Extending morphism between derived functors

Question

Let N,M be objects in a left exact functor $F:A\rightarrow B$ between abelian categoire's source, most importantly say there is an isomorpihsm $\psi: F(M)\rightarrow R^dF(N)$ is it possible to extend this morphism to a morphism of complexes, $\Psi: R^nF(M)\rightarrow R^{n+d}F(M)$, which is an isomorphism of $n \geq 0$?

Answer 1

Your question has so many typos that it's hard to be sure that I'm interpreting it correctly. But if I am, then even for $d=0$ the answer is no.

Let $A$ be the category of $R$-modules for some $k$-algebra $R$, $B$ the category of vector spaces over $k$, and $F$ the functor $\operatorname{Hom}(X,-)$ for some $R$-modules $X$.

Then all that an isomorphism between $F(M)$ and $F(N)$ gives you is an equality of dimensions of $\operatorname{Hom}(X,M)$ and $\operatorname{Hom}(X,N)$, which is certainly not enough to deduce an isomorphism between Ext-spaces.

For example, take $R=k[x]/(x^2)$, $X=M=k$, and $N=R$.