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Does the inverse mapping theorem in the setting of subsets of $\mathbb{R^n}$ imply the inverse mapping theorem in the setting of manifolds?

Or does one have to plunge back into the original proof and make suitable adjustments there? I'm not seeing an obvious implication.

Any comments will be much appreciated. Thanks!

user93826
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  • judging from how straightforward the 1-dimensional case is and how hard the general case is, I doubt there is an easy generalization. The usual proof for the one dimensional case exploits lots of facts about the real line that aren't true in higher dimensions. E.g. nonempty connected sets contain open subsets. – Tim kinsella Nov 16 '13 at 23:46
  • Sorry Tim! I meant $\mathbb{R^n}$! – user93826 Nov 16 '13 at 23:47
  • Ohh. Funny that your mistake made for a completely different yet still interesting question. In that case, I think Kevin gives a sketch of the answer you want: The case for manifolds follows immediately because the inverse function theorem is a local statement, and manifolds are locally Euclidean. – Tim kinsella Nov 16 '13 at 23:49

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Yes, assuming all the conditions are satisfied, the inverse mapping theorem holds for manifolds.

The point is that both the assumptions and conclusion of the inverse mapping theorem are about some neighborhood of a given point, and any point in a manifold has a neighborhood which looks like an open subset of $\mathbb{R}^n$. Thus, upon restricting to this open subset, we can just appeal to the inverse mapping theorem for $\mathbb{R}^n$.

A shorthand way of saying this is that the inverse mapping theorem is a "local" statement.

Edit: Here is an outline of the proof.

Suppose $f:X\rightarrow Y$ is a smooth map between $n$-manifolds, $f(x)=y$, and the Jacobian of $f$ is invertible at $x$ in some choice (hence any choice) of local coordinates.

There exists an open subset $V\ni y$ such that $V$ looks like an open subset of $\mathbb{R}^n$. Similarly, there exists an open subset $U\subset f^{-1}(V)$ such that $U\ni x$ and $U$ looks like an open subset of $\mathbb{R}^n$. Then we can apply the usual inverse mapping theorem to $f:U\rightarrow V$.

Kevin Ventullo
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  • Hi, sorry Kevin. I was not being clear with my wording! I know the corresponding theorem for manifolds is true, it is just how one deduces it that I would like some clarity on. I.e my question is, can the statement for manifolds be directly deduced by the one for Euclidean space? – user93826 Nov 16 '13 at 23:43
  • Hi, so I included a sketch of the proof. I think this is pretty direct, but any time you prove something about manifolds there is going to be some mucking around with local coordinate charts. – Kevin Ventullo Nov 16 '13 at 23:52