Colimits of a graded additive functor equal right exactness plus coproduct preservation

By [L1] and [L9] the category ${\mathrm{GrMod}}_0(A)$ is abelian, hence additive [L10] and preadditive [L11], and by [L2] it has all small coproducts; for a small diagram $D$ the coproducts $R$ and $S$ and the coequalizer of $(d,c)$ therefore exist, and by [L5] that coequalizer, which by [L6] is the cokernel of $c-d$, is a colimit of $D$. In particular every small diagram has a colimit in ${\mathrm{GrMod}}_0(A)$.

Assume $F$ preserves cokernels and all coproducts, and let $Q=\mathrm{coker}(c-d)$ be the colimit of $D$ from step 1.1 with its canonical cocone. Since $F$ preserves coproducts, the maps $F({\mathrm{ȷ}}_u)$ exhibit $F(R)$ as a coproduct of the $F(D(j))$ and the maps $F({\mathrm{ȷ}}_j)$ exhibit $F(S)$ as a coproduct of the $F(D(j))$, so $F(d)$ and $F(c)$ are the canonical maps of the same recipe for the composite $FD$; since $F$ preserves cokernels, $F(Q)$ with $F$ of the canonical map is a cokernel of $F(c-d)$, and $F(c-d)=F(c)-F(d)$ by additivity [L12]; by [L6] that cokernel is a coequalizer of $(F(d),F(c))$, so by [L5] applied to $FD$ the object $F(Q)$ with the image cocone is a colimit of $FD$.

Conversely, if $F$ is cocontinuous then it preserves every coproduct, because a coproduct of a family is the colimit of the discrete diagram on its index set, and it preserves every cokernel, because the cokernel of a morphism $f$ is the coequalizer of $(f,0)$ by [L6] and hence a colimit over a parallel pair; both index categories are small.

Since the small diagram $D$ of step 2.1 was arbitrary, $F$ preserves every small colimit, that is, $F$ is cocontinuous; together with step 2.2 this shows that preservation of all small colimits is equivalent to preservation of cokernels and all coproducts.

For the right-exact reformulation: a right exact functor preserves cokernels, since a cokernel is a finite colimit [L4]; conversely, if $F$ preserves cokernels and all coproducts, then it preserves the colimit of every finite diagram, because for finite $\mathcal{J}$ the coproducts $R$ and $S$ are finite and are finite biproducts of ${\mathrm{GrMod}}_0(A)$ [L2, L10] and all finite colimits of the source exist [L8], so the computation of step 2.1 with finite index sets applies verbatim; hence such an $F$ is right exact.

Combining steps 3.1 and 3.2: $F$ preserves all small colimits if and only if $F$ preserves cokernels and all coproducts if and only if $F$ is right exact and coproduct preserving, so cocontinuity of $F$ is exactly right exactness together with coproduct preservation; the coproducts and cokernels used are the canonical ones of step 1.1, so no choice is made.