Let $F:K^+(\mathcal{A}) \to K^+(\mathcal{B})$ be the left exact functor at the level of triangulated homotopy category. Assume that $F$ exist derived functor $R^+F$, and $\mathcal{A}$ is enough injective, with full subcategory of injective objects $\mathcal{I}$.
The question is does $R^kF(X) = H^k\circ R^+F(X) \cong H^k F(X)$? If not does this holds for $X \in K^+(\mathcal{I})$?
My attempt: since cohomology functor $K^+(\mathcal{B}) \to \mathcal{B}$ respect quasi-isomorphism, therefore it factor through the $D^+(\mathcal{B}{})$. However the problem is the derived functor does not commute the diagram that is:
$$Q_B \circ F = R^+F \circ Q_A$$
does not hold in general( where $Q_A : K^+(\mathcal{A}) \to D^+(\mathcal{A})$), therefore it's not reasonable to say $R^kF(X) = H^k\circ R^+F(X) \cong H^k F(X)$ in general, however it's possible that $$R^kF(I) = H^k\circ R^+F(I) \cong H^k F(I)$$ holds true for complex of injective objects $I \in K^+(\mathcal{I})$?
(Maybe I have an idea, but I am not sure is my understanding is correct, any comment are appreciated)
We have the following diagram:
Since injective implies $F-$injective, which implies that $F(I)$ is exact once $I$ is exact, that is $F$ is exact functor $K^+(\mathcal{I})\to K^+(\mathcal{B})$, therefore the derived functor at this level will commute the LHS square, and $H^k$ commute the RHS triangle, which implies $$R^kF(I) \cong H^kF(I)$$
(I was a bit sloppy here, the $F$ here really is the $F\circ i$ however as $R^+(G\circ F) \cong R^+G \circ R^+ F$ ,we can ignore the inclusion here)