Does acyclic resolution induces morphism of cohomologies? Consider sheaves $F,G$ of abelian groups on a topological space $X$. Fix some $f^0\colon F\to G$.
Given chain map between injective resolutions $0\to F\to I^*$, $0\to G\to J^*$, denoted by $f^0\colon F\to G,f^i\colon I^i\to J^i $.(Chain map says $f\circ d=d\circ f$)
Taking global section gives us uniquely determined maps $h^i\colon H^i(X,F)\to H^i(X,G)$.
My question is, if the resolutions $I^*,J^*$were only acyclic or flasque or soft(not necessarily injective), does the operation of taking global section and cohomology induce the same $h^i$?
[The question arise when I tried to understand the proof of Lefschetz (1,1) in Huybrechts's "Complex geometry" p132 Lemma 3.3, which claims that 
for $0\to \mathbf{C}\to A_{\mathbf{C}}^i(X)$ and  $0\to \mathcal{O}_X\to A_{\mathbf{C}}^{0,i}(X)$, the differentials being $d,\bar{\partial}$, the $f^0$ being injection, $f^i$ being projection onto $(0,i)$ part, 
"the chain map induces same $h^i\colon H^i(X,\mathbf{C})\to H^i(X,\mathcal{O}_X)$ induced by sheaf inclusion $\mathbf{C}\to \mathcal{O}_X$"]
 A: Yes this is true. I quickly recall how you show that the derived functor can be computed by choice of acyclic resolutions, because the explicit description of this isomorphism is important. Suppose you have any effecable $\delta$-functor ${\mathscr T}^{\ast}$ in non-negative degrees (e.g. the right derived functors of a left exact functor defined by injective resolutions), and $0\to X\to F^{\ast}$ is a resolution of some $X$ by a complex of $\mathscr{T}^{\ast}$-acyclic objects. Further, denote $$(\dagger^n):\qquad 0\to \text{Z}^n F\to F^n\to\text{Z}^{n+1}F\to 0$$ the short exact sequences induced from $F$ (with $\text{Z}^0 := X$). Then, for any $n\geq 0$, you get isomorphisms $$(\ddagger)\qquad{\mathscr T}^n X = {\mathscr T}^n (\text{Z}^0{\mathscr F})\cong{\mathscr T}^{n-1}(\text{Z}^1 F)\cong\ldots\cong{\mathscr T}^1(\text{Z}^{n-1} F)\\\ \quad\quad\ \ \ \ \, \ \  \ \ \ \cong\text{coker}\left({\mathscr T}^0 F^n\to {\mathscr T}^0\text{Z}^n F^n\right)\cong\text{H}^n({\mathscr T}^0F^{\ast}).$$
where the isomorphisms are the connecting morphisms for ${\mathscr T}$ induced by the exact sequences $(\dagger^n)$. 
Now, if you have a morphism from $0\to X\to F^{\ast}$ to another resolution $0\to Y\to G^{\ast}$, you also get an isomorphism ${\mathscr T}^nY\cong\text{H}^n({\mathscr T}^0 G^{\ast})$, and by using the naturality of the connecting morphism for ${\mathscr T}^{\ast}$ at each step of the chain of isomorphisms $(\ddagger)$ shows that indeed
$$\begin{array}{ccc} {\mathscr T}^n X & \to & {\mathscr T}^n Y \\ \downarrow && \downarrow \\ \text{H}^n({\mathscr T}^0F^{\ast}) & \to & \text{H}^n({\mathscr T}^0 G^{\ast})\end{array}$$ commutes.
