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Does the Implicit mapping theorem imply the inverse mapping theorem?

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What are your thoughts on this question? –  amWhy Nov 10 '12 at 19:10
    
Consider the function $\phi(x,y) = f(x)-y$. –  copper.hat Nov 10 '12 at 19:18
    
It might help if you could provide the precise statement of each theorem, as you have encountered them. –  amWhy Nov 10 '12 at 20:05
    
@copper.hat I know about the equivalence, but I realized that I dont know how to prove it. With your suggestion, I found an application $C^1$ such that $f(g(x))=x$ under some conditions. I dont see why $g$ is the inverse of $f$, i.e $g(f(x))=x$. How should we proceed with your suggestion? –  Binai Feb 15 '13 at 12:58
    
@MattS.: Well, without extra info., I'm not sure what conditions you have on $f,g$, but presumably there is some open set $U$ such that $f(g(x)) = x$ for $x \in U$ and $g$ has some continuity properties that ensure that $g(U)$ is open. Then apply $g$ to both sides to get $g(f(g(x)) = g(x)$. If $g(U)$ is open, then this says, for $y \in g(U)$, that $g(f(y)) = y$. –  copper.hat Feb 15 '13 at 15:59

2 Answers 2

up vote 1 down vote accepted

This is not an answer, it is the converse.

Suppose you want to 'solve' the implicit system $\phi(x,y) = 0$, ie, given $y$, find an $x$ satisfying the equation.

Create the extended system $$f((x,y)) = \pmatrix{ \phi(x,y) \\ y }$$ Noting that this is a 'square' system, and apply the inverse function theorem to the equation $$f((x,y)) = \pmatrix{u_1 \\ u_2}$$ This gives functions $(u_1,u_2) \mapsto x((u_1,u_2))$, $(u_1,u_2) \mapsto y((u_1,u_2))$ satisfying the equation. From the form of the equation, we have $y((u_1,u_2)) = u_2$, hence by setting $u_1 = 0$ we have 'solved' the equation $\phi(x((0,u_2)),u_2) = 0$.

In many situations, I prefer to apply Newton's method rather than use the inverse function theorem directly. They are equivalent, of course, but Newton's method makes some details explicit that only appear in the proof of the inverse function theorem. In particular, Newton's method shows that the solution can be computed using the Newton iteration $x_{n+1} = x_n -Df(x_n)^{-1} (f(x_n))$. This equation can often be used to make properties of the solution 'obvious' (eg, symmetry in the case of matrices, belonging to a closed space, etc.).

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Thanks! Note: I didn't actually answer your question - amWhy actually answered the question. I just added info to show how the converse works... –  copper.hat Nov 11 '12 at 0:16
    
So goes it, (sigh)...You must have provided the "nudge" the OP needed! –  amWhy Nov 11 '12 at 21:34
    
Maybe it was nearer the top? –  copper.hat Nov 11 '12 at 21:49

Yes.

The implicit function theorem implies the inverse function theorem.

Consider $f : \mathbb{R}^3 \to \mathbb{R}^3$, as an example. Saying that $f$ has an inverse is precisely saying that the system $$f1(x_1, x_2, x_3) = u_1$$ $$f2(x_1, x_2, x_3) = u_2$$ $$f3(x_1, x_2, x_3) = u_3$$ determines the values of the $x_i$’s if you’re given the $u_i$’s.

But this is a system of three equations in six unknowns, and one can apply the implicit function theorem to answer this question. Of course, the result is precisely the inverse function theorem, that one can do this if $D$f is invertible.

Can you generalize?

The converse is also true, that is: the Implicit and Inverse Function Theorems are equivalent;
so each theorem implies the other.


Note: The implicit and inverse function theorems are sometimes referred to as the implicit and inverse mapping theorems, respectively.

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I know that this theorem is a bit difficult to understand :+) –  Babak S. Aug 8 '13 at 9:33

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