Let $A = Df(a)$, which is invertible since $Df(a)$ is invertible. Consider the auxiliary map

Then $g$ is continuously differentiable and $Dg(a) = A^{-1}Df(a) = I$. If we can prove the Inverse Function Theorem for $g$, then the result for $f$ follows immediately (since invertibility of $g$ implies invertibility of $f$ via composition with the linear isomorphism $A$). Thus we may assume without loss of generality that

Since $Df$ is continuous and $Df(a)$ is invertible, we can choose a small closed rectangle $U$ around $a$ such that the following hold:

1. $f(x) \neq f(a)$ for all $x \in U$, $x \neq a$: This ensures that $f(a)$ has no other preimages in $U$. In other words, within $U$, $f$ is injective at $a$.
2. $Df(x)$ is invertible for all $x \in U$: Invertibility of $Df(a)$ extends to a small neighborhood by continuity. This guarantees that the function behaves “locally like a linear isomorphism.”
3. Derivatives are uniformly close to the identity: $$ |D_j f^i(x) - \delta_{ij}| < \frac{1}{2n^2} \quad \text{for all } x \in U. $$ This technical condition ensures that $f$ is a small perturbation of the identity map. It allows us to control the differences $f(x_1)-f(x_2)$ quantitatively in Step 3.

Intuition: Inside this rectangle, $f$ behaves like the identity plus a small correction. This setup is crucial to establish injectivity and to eventually construct the local inverse.

To control the differences $f(x_1) - f(x_2)$, we first prove the following standard inequality for rectangles in $\mathbf{R}^n$:

Lemma (Inequality for continuously differentiable maps)

Let $A \subset \mathbf{R}^n$ be a rectangle and $f:A \to \mathbf{R}^n$ be continuously differentiable. Suppose there is a number $M$ such that $|D_j f^i(x)| \le M$ for all $x \in A$ and all $i,j$. Then for any $x, y \in A$:

Proof:

For each component $f^i$, write

By the mean value theorem in one variable, for each term in the sum there exists some $z_{ij}$ such that

Taking absolute values and using $|D_j f^i(z_{ij})| \le M$:

Finally, summing over $i = 1, \dots, n$, we get

Now apply this lemma to $g(x) = f(x) - x$ on the rectangle $U$ from Step 2. Since $|D_j g^i(x)| = |D_j f^i(x) - \delta_{ij}| < 1/(2n^2)$ for all $i,j$, the lemma implies:

Using the triangle inequality,

Rearranging, we obtain the key injectivity inequality:

This shows that $f$ is injective on $U$. Intuitively, $f$ is a small perturbation of the identity, so points cannot collapse together.

By injectivity and continuity, $f$ maps $U$ homeomorphically onto its image. To show that $f(U)$ actually contains a neighborhood of $f(a)$, we proceed as follows.

Let $\partial U$ denote the boundary of $U$. Recall that by construction, $f(x) \neq f(a)$ for all $x \in \partial U$. Therefore, the distance from $f(a)$ to the boundary image is positive:

Define the open ball

For any $y \in W$ and any $x \in \partial U$, we have

ensuring that the function

attains its minimum inside the interior of $U$. At the minimum point $x \in \text{interior } U$, the derivative vanishes:

Since $Df(x)$ is invertible for $x \in U$, the only solution is $y - f(x) = 0$, i.e., $y = f(x)$. Uniqueness follows from the injectivity inequality in Step 3. Therefore, each $y \in W$ has a unique preimage in $U$.

We conclude that

is a bijection, giving a local inverse around $a$.

Let $g = f^{-1}$ on $W$ (constructed in Step 4). To show that $g$ is differentiable, take any $x \in V = (\text{interior } U) \cap f^{-1}(W)$. Let $\mu = Df(x)$, which is invertible by Step 2. For $x_1 \in V$ close to $x$, we can write

where the remainder $\phi(h)$ satisfies

Solving for $x_1$ in terms of $y_1 = f(x_1)$ gives

Since $x_1 = g(y_1)$ and $x = g(y)$ with $y = f(x)$, we can write

To check differentiability, consider the limit

Using the inequality from Step 3, $g$ is Lipschitz continuous:

and by the property of $\phi$, the numerator approaches zero faster than the denominator. Therefore, the limit is zero, proving that $g$ is differentiable at $y$ with

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