How to show that a locally diffeomorphic mapping from R to R is open?

[Relative0]

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 $\forall$ 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

 

[micromass]

You're only dealing with $\mathbb{R}$ 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".

 

[Bacle2]

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 $\forall$ 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

Think about the standard result that tells you when you have a local diffeomorphism, and see the obstacle to having the local diffeomorphism extended to a global diffeomorphism.

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

 

[Bacle2]

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).

 

[blue_raver22]

Hi Brian,

First of all, great job on starting to think about this problem! You are definitely on the right track.

To show that the image of R under f is an open interval, we can use the fact that f is a local diffeomorphism. This means that for every point x in R, there is a neighborhood U of x such that the restriction of f to U is a diffeomorphism onto its image. In other words, f(U) is an open set in f(R).

Now, since this holds for every point x in R, we can take the union of all such open sets f(U) to get an open interval (since R is connected and we are dealing with real numbers). Therefore, the image of R under f is an open interval.

For the second part, you are correct in saying that f is going from R to f(R) which goes back to R. However, the important thing to note here is that f is a bijection (since it is a diffeomorphism) and therefore, it has an inverse mapping f^-1. This means that the points in f(R) can also be mapped back to R.

In other words, f is a one-to-one and onto mapping between R and f(R), and therefore, it is a diffeomorphism of R onto f(R).

I hope this helps clarify things a bit. Keep up the good work!

Best,