x_0$. Since $Q = \bigcup_{t\geq 0}\Phi^{-t}(U)$ by asymptotic stability of $x_0$, and the chain rule imply that ${\mathsf{D}}_{x}\psi$ is invertible for all $x\in Q$. Hence $\psi$ is a local diffeomorphism.

To see that $\psi$ is injective, let $U$ be a neighborhood of $x_0$ such that $\psi|_U\colon U\to \psi(U)$ is a diffeomorphism, and let $x,y \in Q$ be such that $\psi(x) = \psi(y)$. By asymptotic stability of $x_0$, there is $T > 0$ such that $\Phi^T(x),\Phi^T(y)\in U$, and implies that $\psi\circ \Phi^T(x) = \psi \circ \Phi^T(y)$. Injectivity of $\psi|_U$ then implies that $\Phi^T(x) = \Phi^T(y)$, and injectivity of $\Phi^T$ then implies that $x = y$. Hence $\psi$ is injective.

To see that $\psi$ is surjective, fix any $y\in {\mathsf{T}}_{x_0}Q$ and let the neighborhood $U$ be as in the last paragraph. Asymptotic stability of $0$ for ${\mathsf{D}}_{x_0}\Phi \colon {\mathsf{T}}_{x_0}Q\times {\mathbb{T}}\to {\mathsf{T}}_{x_0}Q$ implies that there is $T > 0$ such that ${\mathsf{D}}_{x_0} \Phi^T y \in \psi(U)$, so there exists $x\in U$ with ${\mathsf{D}}_{x_0}\Phi^T y = \psi(x)$. Hence $y = {\mathsf{D}}_{x_0} \Phi^{-T}\psi(x) = \psi \circ \Phi^{-T}(x)$, where we have used . It follows that $\psi$ is surjective. This completes the proof.

The following is an existence and uniqueness result for the ${C_{\textnormal{loc}}^{k,\alpha}}$ Floquet normal 