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

In summary, the set {(x,y)∈Ω×R|y=f(x)} is a manifold because it can be locally described by smooth charts that satisfy the properties of differentiability and continuity. This set, representing the graph of a function f defined on a manifold Ω, inherits the manifold structure from Ω, ensuring that it is locally Euclidean and satisfies the conditions necessary for manifold classification. Thus, it can be treated as a smooth manifold in its own right.
  • #1
SaschaSIGI
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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:)
 
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  • #2
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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  • #3
SaschaSIGI said:
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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Proving this in the one dimensional case first is probably instructive if you're still lost.
 
  • #5
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.
 
  • #6
IIRC, there are results regarding the Jacobian being nonzero.
 
  • #7
wrobel said:
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?
 

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

What is a manifold?

A manifold is a topological space that locally resembles Euclidean space near each point. More formally, it is a space that can be covered by a collection of open sets, each of which is homeomorphic to an open subset of Euclidean space. Manifolds can have various dimensions and can be equipped with additional structures, such as differentiable or Riemannian structures.

What does it mean for a set to be of the form {(x,y)∈Ω×R|y=f(x)}?

This set represents the graph of a function f: Ω → R, where Ω is a subset of the domain of f. The points (x, y) in the set correspond to pairs where y is the value of the function f at x. Essentially, it captures all the points in the Cartesian plane that lie on the curve defined by the function f.

What conditions must f satisfy for its graph to be a manifold?

For the graph of a function f to be a manifold, it typically needs to be continuous and differentiable. In many cases, if f is a smooth function (infinitely differentiable), then its graph will form a smooth manifold. Additionally, the domain Ω should be an open subset of R, which helps ensure the necessary local properties for the manifold structure.

How do we verify that the graph of f is a manifold?

To verify that the graph of f is a manifold, we can use the implicit function theorem or the regular value theorem. These theorems provide conditions under which the graph can be locally described as a level set of a smooth function. Specifically, if the function f is smooth and its derivative is non-zero, we can show that around each point, the graph can be represented in a way that satisfies the manifold criteria.

What is the dimension of the manifold formed by the graph of f?

The manifold formed by the graph of a function f: Ω → R is a 1-dimensional manifold embedded in a 2-dimensional space (R²). This is because for each point x in the domain Ω, there is a corresponding unique point (x, f(x)) in the graph, which can be parameterized by the single variable x. Thus, it has the same dimension as the domain of the function, which is one dimension lower than the ambient space in which it resides.

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