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Below are theorems from Munkres' "Analysis on Manifolds". The proof of Theorem 7.4 on the right invokes the chain rule, stated on the left. The conditions of Theorem are somewhat strange and appear only to be needed so the chain rule's conditions are satisfied in the proof. However, even when I find a convoluted to make the appropriate substitutions to show why the chain rule can be invoked, I am left with many conditions stated that are never actually used, or are redundant. Take a look:

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What's going on? The theorem seems like it's badly botched, especially considering that the conditions can actually just be reduced to:

  1. f and g are functions between euclidean spaces.
  2. f and g are differentiable at a and f(a), respectively
  3. g is f's left inverse on a neighborhood of a in f's domain.
  4. ∴ Dg(f(a)) = [Df(a)]⁻¹

Is this a correct way to produce a simplified statement of the theorem which is equivalent to Munkres'?

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    @Dylan: It's better to have fewer conditions when possible, because then there are fewer conditions to check each time, and fewer conditions to remember. If you already have diffability you don't need to go checking that the set is open, so why have the condition there?2011-08-08

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Your simplified statement is missing the premise that the Euclidean spaces are of the same dimension; else the inverse in the conclusion makes no sense.

Other than that, I think you're mostly right:

  • $A$ doesn't have to be open, it just has to contain a neighbourhood of $\mathbf a$ (which, as Dylan pointed out, is already implicit in $f$ being differentiable at $\mathbf a$, and thus also in the conditions of Theorem 7.1).
  • $g(\mathbf b)=\mathbf a$ is just a special case of $g(f(\mathbf x))=\mathbf x$ for $\mathbf x=\mathbf a$.
  • $g$ mapping a neighbourhood of $\mathbf b$ into $\mathbb R^n$ is implied by $g$ being differentiable at $\mathbf b$. (The proof of Theorem 7.1 uses this in saying "By hypothesis, $g$ is defined in a neighbourhood of $\mathbf b$".)
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    is requiring i$t$ to be a right inverse too strong though? How do we know there aren't cases of left-only inverses this theorem applies to?2011-08-12