[proofplan]
The proof is the formal verification of exactness for the quotient of the symbol space $S(m)$ by the subspace $hS(m)$. We first check that the inclusion of $hS(m)$ into $S(m)$ has zero kernel. Then we identify the kernel of the quotient map with precisely the symbols divisible by $h$. Finally, surjectivity of the quotient map gives exactness at the quotient, and the principal-symbol criterion is exactly the same kernel statement written in symbol language.
[/proofplan]
[step:Verify exactness at $hS(m)$ by checking that the inclusion has zero kernel]
The map $\iota: hS(m) \to S(m)$, defined by $\iota(c)=c$, is the inclusion map. Hence, for $c \in hS(m)$, the equality $\iota(c)=0$ in $S(m)$ holds if and only if $c=0$ in $hS(m)$. Therefore
\begin{align*}
\ker \iota = \{0\}.
\end{align*}
Since the image of the initial zero map $0 \to hS(m)$ is also $\{0\}$, the sequence is exact at $hS(m)$.
[guided]
We begin with the first nonzero map in the sequence. The symbol $\iota: hS(m) \to S(m)$, defined by $\iota(c)=c$, denotes the inclusion of the subspace $hS(m)$ into the ambient symbol space $S(m)$. This means that $\iota$ does not change the element; it only regards the same symbol $c$ as an element of the larger space $S(m)$.
To compute the kernel, take an arbitrary element $c \in hS(m)$. By definition of kernel,
\begin{align*}
c \in \ker \iota \iff \iota(c)=0 \text{ in } S(m).
\end{align*}
Since $\iota(c)=c$, this is equivalent to
\begin{align*}
c=0 \text{ in } S(m).
\end{align*}
The zero element of the subspace $hS(m)$ is the same symbol as the zero element of $S(m)$, so this gives
\begin{align*}
\ker \iota = \{0\}.
\end{align*}
Exactness at $hS(m)$ asks that the image of the preceding map equal $\ker \iota$. The preceding map is the zero map $0 \to hS(m)$, whose image is $\{0\}$. Thus
\begin{align*}
\operatorname{im}(0 \to hS(m))=\{0\}=\ker \iota.
\end{align*}
This proves exactness at $hS(m)$.
[/guided]
[/step]
[step:Identify the kernel of the quotient map with $hS(m)$]
The quotient map is $\pi: S(m) \to S(m)/hS(m)$, defined by $\pi(a)=a+hS(m)$. For $a \in S(m)$, the equality $\pi(a)=0$ in $S(m)/hS(m)$ means
\begin{align*}
a+hS(m)=hS(m).
\end{align*}
This holds exactly when $a \in hS(m)$. Hence
\begin{align*}
\ker \pi = hS(m).
\end{align*}
Because $\operatorname{im}\iota=hS(m)$, we obtain
\begin{align*}
\operatorname{im}\iota=\ker \pi.
\end{align*}
Thus the sequence is exact at $S(m)$.
[guided]
The central point is to translate the meaning of the zero element in the quotient space. The map $\pi: S(m) \to S(m)/hS(m)$, defined by $\pi(a)=a+hS(m)$, sends a symbol $a \in S(m)$ to its equivalence class modulo the subspace $hS(m)$.
The zero element of the quotient $S(m)/hS(m)$ is the coset
\begin{align*}
0+hS(m)=hS(m).
\end{align*}
Therefore, for a symbol $a \in S(m)$,
\begin{align*}
a \in \ker \pi \iff \pi(a)=0 \text{ in } S(m)/hS(m).
\end{align*}
Using the definition of $\pi$, this becomes
\begin{align*}
a \in \ker \pi \iff a+hS(m)=hS(m).
\end{align*}
A coset $a+hS(m)$ equals the zero coset $hS(m)$ precisely when its representative $a$ lies in the subspace $hS(m)$. Hence
\begin{align*}
a \in \ker \pi \iff a \in hS(m).
\end{align*}
Since this equivalence holds for every $a \in S(m)$, we have
\begin{align*}
\ker \pi = hS(m).
\end{align*}
Now compute the image of the inclusion map. Since $\iota(c)=c$ for every $c \in hS(m)$, its image is exactly the subspace $hS(m)$ inside $S(m)$:
\begin{align*}
\operatorname{im}\iota=hS(m).
\end{align*}
Combining the two identities gives
\begin{align*}
\operatorname{im}\iota=\ker \pi.
\end{align*}
This is precisely exactness at the middle term $S(m)$.
[/guided]
[/step]
[step:Use the quotient construction to prove exactness at $S(m)/hS(m)$]
Let $q \in S(m)/hS(m)$. By definition of the quotient space, there exists $a \in S(m)$ such that
\begin{align*}
q=a+hS(m).
\end{align*}
Thus $\pi(a)=q$, so $\pi$ is surjective. Therefore
\begin{align*}
\operatorname{im}\pi=S(m)/hS(m).
\end{align*}
The final map $S(m)/hS(m)\to 0$ has kernel all of $S(m)/hS(m)$, so the sequence is exact at $S(m)/hS(m)$.
[/step]
[step:Translate the kernel statement into the principal-symbol criterion]
By definition, the principal symbol map is $\sigma_{\mathrm{pr}}: S(m) \to S(m)/hS(m)$, defined by $\sigma_{\mathrm{pr}}(a)=a+hS(m)$. This is the same map as $\pi$. From the previous computation,
\begin{align*}
\ker \sigma_{\mathrm{pr}}=\ker \pi=hS(m).
\end{align*}
Therefore, for every $a \in S(m)$,
\begin{align*}
\sigma_{\mathrm{pr}}(a)=0 \iff a \in hS(m).
\end{align*}
This proves both the exactness of the sequence and the stated characterization of symbols with zero principal symbol.
[/step]
