Let $U \subset \mathbb{R}^n$ be open, let $m \in \mathbb{R}$, and use the local convention that every element of $\Psi^m(U)$ is represented by a left quantisation $\operatorname{Op}_L(a)$ with $a \in S^m(U \times \mathbb{R}^n)$, modulo smoothing operators. Assume the standard local left-symbol kernel property: for every $r \in \mathbb{R}$ and every $c \in S^r(U \times \mathbb{R}^n)$, if $\operatorname{Op}_L(c)$ is smoothing, then $c \in S^{-\infty}(U \times \mathbb{R}^n) := \bigcap_{s \in \mathbb{R}} S^s(U \times \mathbb{R}^n)$. Define the principal symbol map

\begin{align*} \sigma_m: \Psi^m(U) \to S^m(U \times \mathbb{R}^n) / S^{m-1}(U \times \mathbb{R}^n) \end{align*}

by

\begin{align*} \sigma_m(P) = [a] \end{align*}

whenever $P = \operatorname{Op}_L(a)$ modulo a smoothing operator, where $[a]$ denotes the class of $a$ in the quotient $S^m(U \times \mathbb{R}^n) / S^{m-1}(U \times \mathbb{R}^n)$. Then the sequence

\begin{align*} 0 \longrightarrow \Psi^{m-1}(U) \longrightarrow \Psi^m(U) \xrightarrow{\sigma_m} S^m(U \times \mathbb{R}^n) / S^{m-1}(U \times \mathbb{R}^n) \longrightarrow 0 \end{align*}

is exact, where the first nonzero arrow is the order-filtration inclusion.