Can we always find a chart $(U,f)$ such that $f(U)=\mathbb{R}^n$

Question

Let $M$ be an n-dimensional differentiable manifold. Can we always find a coordinate system $(U,f) $ such that $f(U)= \mathbb{R}^n$?

I can see that this is indeed true for the examples I know- the sphere, finite dimensional vector space, the torus, the cylinder. But in general I am not able to say what that $U$ will be. All we have is that for each point $m\in M$, there is an open set homeomorphic to an open subset of $\mathbb{R}^n$. Any help will be appreciated!

Taladris · Accepted Answer · 2014-03-13T05:01:56.070

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I assume that the atlas on $M$ is maximal. Let $m$ be a point of $M$.

Take any local chart $(V,g)$ at $m$. Then $g(V)$ is a non empty open set in ${\mathbb R}^n$. Hence $g(V)$ contains an open ball $B$ with center $g(m)$, which is diffeomorphic to ${\mathbb R}^n$ via an diffeomorphism $\phi:B\to {\mathbb R}^n$.

Now, if $U$ is $g^{-1}(B)$ and $f=\phi\circ g$, $(U,f)$ is a local chart at $m$ on $M$, and $f(U)={\mathbb R}^n$.

edited Mar 13 '14 at 05:01

answered Mar 13 '14 at 04:51

Taladris

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If $\phi$ is chosen to only be a homeomorphism the transition functions are not necessarily smooth. – PVAL-inactive Mar 13 '14 at 04:54
I didn't read "differentiable" manifold. I'll edit, thank you. – Taladris Mar 13 '14 at 05:00
Thank you! I should have been able to do this myself, but was lost. Your answer is very useful. Thanks again! – gradstudent Mar 13 '14 at 05:19

PVAL-inactive · Answer 2 · 2014-03-13T04:43:09.990

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For a a general chart $(U,f)$ consider the chart $(f^{-1}(B),g\circ f)$ where $B$ is a ball containing the image of the desired point and $g$ is an appropriate diffeomorphism from $B$ to $\Bbb R^n$. (Leaving it to you to find $g$.)

edited Mar 13 '14 at 04:43

answered Mar 13 '14 at 04:37

PVAL-inactive

8,551

Can we always find a chart $(U,f)$ such that $f(U)=\mathbb{R}^n$

2 Answers2