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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:

enter image description here

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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    The image is sort of hard to read. You might want to just type up your observations here, although I grant that's a fair amount of extra work.2011-08-08
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    Suppose $f\colon X \to \mathbf R$, where $X \subset \mathbf{R}^n$ is some arbitrary subset. How do you make sense of $f$ being differentiable at a point without extending to an open neighborhood? So I don't see any big gain in removing this hypothesis. Maybe I don't understand your qualms.2011-08-08
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    @Dylan: You can open the image in a new tab/window and display it in full resolution there, then it's fine; it's just being scaled down to fit the column width here.2011-08-08
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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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    Isn't their being the same dimension entailed by local invertibility? And diffability at b just gets you that g is defined in a neighborhood of b--not that its range is in Rn.. Moreover, neither fact is actually used in the proof! Am I right?2011-08-08
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    @Quine42: No, $g$ could be a left inverse of $f$ without the spaces having the same dimensions, as long as $f$ is injective. Requiring it to be a left and right inverse would do. On your second point: You're right, I only meant the "is defined on a neighbourhood of $\mathbf b$" part; I thought that was what you were referring to by "I cannot prove that $f(U_{\mathbf a})$ has to be open" in the question. The $\mathbb R^n$ part is required for the matrix inversion.2011-08-09
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    is requiring it 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