Here is the theorem I am trying to prove:
$\mathcal{A}$ and $\mathcal{B}$ are abelian categories. Suppose $F: \mathcal{A} \to \mathcal{B}$ is a left exact functor. If $\mathcal{A}$ has enough injectives, then the right derived functors $\{R^iF: \mathcal{A}\to \mathcal{B}\}_{i \geq 0}$ consist a $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$.
To verify the condition that if a SES $0\to A \to A^\prime \to A^{\prime\prime} \to 0$ is given in $\mathcal{A}$, there exists a morphism $\delta^i : T^i(A^{\prime\prime}) \to T^{i+1}(A)$ for all $i \geq 0$ such that $0 \to T^0A \to T^0A^\prime \to T^0A^{\prime\prime} \xrightarrow{\delta^0}T^1A \to \cdots$ is exact, we consider the following commutative diagram
where vertical arrows are quasi-isomorphisms of complexes, from $A\to 0 \to \cdots$ to $I^0(A) \to I^1(A) \to \cdots$.
Here are the notes I have taken that I really cannot understand:
Note that $0 \to F(I(A)) \to F(I(A)\oplus I(A^{\prime\prime})) \to F(I(A^{\prime\prime})) \to 0$ is split exact. Then $R^iF(A^\prime) = H^iF(I(A^\prime)) \simeq H^iF(I(A) \oplus I(A^{\prime\prime}))$ gives us the result.
Doesn't the split exact part need the assumption that $F$ is an additive functor? Or is there some theorem that tells us something about functors between abelian categories being additive? Also, how does this imply the existence of such $\delta^i$ described above? Thanks in advance.