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

[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:)

 

[andrewkirk]

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.

 

[wrobel]

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

 

[Office_Shredder]

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

 

[mathwonk]

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.

 

[WWGD]

IIRC, there are results regarding the Jacobian being nonzero.

 

[WWGD]

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?