# Why is the set {(x,y)∈Ω×R|y=f(x)} a manifold?

• B
• SaschaSIGI

#### SaschaSIGI

I am thinking why the following holds: Let f be a smooth function with f: Ω⊂R^m→R. Why is the set {(x,y)∈Ω×R|y=f(x)} a manifold?
Would be helpful if you are providing me some guidance or tips:)

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.

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• topsquark
I am thinking why the following holds: Let f be a smooth function with f: Ω⊂R^m→R. Why is the set {(x,y)∈Ω×R|y=f(x)} a manifold?
Would be helpful if you are providing me some guidance or tips:)
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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• topsquark
Proving this in the one dimensional case first is probably instructive if you're still lost.

Basic principle: a graph is isomorphic to its domain. i.e. map x in the domain to (x,f(x)), and map back the point (x,f(x)) to x, via projection onto the "x axis".

remember this, it is very useful.

IIRC, there are results regarding the Jacobian being nonzero.

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
Why do you assume ##\Omega## is open?