# Given a chart $(U,\phi)$ find a chart $(V,\psi)$ such that $(U,\phi)$ and $(V,\psi)$ are $C^\infty$-compatible and $\psi(V)=\mathbb{R}^n$?

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.

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 $$\psi(V)= \mathbb{R}^n$$.
• One question: is there a $C^\infty$-diffeomorphism $\alpha:B_r(0)\to\mathbb{R}^n$? Jun 10 '20 at 2:48
• @rfloc The accepted answer here math.stackexchange.com/questions/1025308/… provide a diffeomorphism from unit ball to $\mathbb{R}^n$. You can tweak it easily by scaling to any non-unit radius ball. Jun 10 '20 at 2:51