Cartan-Eilenberg injective resolution of a bounded-below complex

Let $K$ be a cochain complex in an abelian category, with $K^p=0$ for $p<b$. We use commuting arrows $h:I^{p,q}\to I^{p+1,q}$ and $v:I^{p,q}\to I^{p,q+1}$, with $h^2=v^2=0$ and $hv=vh$. A Cartan–Eilenberg injective resolution is this bicomplex, zero for $p<b$ or $q<0$, and an augmentation $ϵ:K\to I^{\bullet ,0}$ satisfying $hϵ=ϵd_K$ and $vϵ=0$, with the following data.

Write $Z^{p,q}=\ker h^{p,q}$, $B^{p,q}=\mathrm{im}{h}^{p-1,q}$ and $H^{p,q}=Z^{p,q}\mathrm{/}{B}^{p,q}$. The induced vertical complexes, with the augmentations induced by $ϵ$, are injective resolutions of $K^p$, $Z^{p}K$, $B^{p}K$ and $H^{p}K$, respectively: $0\to K^p\to I^{p,0}\to I^{p,1}\to \cdots ,$ $0\to Z^{p}K\to Z^{p,0}\to Z^{p,1}\to \cdots ,$ $0\to B^{p}K\to B^{p,0}\to B^{p,1}\to \cdots ,$ $0\to H^{p}K\to H^{p,0}\to H^{p,1}\to \cdots \mathrm{.}$ In particular every displayed unaugmented term is injective. The sequences $0\to B^{p,q}\to Z^{p,q}\to H^{p,q}\to 0$ and $0\to Z^{p,q}\to I^{p,q}\to B^{p+1,q}\to 0$ and their augmentation squares are part of the compatibility. Require these two sequences to be split exact in each bidegree; splittings need not commute with $v$ and are not distinguished data.

For comparison with Homological double complex, set $C_{r,s}=I^{-r,-s}$ and twist the vertical arrow by $(-1{)}^r$, as in Commuting versus anticommuting double complex conventions. The cohomological version of Direct sum total complex of a double complex is therefore ${\mathrm{Tot}}^{n}I={⨁}_{p+q=n}{I}^{p,q},D{\mathrm{\mid }}_{I^{p,q}}=h+(-1{)}^{p}v\mathrm{.}$ The identity $D^2=0$ follows from the two square-zero identities and cancellation of the mixed terms. Only $b\le p\le n$ occur. Replacing $p$ by $p-b$ makes the support first quadrant and shifts total degree by $b$; the signed differential is the displayed one with the original $p$.

This definition asks for supplied resolution data and makes no existence or choice claim. The zero bicomplex resolves the zero complex. A complex concentrated in one degree may use a single column resolving that object. Empty diagonals are zero; a one-term diagonal is that term. The lower bound $b$ may be negative.