Two charts $(U,\phi : U → R^n)$, $(V,\psi: V → R^n)$ of a topological manifold are $C^{\infty}$ compatible if the two maps: $\phi \circ \psi^{-1}$ and $\psi \circ \phi^{-1}$ are $C^{\infty}$ class.
Now, if we consider a circle $S^1$ in a complex plane and define two open set as: \begin{align} U_1 &= \{e^{it} \in \mathbb{C}| -\pi <t < \pi \}, \\ U_2 &= \{e^{it} \in \mathbb{C}| 0 <t < 2\pi \} \end{align} Define: $\phi_1(e^{it}) = t, -\pi < t < \pi$ and $\phi_2(e^{it}) = t, 0 < t < 2\pi$. Then we can show that:$(\phi_2 \circ \phi_1^{-1})(t) = t \text{ for }t \in (0,\pi)$ and $(\phi_1 \circ \phi_2^{-1})(t) = t \text{ for }t \in (0,\pi)$.
Therefore $(U_1,\phi_1)$ and $(U_2,\phi_2)$ are $C^{\infty}$ compatible charts and form a $C^{\infty}$ atlas. This example is taken from "An introduction to manifolds" by Loring W.Tu (chapter-2, pg-50, example:5.7). My question is I think $\phi_1 \circ \phi_2^{-1}$ and $\phi_2 \circ \phi_1^{-1}$ are in $C^1$ class, if then how can it is $C^{\infty}$ compatible.
I am quite new in this topic. Please forgive me, if it is a very silly question.