On page 4 of the book "Differential Topology" (written by Amiya Mukherjee) the following is written:
[…] observe that the charts $(U,\phi)$ and $(U,\alpha\circ \phi )$,
where $\alpha:\mathbb{R}^n\to \mathbb{R}^n$ is a diffeomorphism,
are always compatible. In particular, taking $\alpha$ to be the
translation which sends $\phi(p)$ to $0$, we can always suppose that
every point $p\in M$ admits a coordinate chart $(U,\phi)$ such that
$\phi(p)=0$. We may also suppose that $\phi(U)$ is a convex set, or
the whole of $\mathbb{R}^n$.
In that book the word "diffeomorphism" means "$C^\infty$-diffeomorphism" and two charts $(U,\phi)$, $(V,\psi)$ are said to be compatible if $\psi \circ \phi ^{-1}:\phi(U\cap V)\to\psi(U\cap V)$ is a $C^\infty$-diffeomorphism.
My question is about the end of the above quote: "We may also suppose that $\phi(U)$ is a convex set, or the whole of $\mathbb{R}^n$".
Question: Given a chart $(U,\phi)$ how can I prove that exists a chart $(V,\psi)$ such that $(U,\phi)$ and $(V,\psi)$ are $C^\infty$-compatible and $\psi(V)=\mathbb{R}^n$?
I tried to use the questions below to answer my question but I couldn't.
Best Answer
The thing is that we need only to find a subset of $U$ such that the image under chart map $\phi$ is an open ball $B_r(0)$ in $\mathbb{R}^n$. After this, we can blow up the ball to $\mathbb{R}^n$ by a diffeomorphism. Suppose we have a chart $(U,\phi)$ with a point $p \in U$ having $\phi(p)=0 \in \phi(U)\subset \mathbb{R}^n$.
Let $V=\phi^{-1}(B_r(0))$ for some $r>0$ and $\psi :=\phi|_V$. Then $(V,\psi)$ is $C^{\infty}$-compatible with $(U,\phi)$ since it is just restriction of the larger chart.
Choose your favorite diffeomorphism $\alpha : B_r(0) \to \mathbb{R}^n$, we have new chart $(V,\alpha \circ \psi)$ with $(\alpha \circ \psi)(V) = \mathbb{R}^n$ and $C^{\infty}$-compatible with $(V,\psi)$ as you can verify it by yourself.
Therefore $(V,\alpha \circ \psi)$ $C^{\infty}$-compatible with $(U,\phi)$ with $\psi(V)= \mathbb{R}^n$.