[proofplan]
We first invert the principal symbol of $A$ on $T^*M \setminus 0$, obtaining the leading symbol of a candidate inverse of order $-m$. Using the principal symbol exact sequence and the composition formula in the classical pseudodifferential calculus, we recursively add lower-order corrections so that the left error $I - BA$ has successively lower order. Asymptotic summation realizes the resulting formal inverse symbol as a genuine classical operator $B_L$ with $B_LA - I$ smoothing. Repeating the same construction on the right gives $B_R$ with $AB_R - I$ smoothing, and the two parametrices differ by a smoothing operator, which upgrades $B_L$ to a two-sided parametrix.
[/proofplan]
[step:Invert the principal symbol to start the left inverse]
Let
$\pi: T^*M \setminus 0 \to M$ denote the cotangent projection restricted to the punctured cotangent bundle. Define
\begin{align*}
a_m: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
by
\begin{align*}
a_m(x,\xi) = \sigma_m(A)(x,\xi).
\end{align*}
By ellipticity, $a_m(x,\xi) \neq 0$ for every $(x,\xi) \in T^*M \setminus 0$. Therefore the map
\begin{align*}
b_{-m}: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
defined by
\begin{align*}
b_{-m}(x,\xi) = a_m(x,\xi)^{-1}
\end{align*}
is smooth and homogeneous of degree $-m$ in $\xi$.
By the quantization part of the classical pseudodifferential calculus, choose an operator
$B_0 \in \Psi^{-m}_{\mathrm{cl}}(M)$ whose principal symbol is $b_{-m}$. The composition formula for classical pseudodifferential operators gives
\begin{align*}
\sigma_0(B_0A) = \sigma_{-m}(B_0)\sigma_m(A) = b_{-m}a_m = 1.
\end{align*}
Hence the order-zero principal symbol of $I - B_0A$ vanishes. By the principal symbol exact sequence for classical pseudodifferential operators, citing a result not yet in the wiki: Principal Symbol Exact Sequence for Classical Pseudodifferential Operators, the operator
$R_1 := I - B_0A$ lies in $\Psi^{-1}_{\mathrm{cl}}(M)$.
[/step]
[step:Recursively remove each homogeneous term of the left error]
Assume that for some integer $N \geq 1$ we have constructed an operator
$B_{N-1} \in \Psi^{-m}_{\mathrm{cl}}(M)$ such that
\begin{align*}
R_N := I - B_{N-1}A \in \Psi^{-N}_{\mathrm{cl}}(M).
\end{align*}
Let
\begin{align*}
r_{-N}: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
denote the principal symbol $\sigma_{-N}(R_N)$, homogeneous of degree $-N$ in $\xi$. Define
\begin{align*}
c_{-m-N}: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
by
\begin{align*}
c_{-m-N}(x,\xi) = r_{-N}(x,\xi)a_m(x,\xi)^{-1}.
\end{align*}
Since $r_{-N}$ has degree $-N$ and $a_m^{-1}$ has degree $-m$, the function $c_{-m-N}$ is homogeneous of degree $-m-N$.
Choose $C_N \in \Psi^{-m-N}_{\mathrm{cl}}(M)$ with principal symbol $c_{-m-N}$. Define
\begin{align*}
B_N := B_{N-1} + C_N.
\end{align*}
Then
\begin{align*}
I - B_NA = R_N - C_NA.
\end{align*}
The composition formula gives
\begin{align*}
\sigma_{-N}(C_NA) = \sigma_{-m-N}(C_N)\sigma_m(A) = c_{-m-N}a_m = r_{-N}.
\end{align*}
Thus the order $-N$ principal symbol of $R_N - C_NA$ vanishes. By the principal symbol exact sequence, $I - B_NA \in \Psi^{-N-1}_{\mathrm{cl}}(M)$.
[guided]
The recursive step has one purpose: if the current left error has leading term of order $-N$, we add a correction to the inverse whose product with $A$ has exactly that same leading term.
Assume that $B_{N-1} \in \Psi^{-m}_{\mathrm{cl}}(M)$ has already been chosen and that its left error
\begin{align*}
R_N := I - B_{N-1}A
\end{align*}
belongs to $\Psi^{-N}_{\mathrm{cl}}(M)$. The leading part of this error is its principal symbol
\begin{align*}
r_{-N} := \sigma_{-N}(R_N): T^*M \setminus 0 \to \mathbb{C}.
\end{align*}
This symbol is homogeneous of degree $-N$ in the covariable $\xi$.
We want to find a correction $C_N$ such that $C_NA$ has principal symbol $r_{-N}$. Since $A$ has principal symbol
\begin{align*}
a_m := \sigma_m(A): T^*M \setminus 0 \to \mathbb{C}
\end{align*}
and $a_m$ is nonzero by ellipticity, the only possible leading symbol for $C_N$ is
\begin{align*}
c_{-m-N}: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
defined by
\begin{align*}
c_{-m-N}(x,\xi) = r_{-N}(x,\xi)a_m(x,\xi)^{-1}.
\end{align*}
The degree check is essential: $r_{-N}$ has degree $-N$, while $a_m^{-1}$ has degree $-m$, so $c_{-m-N}$ has degree $-m-N$. Therefore it is the correct principal symbol for an operator of order $-m-N$.
By the quantization theorem for classical symbols, citing a result not yet in the wiki: Quantization of Classical Symbols on a Compact Manifold, choose
$C_N \in \Psi^{-m-N}_{\mathrm{cl}}(M)$ with
\begin{align*}
\sigma_{-m-N}(C_N) = c_{-m-N}.
\end{align*}
Set
\begin{align*}
B_N := B_{N-1} + C_N.
\end{align*}
Then the new error is
\begin{align*}
I - B_NA = I - B_{N-1}A - C_NA = R_N - C_NA.
\end{align*}
The composition formula for classical pseudodifferential operators, citing a result not yet in the wiki: Composition Formula for Classical Pseudodifferential Operators, gives
\begin{align*}
\sigma_{-N}(C_NA) = \sigma_{-m-N}(C_N)\sigma_m(A).
\end{align*}
Substituting the definitions gives
\begin{align*}
\sigma_{-N}(C_NA) = c_{-m-N}a_m = r_{-N}.
\end{align*}
Hence the principal symbol of $R_N - C_NA$ at order $-N$ is zero. The principal symbol exact sequence then lowers the order of the error by one:
\begin{align*}
I - B_NA \in \Psi^{-N-1}_{\mathrm{cl}}(M).
\end{align*}
This is the induction mechanism: each correction removes exactly one homogeneous layer of the error.
[/guided]
[/step]
[step:Sum the left inverse symbol asymptotically]
The preceding recursion produces a sequence of correction operators
\begin{align*}
C_N \in \Psi^{-m-N}_{\mathrm{cl}}(M)
\end{align*}
for $N \geq 1$. By asymptotic summation for classical symbols, citing a result not yet in the wiki: Asymptotic Summation of Classical Symbols, there exists an operator
$B_L \in \Psi^{-m}_{\mathrm{cl}}(M)$ such that
\begin{align*}
B_L - B_0 - \sum_{N=1}^{K-1} C_N \in \Psi^{-m-K}_{\mathrm{cl}}(M)
\end{align*}
for every integer $K \geq 1$.
Fix $K \geq 1$. Since
\begin{align*}
I - B_{K-1}A \in \Psi^{-K}_{\mathrm{cl}}(M)
\end{align*}
and
\begin{align*}
B_L - B_{K-1} \in \Psi^{-m-K}_{\mathrm{cl}}(M),
\end{align*}
the composition formula gives
\begin{align*}
(B_L - B_{K-1})A \in \Psi^{-K}_{\mathrm{cl}}(M).
\end{align*}
Therefore
\begin{align*}
I - B_LA = I - B_{K-1}A - (B_L - B_{K-1})A \in \Psi^{-K}_{\mathrm{cl}}(M).
\end{align*}
Because this holds for every $K \geq 1$, we obtain
\begin{align*}
I - B_LA \in \bigcap_{K=1}^{\infty}\Psi^{-K}_{\mathrm{cl}}(M) = \Psi^{-\infty}(M).
\end{align*}
Thus
\begin{align*}
B_LA - I \in \Psi^{-\infty}(M).
\end{align*}
[/step]
[step:Construct a right inverse modulo smoothing operators]
Apply the same construction to the right error. Start with an operator
$D_0 \in \Psi^{-m}_{\mathrm{cl}}(M)$ with principal symbol $a_m^{-1}$. Since scalar principal symbols commute, the composition formula gives
\begin{align*}
\sigma_0(AD_0) = a_m a_m^{-1} = 1.
\end{align*}
Hence $I - AD_0 \in \Psi^{-1}_{\mathrm{cl}}(M)$.
Inductively, suppose
$D_{N-1} \in \Psi^{-m}_{\mathrm{cl}}(M)$ satisfies
\begin{align*}
S_N := I - AD_{N-1} \in \Psi^{-N}_{\mathrm{cl}}(M).
\end{align*}
Let
\begin{align*}
s_{-N}: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
be $\sigma_{-N}(S_N)$. Define
\begin{align*}
d_{-m-N}: T^*M \setminus 0 \to \mathbb{C}
\end{align*}
by
\begin{align*}
d_{-m-N}(x,\xi) = a_m(x,\xi)^{-1}s_{-N}(x,\xi).
\end{align*}
Choose $D_N' \in \Psi^{-m-N}_{\mathrm{cl}}(M)$ with principal symbol $d_{-m-N}$, and set
\begin{align*}
D_N := D_{N-1} + D_N'.
\end{align*}
Then
\begin{align*}
I - AD_N = S_N - AD_N'.
\end{align*}
The composition formula gives
\begin{align*}
\sigma_{-N}(AD_N') = a_m d_{-m-N} = s_{-N},
\end{align*}
so $I - AD_N \in \Psi^{-N-1}_{\mathrm{cl}}(M)$ by the principal symbol exact sequence.
Asymptotic summation gives $B_R \in \Psi^{-m}_{\mathrm{cl}}(M)$ with
\begin{align*}
AB_R - I \in \Psi^{-\infty}(M).
\end{align*}
[/step]
[step:Show that the left parametrix is also a right parametrix]
We have constructed operators
$B_L, B_R \in \Psi^{-m}_{\mathrm{cl}}(M)$ such that
\begin{align*}
B_LA - I \in \Psi^{-\infty}(M)
\end{align*}
and
\begin{align*}
AB_R - I \in \Psi^{-\infty}(M).
\end{align*}
Compute
\begin{align*}
B_L - B_R = B_L(I - AB_R) + (B_LA - I)B_R.
\end{align*}
The class $\Psi^{-\infty}(M)$ is a two-sided ideal in the pseudodifferential calculus, citing a result not yet in the wiki: Smoothing Operators Form a Two-Sided Ideal in the Pseudodifferential Calculus. Therefore both terms on the right-hand side are smoothing, and hence
\begin{align*}
B_L - B_R \in \Psi^{-\infty}(M).
\end{align*}
Using this identity,
\begin{align*}
AB_L - I = A(B_L - B_R) + (AB_R - I).
\end{align*}
Again the smoothing ideal property implies $A(B_L - B_R) \in \Psi^{-\infty}(M)$, and therefore
\begin{align*}
AB_L - I \in \Psi^{-\infty}(M).
\end{align*}
[/step]
[step:Name the resulting operator as the parametrix]
Set
\begin{align*}
B := B_L.
\end{align*}
From the left-parametrix construction,
\begin{align*}
BA - I = B_LA - I \in \Psi^{-\infty}(M).
\end{align*}
From the comparison with the right parametrix,
\begin{align*}
AB - I = AB_L - I \in \Psi^{-\infty}(M).
\end{align*}
Thus $B \in \Psi^{-m}_{\mathrm{cl}}(M)$ is a two-sided inverse for $A$ modulo smoothing operators. This is precisely a parametrix for the [elliptic operator](/page/Elliptic%20Operator) $A$.
[/step]
