Smoothness of a manifold implies all maps of its altas' charts are diffeomorphism?

I need to show that :

On any smooth manifold $(M,A)$ all chart maps are $C^{\infty}$-diffeomorphisms.

Definitions : Let $M$ be a Hausdorff second countable topological space. Then a pair $(U, \phi)$ is called a chart for $M$ if $U \in M$ is open and $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ be a homeomorphism. The set $\{\ (U_{\alpha}, \phi_{\alpha}) \}$ is called a topological atlas for $M$ if $\cup U_{\alpha} = M$ and then $(M,A)$ will be called a topological manifold. For $(M,A)$, two charts $(U, \phi), (V, \psi) \in A$ are called $C^{\infty}$-related if $U \cap V = \emptyset$ or if $U \cap V \ne \emptyset$ then $\psi \circ \phi^{-1} : \phi(U \cap V) \subset \mathbb{R}^n \to \psi(U \cap V) \subset \mathbb{R}^n$ be $C^{\infty}$-diffeomorphism. If all charts in $(M,A)$ are pairwise $C^{\infty}$-related then $(M,A)$ is called a smooth manifold.

My attempt based on the definitions above to answer the question in the first paragraph : , The question asks that if for any $(U, \phi) \in A$, $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$-diffeomorphism. So because $\phi \circ \phi^{-1} = \operatorname{id}$ is $C^{\infty}$-diffeomorphism so is $\phi$.

Am I in a right track and if so how to make this rigorous?

Added : Can I go with this approach that I add $\operatorname{id}$ to atlas and say that $\phi = \phi \circ id$ so $\phi$ is diffeomorphism?

Consider $\mathcal{A}_U = \{(U,\phi)\}$. It is a smooth atlas on $U$ (show it), and therefore, $(U,\mathcal{A}_U)$ is a smooth manifold.

Similarly, $\mathcal{A}_{\phi(U)} = \{\phi(U),\mathrm{id}_{\phi(U)}\}$, is a smooth atlas on $\phi(U)$ and it follows that $(\phi(U),\mathcal{A}_{\phi(U)})$ is a smooth manifold.

Let us look at the homeomorphism $\phi \colon U \to \phi(U)$ through the only charts of our atlases. It is given by $$ \mathrm{id}_{\phi(U)}\circ \phi \circ (\phi)^{-1} = \mathrm{id}_{\phi(U)} $$ and is therefore smooth. Similarly, the inverse map $\phi^{-1} \colon \phi(U) \to U$ reads, in the charts, as follows $$ \phi \circ \phi^{-1} \circ (\mathrm{id}_{\phi(U)})^{-1} = \mathrm{id}_{\phi(U)} $$ which is again smooth. It follows that $\phi\colon U \to \phi(U)$ is a smooth homeomorphism with smooth inverse between two smooth manifolds, and is hence a diffeomorphism.

Edit

Smoothness is a notion defined for functions from some open subset of $\Bbb R^p$ with range into some other open subset of $\Bbb R^q$. In order to extend this definition for maps between manifolds, one has to find some trick to go back to the Euclidean setting. The trick is the following.

Let $f\colon M \to N$ be a map between smooth manifolds. If $(U,\phi)$ is a chart of $M$, $(V,\psi)$ a chart of $N$, and if $f(U) \subset V$, then the map $$ \psi \circ f \circ \phi^{-1} \colon \phi(U) \to \psi(V) $$ goes from an open subset of an Euclidean space onto another one. Here, we have a notion of smoothness.

Definition. A map $f\colon M\to N$ is said to be smooth if for any charts $(U,\phi)$ and $(V,\psi)$ as above, the map $\psi\circ f \circ \phi^{-1} \colon \phi(U) \to \psi(V)$ is smooth.