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!

Answer 1

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 $f(U)={\mathbb R}^n$.

Answer 2

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 $g$.)