Every differentiable atlas is in an unique maximal atlas I want to proof the following statement:

Every differentiable atlas is in an unique maximal atlas.

Proof:
Let $\mathcal{A}=\{h_\alpha: U_\alpha\to U'_\alpha\subset\mathbb{R}^n|\alpha\in A\}$ be an arbitrary differentiable atlas of $M$.
Define an atlas $\mathcal{A}':=\mathcal{A}\cup\{\underbrace{h_{\beta\alpha}}_{=:h_\beta\circ h^{-1}_\alpha}:h_\alpha(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)\}$.
I have to show, that $\mathcal{A}'$ is maximal. 
Therefore every map of $M$ in $\mathcal{A}'$ has to be compatible with every map of $\mathcal{A}'$.
The transition map $h_{\beta\alpha}:h_\alpha(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)$ is a diffeomorphism, since $\mathcal{A}$ is a differentiable atlas.
Hence two arbitrary maps are compatible, and $\mathcal{A}'$ is maximal.
By construction $\mathcal{A}'$ is unique.
What do you think?
Is my proof correct?
Thanks in advance for any comment.
 A: The set $\mathcal{A}'$ is not an atlas because the maps $h_{\beta \alpha}$ are not charts, they're transition maps (you can see that they are maps from subsets of $\mathbb{R}^n$ to $\mathbb{R}^n$ rather from subsets of $M$ to $\mathbb{R}^n$). Additionally, saying "by construction $\mathcal{A}'$ is unique" only works if you show that somehow you were forced to construct $\mathcal{A}'$ in that way.
What you want to show is that the maximal atlas containing $\mathcal{A}$ is the set of all charts that are compatible with $\mathcal{A}$ (all the transition maps are diffeomorphisms).
A: 
Every differentiable atlas is in an unique maximal atlas.

Proof:
Let $\mathcal{A}$ be a differentiable atlas. For every homeomorphism $h_\beta: U_\beta\to U'_\beta\subset\mathbb{R}^n$ define 
$\mathcal{A}':=\{h_\beta|\, h_\beta\circ h^{-1}_\alpha\colon h_\alpha(U_\alpha\cap U_\beta)\to h_\beta(U_\alpha\cap U_\beta)\,\,\text{is a diffeomorphism for all}\, h_\alpha\in\mathcal{A}\}$.
Obviously $\mathcal{A}'$ is a differentiable atlas, and every $h_\beta\in\mathcal{A}'$ is compatible with every $h_\alpha$ by definition.
Therefore $\mathcal{A}'$ is a maximial atlas.
It seems like with this definition of $\mathcal{A}'$ everything becomes trivial.
Thanks in advance for every comment.
