Cartan-Eilenberg only resolves terms

Take $K^0=\mathbb{Q}$, $K^1=\mathbb{Q}\mathrm{/}\mathbb{Z}$, $d$ the quotient map, zero elsewhere. Both terms are divisible, hence injective by F3. Set $I^{p,0}=K^p$ and $I^{p,q}=0$ for $q>0$, with horizontal differential $d$, vertical differential zero and identity augmentation. Each column is an injective resolution of the corresponding term, and all squares commute.

But horizontal $Z^{0,0}=H^{0,0}=\mathbb{Z}$. This is not injective: the identity map on the subgroup $\mathbb{Z}\subset \mathbb{Q}$ cannot extend to $f:\mathbb{Q}\to \mathbb{Z}$, since $2f(1\mathrm{/}2)=f(1)=1$ has no integer solution. Thus the induced cycle and cohomology columns are not injective resolutions. Also $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}\mathrm{/}\mathbb{Z}\to 0$ cannot split, since a splitting would retract $\mathbb{Q}$ onto $\mathbb{Z}$ and give the same impossible extension. F1 therefore excludes these termwise data. F2 needs exactly the missing clauses to compute $R^{p}F(H^{q}K)$; commuting term resolutions alone do not provide that computation. The example is bounded, has only one resolution row, and uses AC only for the two divisible terms.