If two charts are smoothly compatible with the atlas, then they are naturally compatible I have been enrolled in an course on Riemannian Geometry and was struck on this exercise which was given as homework.

Question: If two charts are compatible with the atlas, then show that they are naturally compatible.

Attempt: Two charts $( U, \phi)$ and $(V, \psi)$ are said to be smoothly compatible if either $U\cap  V = \varnothing $  or $ \psi \circ \phi^{-1} $ is a diffeo.
Let $( U, \phi)$ be a chart and {$(W_{\alpha} ,X_{\alpha})$} be an atlas. Let $(V,\phi)$be another chart.
I couldn't understand what natural means, to me it's not a mathematical term, so, I decided to omit it anyway and try moving towards a rigorious proof.
Case 1: $U\cap W_{\alpha} = \varnothing$ and $ V \cap W_{\alpha} = \varnothing$. To show that either $U\cap V=\varnothing$ or $ \psi \circ \phi^{-1} $ is  a diffeo.
I can't prove $U\cap V=\varnothing$ ( It is easy to find a counterexample). So, I must show that $\psi \circ \phi^{-1}$ is a diffeo. ( But I don't have any ideas on how to do it.) I know that $\phi$ and $\psi$ are homeomorphisms but I am not able to relate them to diffeomorphisms.
(b) $U \cap W_{\alpha} =\varnothing $ and $X_{\alpha} \circ \psi^{-1}$ is diffeo. I have to prove that either $U \cap V =\varnothing$ or $\phi \circ \psi^{-1}$ is diffeo.
Let $ U \cap V \neq \varnothing$  because if the previous expression is $\phi$ then there is nothing to prove. It is given that $ \psi \circ X_{\alpha}^{-1} $ is $C^{\infty}$. To prove that  $  \phi \circ \psi^{-1} $ is a diffeo. But again , I am not able to relate to what is given to what is proved.
(iii) is similar to (ii).
(iv) Let $U \cap W_{\alpha}\neq \varnothing$ and $ V \cap W_{\alpha} \neq \varnothing$. In this case I have, $X_{\alpha} \circ \phi^{-1} $ is diffeo. and $X_{\alpha} \circ \psi^{-1}$ is diffeo. But again , I am not able to prove $\phi \circ \psi^{-1} $ is diffeo.

The problem I am having can be boiled down to how to use the given diffeomorphisms to prove a function is diffeo. but I am unable to. I am struck on it for the past 6 hrs.

Kindly help me by giving hints.
Thanks!
 A: Let $(U_1,\varphi_1)$ and $(U_2,\varphi_2)$ be two charts compatible with some atlas $\mathcal{A}=\{(W_{\alpha},\psi_{\alpha})\}_{\alpha}$.
Let $U:=U_1\cap U_2$.
Case 1.
Suppose that $U = \varnothing$.
Then, by definition, $(U_1,\varphi_1)$ and $(U_2,\varphi_2)$ are compatible.
Case 2.
Suppose that $U\neq \varnothing$, and let us show that $(\varphi_1|_U)\circ (\varphi_2|_U)^{-1}$ is a diffeomorphism.
First of all, we know that both $\varphi_1\colon U_1 \to \varphi_1(U_1)$ and $\varphi_2\colon U_2\to \varphi_2(U_2)$ are homeomorphisms (they are charts).
Then so are $\varphi_1|_{U}\colon U\to \varphi_1(U)$ and $\varphi_2|_U\colon U \to \varphi_2(U)$.
To conclude, it suffices to show that both $(\varphi_1|_U)\circ (\varphi_2|_U)^{-1}$ and $(\varphi_2|_U)\circ (\varphi_1|_U)^{-1}$ are smooth.
Let $y\in \varphi_2(U)$ and let $x\in U$ such that $\varphi_2(x)=y$.
Let $(W,\psi)$ be any chart of the atlas $\mathcal{A}$ with $x\in W$.
Consider $V=U\cap W\neq$, which is non empty since $x\in V$.
It is clear that $\left((\varphi_1|_U)\circ(\varphi_2|_U)^{-1}\right)|_{\varphi_2(V)}= (\varphi_1|_V)\circ (\varphi_2|_V)^{-1}$, and moreover, we have
$$
(\varphi_1|_V)\circ (\varphi_2|_V)^{-1} = (\varphi_1|_V)\circ \operatorname{Id}_V\circ (\varphi_2|_V)^{-1} = (\varphi_1|_V)\circ(\psi|_V)^{-1}\circ (\psi|_V)\circ  (\varphi_2|_V)^{-1}.
$$
Since $\varphi_1$ is compatible with $\psi$, as well as $\varphi_2$, the functions $(\varphi_1|_V)\circ (\psi|_V)^{-1}$ and $(\psi|_V)\circ (\varphi_2|_V)^{-1}$ are smooth (between open subsets of $\Bbb R^n$).
It follows that their composition is smooth, and we have already shown that it is $\left((\varphi_1|_U)\circ(\varphi_2|_U)^{-1}\right)|_{\varphi(V)}$.
Finally, $(\varphi_1|_U)\circ(\varphi_2|_U)^{-1}$ is smooth when restricted to $\varphi_2(V)$, so in particular, it is smooth at the point $y$.
This being true for all $y\in \varphi_2(U)$, we have shown that $(\varphi_1|U)\circ (\varphi_2|_U)^{-1}\colon \varphi_2(U)\to \varphi_1(U)$ is smooth.
Note that there was nothing particular with picking $\varphi_1$ first, so that the same proof shows that $(\varphi_2|_U)\circ (\varphi_1|_U)^{-1}$ is also smooth.
We have then shown that the charts $(U_1,\varphi_1)$ and $(U_2,\varphi_2)$ are compatible.
