Theorem-1
If for a given operator $A$, both left inverse $A_l^{-1}$ and right inverse $A_r^{-1}$ exists, they are unique and identical.
Proof -
$A_l^{-1}=A_l^{-1}I=A_l^{-1}(AA_r^{-1})=(A_l^{-1}A)A_r^{-1}=IA_r^{-1}=A_r^{-1}$.
Suppose there are two left inverses then they should be equal to the right inverse hence they are also equal.
Similar argument for right inverse.
Theorem-2
If a linear operator $A$ has a uniques left inverse then its left inverse is also the (unique) right inverse.
Proof -
$A(A_l^{-1}A)=A$
$\implies (AA_l^{-1})A=IA\tag{1}$
$\implies (AA_l^{-1}-I)A=0=I-A_l^{-1}A$.
$\implies (AA_l^{-1}-I+A_l^{-1})A=I$
So, $AA_l^{-1}-I+A_l^{-1}=A_l^{-1}$.
$\implies AA_l^{-1}=I$.
Thus $A_l^{-1}$ is the right inverse of $A$ and by Theorem-1, it is unique.
I am thinking of proving Theorem-2 in another way.
We have $(AA_l^{-1})A=IA$.
$\implies AA_l^{-1}=I$.
Thus $A_l^{-1}$ is the right inverse of $A$ and by Theorem-1, it is unique.
Instead of the long proof to Theorem-2 we can directly prove it this way.
Am I correct in doing so or have I missed something in my proof?