Let $F^\bullet : \mathcal{A} \to \mathcal{B}$ be a cohomological delta-functor which vanishes in degree strictly greater than $d$.
Thus, $F^{d-\bullet}$ is a homological delta-functor.


Now assume that $F^n$ is effaceable (by injectives) for all $n \geq 1$, so that $F^\bullet$ is a universal delta-functor.


Then what can be said about $F^{d-\bullet}$ ? Is it coeffaceable (even if $\mathcal{A}$ does not necessarily have enough projectives) ? Is it universal ?



Finite universal delta-functors
I don't think $F^{d-\bullet}$ will often be universal, even when $\mathcal{A}$ does have enough projectives.


Let $A$ be any ring with finite global dimension that is not hereditary, so that there is an ideal $I$ with $0<\operatorname{projdim}(I)<\infty$, and let $d=\operatorname{projdim}(I)$. Let $\mathcal{A}$ be the category of $A$-modules, and $F^0=\operatorname{Hom}\_A(I,-)$ , so that $F^d=\operatorname{Ext}\_A^d(I,-)\neq0$.


If $F^{d-\bullet}$ were universal, then $F^0$ would be the $d$-th left derived functor of $F^d$. But it can't be, since it doesn't vanish on projective modules.


