Grothendieck collapse when one functor is exact

If $F$ is exact, F2 makes $R^{q}F(A)=0$ for $q>0$. Thus F1 is supported on $q=0$ and $E_2^{p,0}=R^{p}G(F(A))$. If $G$ is exact, F2 instead makes $R^{p}G(R^{q}F(A))=0$ for $p>0$, leaving $E_2^{0,q}=G(R^{q}F(A))$. In either alternative, a differential of bidegree $(r,1-r)$ with $r\ge 2$ cannot have both source and target on that one axis. All such differentials vanish, and repeated page homology preserves this support, proving collapse in the sense of F3.

In degree $n$ the first alternative has only the quotient at $(n,0)$, so all preceding filtration quotients vanish and $F^n{H}^n=H^n$, with $F^{n+1}{H}^n=0$. Its lower edge is therefore the first claimed isomorphism. The second alternative has only $(0,n)$; all positive filtration pieces are zero, so its upper edge is the second claimed isomorphism. Naturality comes from the edge maps in F1. For $n=0$ these are the identification with $GF(A)$; if both functors are exact the positive targets are zero. No splitting or additional choice is used.