The exact projective bundle map in every integer bidegree

Regard $X$ as a qcqs scheme over the prime subfield of its base field; this does not change its underlying scheme or the supplied motivic complexes. For $b\ge 0$, apply [F1] to $E^{\vee }$ of rank $r$ and target weight $b$. Using [A1] to discharge the AC hypothesis of the dual-conventions supplier in [F2], transport its target through that identification. Each component remains ${\xi }^m{\pi }^{\ast }$ for the actual product in [F2]. Taking $H^a$ commutes with this finite direct sum, and $H^a(C_{b-m}[-2m])=H^{a-2m,b-m}(X)$. This gives an isomorphism with those components.

The degree of ${\xi }^m$ is $2m$, even. Graded commutativity therefore identifies ${\xi }^m{\pi }^{\ast }(x_m)$ with ${\pi }^{\ast }(x_m){\xi }^m$ without a sign, giving exactly the displayed map. For $m=0$ the power is the unit, so the component is pullback.

For $b<0$, every $b-m<0$ with $m\ge 0$. All source groups and the target vanish by the explicit zero convention. The unique homomorphism between these zero groups is an isomorphism, and its formula is the same sum with zero entries. This covers all $a\in \mathbb{Z}$, including negative cohomological degrees; no truncation was taken.