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How to show that a locally diffeomorphic mapping from R to R is open? 
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#1
Feb2813, 11:24 AM

P: 11

Let f: R > R be a local diffeomorphism (diffeomrophism in a neighborhood of each point). Show that the image of R under f is an open interval. Furthermore show that f is a diffeomorphism of R on to f(R).
Ok, here is what I am thinking.. that since we are dealing with a "diffeomorphism" we have a bijective differential morphism between manifolds whose inverse is also bijective (a differentiable homomorphism). Now since we are dealing with diffeomorphisms, we are dealing with manifolds (can I say this?). If we are dealing with manifolds then we have some sort of natural Topology on it (locally) as each point on the manifold can be mapped to Euclidean space of some dimension, and because of such we have an infinitely huge Open ball of a neighborhood?? but in this case an infinite line which is open. Now I am not sure what is meant by showing that "the image of R under f is an open interval"  I am guessing we are just dealing with the whole of R? and since R is infinite it is open and since mapping an open set gives an open set we have that f(R) is open? Please forgive the lack of equations in the above  but don't know how this would look. perhaps I should try to say something as f(ε,ε) → ℝ [itex]\forall[/itex] ε > 0. Now for the second part  to show that f is a diffeomorphism of R on to f(R). Now I don't know how to interpret that.. is f somehow going from R to f(R) which goes to R? I understand that both R and f(R) are on the real line and so it seems like I am just mapping it back to itself and can find an infinite amount of points (of the same cardinality of infinity) that map to one another. Any thoughts would be very appreciated, Brian 


#2
Mar113, 03:56 AM

Mentor
P: 18,040

You're only dealing with [itex]\mathbb{R}[/itex] here. This is a very simple space. There is no need to go talking about manifolds since that is a notion that is much more complex.
I would like to hear from you the exact, rigorous definition of "local diffeomorphism". 


#3
Mar113, 08:03 PM

Sci Advisor
P: 1,169

Edit: I am assuming the local diffeo means every point has a 'hood diffeo. to R^n. 


#4
Apr713, 12:33 PM

Sci Advisor
P: 1,169

How to show that a locally diffeomorphic mapping from R to R is open?
I thought of this some more:
Let R be the copy in thedomain and R' in the codomain If every x has a 'hood Ux ( we can take an open 'hood) and a local diffeo f_x : Ux>f(Ux)., then f(Ux) is open in R' . Then we only need to find a cover of f(R) by balls f(Ux_j), with Ux_j associated to xj as Ux is associated with Ux. Each f(Ux_j) is open in R' , and the union of f(Ux_j)'s should cover f(R). 


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