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Basic Idea

The inverse function theorem lets us say when a function is invertible, and gives a few statements about its inverse.

Formal Theorem

$$\text{Preconditions: }$$ Let $U$ be an open subset of $\mathbb{R}^n$ and let $f: U \to \mathbb{R}^n$ be a smooth function. $$\text{Theorem: }$$ If $p \in U$ and $[Df(p)]$ is invertible, then $\exists$ a neighborhood V around p $\in$ U such that f is smoothly invertible. $$\text{A few conditions:}$$ The function restricted to v must be injective. Otherwise, there is no inverse. For example, on the domain $[-5, 5]$, the inverse of $f(x) = x^2$ doesn't exist for f(x) = 4. It's either 2 or -2. However, when you restrict it to $[0, 5]$, we get an answer.

Additionally, $[Dg(f(p)] = [Df(p)]^{-1}$, where g is the inverse function.

Example 1:

Let's start off extremely simple. $$f(x) = x^2$$ Let's choose $U \in \mathbb{R}^n$. U is open. We just have to show that the Jacobian of f(x) is invertible. $$Jf(x) = \begin{bmatrix}2x\end{bmatrix}$$ This is invertible whenever the rank = 1, meaning $x \neq 0$. Why is it not invertible there? In the neighborhood around 0, f(x) is only invertible at the point x = 0! That's because when we're at x = 0, the window $[-\epsilon, \epsilon]$ is not invertible. $f^{-1}(\epsilon^2) = \pm \epsilon$. At every other point, we can restrict the domain to positive numbers or negative numbers, and the inverse will be unique.

Example 2:

$$\text{Let } f(x,y) = \begin{bmatrix}x^2 - y^2\\xy\end{bmatrix}$$ There's an issue: $f(1,1) = f(-1,-1) = \begin{bmatrix}0\\1\end{bmatrix}$, meaning $f$ is not invertible. However, this doesn't mean it's not locally invertible as we'll see: $$Jf(x,y) = \begin{bmatrix}2x & -2y \\y & x\end{bmatrix}$$ This matrix is invertible iff the determinant is nonzero. This happens whenever $2x^2 + 2y^2 \neq 0$, which is true whenever $(x,y) \neq (0,0)$. Therefore, $\forall (x,y) \in \mathbb{R}^2 - {0}, f(x,y)$ is locally invertible.
David Witten