# Understanding the proof of the implicit function theorem given the inverse function theorem

I've been spending a reasonable chunk of time trying to understand the proof, but I have some questions that are getting in the way. This is going to be a pretty long post, sorry about that.

First, let me state the theorem as given in my lecture notes.

Let $$U$$ be an open subset of $$\mathbb{R}^{n+m}$$ and $$f : U \to \mathbb{R}^n$$, $$n \geq 2, 1 \leq m$$ be at least once continuously differentiable. We'll write $$f(x) = f(x_1,\dots,x_m,y_1,\dots,y_n) = f(x,y)$$. Assume that at a point $$(x_0,y_0) \in U$$:

1. $$f(x_0,y_0) = 0$$

2. $$D_y f(x_0,y_0)$$ is invertible.

Then there exist open neighborhoods $$W \subset \mathbb{R}^m$$ and $$W' \subset \mathbb{R}^n$$, where $$W \times W' \subset U$$ and $$(x_0,y_0) \in W \times W'$$, and exactly one $$C^1$$ mapping $$g : W \to W'$$ such that the claims 1 and 2 apply for $$f(x, g(x))$$.

$$\underline{\text{Now for the actual proof:}}$$

$$\textbf{Section A}$$

First we'll define the usual help function, $$F : U \to \mathbb{R}^{n+m}$$, $$F(x,y) = (x, f(x,y))$$. Then $$F \in C^1$$ and $$\det D F(x_0,y_0) = \det D_y f(x_0,y_0) \neq 0$$, based on assumption 2.

By the inverse function theorem, $$F$$ is a local diffeomorphism and thus has a $$C^1$$ inverse in a neighborhood $$V \subset U$$ around the point $$(x_0,y_0)$$.

Now the function $$\bar{F} := F : V \to V' := F(V)$$ has the inverse map:

$$\bar{F}^{-1} = (\eta, \xi) : V' \to V.$$

Especially: $$(x,y) = \bar{F}(\bar{F}^{-1}(x,y)) = (\eta(x,y),f(\eta(x),\xi(y))) \quad \forall (x,y) \in V'.$$

Therefore $$\eta(x,y) = x$$ and $$f(x,\xi(x,y)) = y$$ when $$(x,y) \in V'$$.

$$\textbf{Section B}$$ (The existence of $$g$$)

Fix an open neighborhood of $$(x_0,y_0)$$, $$V_x \times V_y \subset V$$. Then by the continuity of $$\bar{F}^{-1}$$, the point

$$\bar{F}(x_0,y_0) = (x_0,f(x_0,y_0)) = (x_0, 0) \in V'$$

has an open neighborhood, $${V_x}' \times {V_y}' \subset V'$$ such that

$$\bar{F}^{-1}({V_x}' \times {V_y}') = {V_x}' \times \xi({V_x}' \times {V_y}') \subset {V_x}' \times {V_y}'.$$

We'll define $$g : {V_x}' \stackrel{\textrm{into}}{=} V_y, \quad g(x) = \xi(x,0)$$

(Note that we'll define $$W := {V_x}'$$ and $$W' := V_y$$ for consistency of notation with the theorem.)

Then $$g$$ fulfills the criteria (1) and (2) for all $$x \in W$$.

$$\textbf{Section C}$$ (The uniqueness of the solution)

Assume there exists a $$C^1$$ function, $$h : W \to W'$$ such that $$f(x,h(x)) = 0$$ for all $$x \in W$$.

Then we have: $$(x,h(x)) = \bar{F}^{-1}(\bar{F}(x,h(x))) = \bar{F}^{-1}(x,f(x,h(x))=0) = \bar{F}^{-1}(x,f(x,g(x))=0) = \bar{F}^{-1}(\bar{F}(x,g(x))) = (x,g(x)),$$

So $$h(x) = g(x)$$ for all $$x \in W$$, which completes the proof.

$$\textbf{Discussion of the proof}$$

Now, I understand most things up until the use of the inverse function theorem, where we get the local diffeomorphism, and thus a local bijection and inverse.

I don't understand how the functions $$\eta$$ and $$\xi$$ are defined, however, what their motivations are. They're written with $$\eta(x), \xi(y)$$ first, then with both variables. I can't seem to really grasp what their domains and codomains are with the notation, or why we're defining the inverse exactly as two separate functions in the first place.

Section B is just... a mess. Without proper motivation and explanation for $$\eta(x,y), \xi(x,y)$$, I've little hope of seeing what happens there. A rough overview of the proof or something would be very nice.

Appreciate the help, this thing's pretty complicated, somehow.

• by the way, the theorem doesn't require $m < n$; also the target space of $F$ should be $\Bbb{R}^m \times \Bbb{R}^n$. Also, it should be $F^{-1}: V' \to V$ Nov 8, 2019 at 0:42
• I've fixed the (specified) issues. Good call, non-square matrix would ruin the whole argument. Nov 8, 2019 at 0:57
• later I might write up something. But for now, I think you should focus on the following special case: let $A \in M_{n \times m}(\Bbb{R})$, and let $B \in M_{n \times n}(\Bbb{R})$ be invertible, and define $f: \Bbb{R}^m \times \Bbb{R}^n \to \Bbb{R}^n$ by $f(x,y) = Ax + By$. Now, consider the equation $f(x,y) = 0$, and the task is to solve $y$ in terms of $x$. This situation is the linearized version of the implicit function theorem: I suggest you go through every step of the proof written above, and apply it to this specific case. Nov 8, 2019 at 1:03
• The linear case is EXTREMELY important (as I realized after struggling with both inverse and implicit function theorems when I learnt them last year). So, you should really try to understand what the proof is trying to do with this argument. Nov 8, 2019 at 1:06

I don't understand how the functions $$η$$ and $$ξ$$ are defined, however, what their motivations are. They're written with $$η ( x ) , ξ ( y )$$ first, then with both variables. I can't seem to really grasp what their domains and codomains are with the notation, or why we're defining the inverse exactly as two separate functions in the first place.
I don't believe its right to have $$\xi(y)$$ instead of $$\xi(x,y)$$, but part of the proof shows that $$\eta$$ is $$y$$ independent.
Recall that $$F$$ is a map from some subset of $$\mathbb R^m\times\mathbb R^n$$. A local inverse would therefore map into $$\mathbb R^m\times\mathbb R^n$$, and therefore every point $$F^{-1}(x,y)$$ (given by some $$x\in\mathbb R^m,y\in\mathbb R^n$$ in the domain) in the image of the inverse can be written $$F^{-1}(x,y)=(\eta(x,y),\xi(x,y))$$ where $$\eta(x,y)\in\mathbb R^m$$ and $$\xi(x,y)\in\mathbb R^n$$. (I'll continue to avoid precisely stating the co/domains and ignore the tildes; I think this is a good first approximation)
So far it was just Inverse Function Theorem. Then now we use the structure of the helper function in the form $$F F^{-1}=\operatorname{id}$$ i.e. $$FF^{-1}(x,y)=(x,y)$$ to see that $$(x,y)=F(F^{-1}(x,y)) = F(\eta(x,y),\xi(x,y)) = (A, f(A,B))\Big|_{\substack{A=\eta(x,y) \\ B = \xi(x,y)}} = (\eta(x,y),f(\eta(x,y),\xi(x,y))$$ i.e. $$x = \eta(x,y)$$ and $$f(\xi(x,y)) = y$$. Thus, the function $$\eta$$ doesn't depend on $$y$$, and the function $$f(\eta(x,y),\xi(x,y))$$ [which is a posteriori just $$f(x,\xi(x,y))$$] is constant in $$x$$. More than that, for each $$y_0$$ in the image, we have that $$f(x,\xi(x,y_0)) = y_0.$$ By translation this $$y_0$$ is without loss of generality $$0$$, but its not at all needed to replace $$y_0$$ as long as its a fixed object. So we've found a function $$\xi(x,y_0)$$ that does the right thing, if you sweep the details about the domain under the rug:
1. It maps from $$x$$ in the image of $$F$$ (which is just the same $$x$$ in the domain of $$f$$) to $$y$$ in the domain of $$f$$.
2. its $$C^1$$ since its the second component of the restriction of $$F^{-1}\in C^1$$ to $$y=y_0$$.
3. It solves $$f(x,\xi(x,y_0)) = y_0$$ for every $$x$$.