The derived couple maps are well defined

If $a$ is locally $ix$, then $ia=i(ix)$ lies in $\mathrm{im}i$. Descent of this membership shows that $i$ restricted to $D^{\mathrm{\prime }}\subseteq D$ factors through the target $D^{\mathrm{\prime }}$. Its factorization is unique because that inclusion is monic. This defines $i^{\mathrm{\prime }}$ without any preimage choice.

The composite $j:D\to E$ lands in $\ker d$ since $dj=jkj=0$. Follow it by $\ker d\to E^{\mathrm{\prime }}$. This map kills $\ker i$: an arrow into $\ker i=\mathrm{im}k$ locally has form $kz$, and its image is $[jkz]=[dz]=0$. Hence it descends through $D\mathrm{/}\ker i\cong \mathrm{im}i$ to $j^{\mathrm{\prime }}$, uniquely. Explicitly, if $ix=iy$ locally, then $x-y=kz$ after a further epic pullback, so $[jx]-[jy]=[dz]=0$. Thus its formula is independent of the preimage.

On $\ker d$, the map $k$ lands in $\ker j=\mathrm{im}i$, because $jke=de=0$. Changing a cycle representative by a boundary $dz$ changes its image by $kdz=kjkz=0$. Thus this restricted map kills the boundary image and descends uniquely to $k^{\mathrm{\prime }}:E^{\mathrm{\prime }}\to D^{\mathrm{\prime }}$. Equality after the epic cycle quotient also proves independence for arbitrary maps into $E^{\mathrm{\prime }}$, not just element representatives.

For $i^{\mathrm{\prime }}$ the degree remains $(1,-1)$. To compute $j^{\mathrm{\prime }}$ on $D_{p,q}^{\mathrm{\prime }}$, its local $i$-preimage is at $D_{p-1,q+1}$ and $j$ sends it to $E_{p-r,q+r}$, giving degree $(-r,r)$. The cycle restriction and quotient for $k^{\mathrm{\prime }}$ preserve the original degree $(-1,0)$. The constructions above still apply when any image or homology object is zero, and for $r=1$ give $\mathrm{deg}{j}^{\mathrm{\prime }}=(-1,1)$. Every lift was a finite local epic pullback used to prove a canonical factorization; no global representative selection or AC was used.