbf{a}$. The Jacobian of $T_{\mathbf{y}}$ is

$JT_{\mathbf{y}}(\mathbf{x}) = I - A^{-1} Jf(\mathbf{x}).$

At $\mathbf{x} = \mathbf{a}$, $JT_{\mathbf{y}}(\mathbf{a}) = I - A^{-1} A = 0$. Since $Jf$ is continuous, there exists $r > 0$ such that for $\mathbf{x} \in B_r(\mathbf{a})$,

$\|JT_{\mathbf{y}}(\mathbf{x})\| \leq 1/2.$

By the Mean Value Theorem, for $\mathbf{x}_1, \mathbf{x}_2 \in B_r(\mathbf{a})$,

$\|T_{\mathbf{y}}(\mathbf{x}_1) - T_{\mathbf{y}}(\mathbf{x}_2)\| \leq \frac{1}{2} \|\mathbf{x}_1 - \mathbf{x}_2\|.$

So $T_{\mathbf{y}}$ is a contraction on $B_r(\mathbf{a})$ (if we ensure $T_{\mathbf{y}}(B_r(\mathbf{a})) \subseteq B_r(\mathbf{a})$, which holds for $\mathbf{y}$ close to $f(\mathbf{a})$).

Step 3: Existence of local inverse. For $\mathbf{y}$ sufficiently close to $f(\mathbf{a})$, the Banach Fixed-Point Theorem guarantees a unique $\mathbf{x} \in B_r(\mathbf{a})$ with $T_{\mathbf{y}}(\mathbf{x}) = \mathbf{x}$, i.e., $f(\mathbf{x}) = \mathbf{y}$. Define $g(\mathbf{y}) = \mathbf{x}$. Then $f(g(\mathbf{y})) = \mathbf{y}$ for all $\mathbf{y}$ in some neighborhood $W$ of $f(\mathbf{a})$.

Step 4: $g$ is continuous. The fixed point of a contraction depends continuously on parameters (by the implicit function theorem for fixed points). Thus $g$ is continuous.

Step 5: $g$ is differentiable. Differentiating $f(g(\mathbf{y})) = \mathbf{y}$ using the chain rule,

$Jf(g(\mathbf{y})) \cdot Jg(\mathbf{y}) = I,$

so $Jg(\mathbf{y}) = [Jf(g(\mathbf{y}))]^{-1}$. Since $Jf$ is continuous and $g$ is continuous, $Jg$ is continuous. Thus $g$ is $C^1$.