- #1

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Would be helpful if you are providing me some guidance or tips:)

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- #1

- 3

- 0

Would be helpful if you are providing me some guidance or tips:)

- #2

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That set is the locus of function ##f##, and is a subset of ##\Omega\times \mathbb R##, which in turn is a subset of ##\mathbb R^{m+1}##.

By analogy with the locii of function from ##\mathbb R\to\mathbb R## ( a line, ie one-dimensional manifold in a 2D space), and from ##\mathbb R^2\to \mathbb R## (a surface or 2D manifold in a 3D space), we'd expect the set to be a ##m##-dimensional manifold.

Your mission, should you choose to accept it, is to, for any arbitrary point in the set, construct a homeomorphism from an open neighbourhood of the point to an open subset of ##\mathbb R^m##. If you can do that, you've proven the set is a ##m##-dimensional manifold.

Construction of that homeomorphism may involve the function ##f## in some way, or at least use its property of smoothness.

EDIT: Just realised the proposition is not true unless we require ##\Omega## to be a manifold. Consider where ##\Omega = \{0\} \cup [1,2]## and ##f(x)=x##. Then neither the domain nor the locus of ##f## is a manifold. They are each the union of a 0-dimensional manifold with a 1-dimensional manifold with boundary.

By analogy with the locii of function from ##\mathbb R\to\mathbb R## ( a line, ie one-dimensional manifold in a 2D space), and from ##\mathbb R^2\to \mathbb R## (a surface or 2D manifold in a 3D space), we'd expect the set to be a ##m##-dimensional manifold.

Your mission, should you choose to accept it, is to, for any arbitrary point in the set, construct a homeomorphism from an open neighbourhood of the point to an open subset of ##\mathbb R^m##. If you can do that, you've proven the set is a ##m##-dimensional manifold.

Construction of that homeomorphism may involve the function ##f## in some way, or at least use its property of smoothness.

EDIT: Just realised the proposition is not true unless we require ##\Omega## to be a manifold. Consider where ##\Omega = \{0\} \cup [1,2]## and ##f(x)=x##. Then neither the domain nor the locus of ##f## is a manifold. They are each the union of a 0-dimensional manifold with a 1-dimensional manifold with boundary.

Last edited:

- #3

Science Advisor

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The open set ##\Omega\subset\mathbb{R}^m## with the standard coordinates is a coordinate patch in this graph, so the graph is a manifold

Would be helpful if you are providing me some guidance or tips:)

Last edited:

- #4

Staff Emeritus

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Proving this in the one dimensional case first is probably instructive if you're still lost.

- #5

Science Advisor

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remember this, it is very useful.

- #6

Science Advisor

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IIRC, there are results regarding the Jacobian being nonzero.

- #7

Science Advisor

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Why do you assume ##\Omega## is open?The open set ##\Omega\subset\mathbb{R}^m## with the standard coordinates is a coordinate patch in this graph, so the graph is a manifold

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