I got stuck in the proof of III.6.8 of Gelfand, Manin: Methods of Homological Algebra which essentially asserts the existence of a right derived functor of a left exact functor in a category with enough injectives. More precisely (slightly changing the notations of Gelfand, Manin in order to simplify my point):
Theorem: Let $\mathcal{A}$ be an abelian category which has enough injectives. Let $\mathcal{B}$ be another abelian category and $F:A\rightarrow B$ a left exact functor. Then $F$ admits a right derived functor $RF$.
There is a construction of a functor $RF$, but I do not understand why it has the universal property of a right derived functor.
In the proof, a right derived functor is defined as follows: Let $\mathcal{I}$ be the full subcategory of $\mathcal{C}$ of injective objects of $\mathcal{C}$. If we consider the diagram
$$ \require{AMScd} \begin{CD} K^{+}(\mathcal{I})@>>> K^{+}(\mathcal{A})@>>{K^{+}(F)}> K^{+}(\mathcal{B})\\ @VVQ_{\mathcal{I}}V @VVQ_{\mathcal A}V @VVQ_{\mathcal B}V\\ D^{+}(\mathcal{I})@>>\psi> D^{+}(\mathcal{A})@. D^{+}(\mathcal{B}) \end{CD} $$ We the vertial arrows are the canonical localisation maps. We can prove that $\psi$ is an equivalence of categories with quasi inverse $\phi$ and by the universal property of the localization there is an arrow $\overline F:D^{+}(\mathcal{I})\rightarrow D^{+}(\mathcal{B})$ making the big rectangle commute. Now we can define a right derived functor of $F$ by $RF:=\overline F\circ\phi$. It is exact and there is a natural transformation $\epsilon_{F}:Q_{\mathcal B}\circ K^{+}(F)\rightarrow \overline F\circ Q_{A}$. So far so good.
I see why for every functor $G: D^+(\mathcal{A})\rightarrow D^+(\mathcal{B})$, every natural transformation $Q_{\mathcal B}\circ K^{+}(F)\rightarrow G\circ Q_{A}$ factorises over $\epsilon_{F}$, but why does it do uniquely so?