For two charts $(U, \phi)$ and $(V,\psi)$ on a topological manifold that are $C^\infty$-compatible, the transition maps
$$\phi \circ \psi^{-1}:\psi(U \cap V)\rightarrow \phi(U \cap V)$$ $$\psi\circ\phi^{-1}: \phi(U \cap V) \rightarrow \psi (U \cap V)$$
are $C^\infty$. Are they diffeomorphisms? Intuitively I think that they will be bijective and hence diffeomorphisms but I don't know how to prove it.
$\phi : U \to U'$ and $\psi : V \to V'$ are homeomorphisms onto open $U', V' \subset \mathbb R^n$. They restrict to homeomorphisms $\bar \phi : U \cap V \to \phi(U \cap V) \subset U'$ and $\bar \psi : U \cap V \to \psi(U \cap V) \subset V'$. The set $U \cap V$ is open in $M$, the sets $\phi(U \cap V)$ and $\psi(U \cap V)$ are open in $U'$ and $V'$, respectively, and therefore open in $\mathbb R^n$.
This allows us to define $$\phi \circ \psi^{-1}:\psi(U \cap V)\stackrel{\bar \psi^{-1}}{\to} U \cap V \stackrel{\bar \phi}{\to}\phi(U \cap V)$$ $$\psi \circ \phi^{-1}:\phi(U \cap V)\stackrel{\bar \phi^{-1}}{\to} U \cap V \stackrel{\bar \psi}{\to}\psi(U \cap V)$$ These maps are homeomorphisms between open subsets of $\mathbb R^n$. By assumption they are smooth. Since they are inverse to each other, they are diffeomorphisms.