= Solution
Write $U=\ker p(T)$, $W=\ker q(T)$. The <linear operator> $E=a(T)p(T)$ vanishes on $U$. On $W$, the identity $E=I-b(T)q(T)$ gives $Ew=w$.
By the <direct sum> in part (i), every vector has a unique expression $u+w$, and
$$
E(u+w)=w.
$$
Consequently
$$
\boxed{E^2=E,\qquad\operatorname{im}E=\ker q(T),\qquad\ker E=\ker p(T).}
$$
The complementary <projection operator> is $I-E=b(T)q(T)$, onto $U$ along $W$. These are <polynomial projections for a coprime decomposition>. In particular the projection does not depend on which Bézout coefficients are chosen: its action on the two summands uniquely determines it.
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