[proofplan] We prove the cohomological statement by the standard effaceability argument. First, the long exact sequences for right derived functors make $(R^iF)_{i \geq 0}$ a cohomological delta functor and identify $R^0F$ with $F$. Then enough injectives and the vanishing of higher derived functors on injective objects show that every $R^iF$ for $i > 0$ is effaceable. Finally, the effaceability criterion for universal delta functors gives the universal property by an induction on degree, using the short exact sequence $0 \to A \to I \to C \to 0$ with $I$ injective and then guaranteeing compatibility with all short exact sequences. The homological statement is the dual argument with epimorphisms from projective objects replacing monomorphisms into injective objects. [/proofplan]

[step:Identify the derived functors as a cohomological delta functor] For each object $A \in \mathcal A$, choose an injective resolution \begin{align*} 0 \longrightarrow A \longrightarrow I^0 \longrightarrow I^1 \longrightarrow I^2 \longrightarrow \cdots . \end{align*} The right derived object $R^iF(A)$ is, by definition, the $i$-th cohomology object of the cochain complex \begin{align*} 0 \longrightarrow F(I^0) \longrightarrow F(I^1) \longrightarrow F(I^2) \longrightarrow \cdots . \end{align*} Since $F$ is left exact, the zeroth cohomology of this complex is naturally isomorphic to $F(A)$, so $R^0F \cong F$. For every short exact sequence \begin{align*} 0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0 \end{align*} in $\mathcal A$, the standard long exact sequence theorem for right derived functors, obtained from injective resolution comparison together with the snake-lemma construction, gives connecting morphisms \begin{align*} \delta^i: R^iF(A'') \longrightarrow R^{i+1}F(A') \end{align*} such that the associated sequence \begin{align*} 0 \longrightarrow R^0F(A') \longrightarrow R^0F(A) \longrightarrow R^0F(A'') \xrightarrow{\delta^0} R^1F(A') \longrightarrow R^1F(A) \longrightarrow \cdots \end{align*} is exact and natural in short exact sequences. Therefore $(R^iF,\delta^i)_{i \geq 0}$ is a cohomological delta functor. [/step]

[step:Efface the higher right derived functors by embedding into injectives]Let $i > 0$ and let $A \in \mathcal A$. Since $\mathcal A$ has enough injectives, there exists a monomorphism \begin{align*} u: A \longrightarrow I \end{align*} with $I$ injective. Because $I$ is injective, the complex concentrated in degree $0$, \begin{align*} 0 \longrightarrow I \xrightarrow{\operatorname{id}_I} I \longrightarrow 0 \longrightarrow 0 \longrightarrow \cdots , \end{align*} is an injective resolution of $I$. This is the standard vanishing property of right derived functors on injective objects: applying $F$ gives a cochain complex with cohomology zero in every positive degree. Hence \begin{align*} R^iF(I) = 0 \qquad \text{for all } i > 0. \end{align*} Thus the induced morphism \begin{align*} R^iF(u): R^iF(A) \longrightarrow R^iF(I) \end{align*} is the zero morphism, because its target is the zero object. Therefore each $R^iF$ with $i > 0$ is effaceable.[/step]

[guided]Fix a positive degree $i > 0$ and an object $A \in \mathcal A$. To prove effaceability, we must find a monomorphism out of $A$ along which the induced map on $R^iF$ becomes zero. The hypothesis that $\mathcal A$ has enough injectives gives exactly such a monomorphism: \begin{align*} u: A \longrightarrow I \end{align*} where $I$ is injective. Now we use the vanishing property of right derived functors on injective objects, and we verify it in this case. Since $I$ is already injective, it has the injective resolution concentrated in degree $0$: \begin{align*} 0 \longrightarrow I \xrightarrow{\operatorname{id}_I} I \longrightarrow 0 \longrightarrow 0 \longrightarrow \cdots . \end{align*} After applying $F$, the resulting cochain complex has no nonzero terms in positive degree, so its positive cohomology objects vanish: \begin{align*} R^iF(I) = 0 \qquad \text{for every } i > 0. \end{align*} Therefore the morphism induced by $u$, \begin{align*} R^iF(u): R^iF(A) \longrightarrow R^iF(I), \end{align*} has zero target, and hence is the zero morphism. This proves that $R^iF$ is effaceable for every $i > 0$.[/guided]

[step:Extend a degree zero transformation uniquely by induction]Let $(T^i,\partial^i)_{i \geq 0}$ be any cohomological delta functor from $\mathcal A$ to $\mathcal B$, and let \begin{align*} \alpha^0: R^0F \longrightarrow T^0 \end{align*} be a natural transformation. A cohomological delta functor is universal precisely when every such $\alpha^0$ extends uniquely to a morphism of cohomological delta functors. We construct this extension by induction on $i$. Assume $i>0$, and suppose $\alpha^j: R^jF \to T^j$ has already been constructed for every $j<i$ and is compatible with connecting morphisms in all degrees $<i-1$. For an object $A \in \mathcal A$, choose a short exact sequence \begin{align*} 0 \longrightarrow A \xrightarrow{u} I \xrightarrow{q} C \longrightarrow 0 \end{align*} with $I$ injective. The derived-functor long exact sequence contains \begin{align*} R^{i-1}F(I) \longrightarrow R^{i-1}F(C) \xrightarrow{\delta_A^{i-1}} R^iF(A) \longrightarrow R^iF(I). \end{align*} Since $R^mF(I)=0$ for every $m>0$, exactness implies that $\delta_A^{i-1}$ is an epimorphism when $i>1$. When $i=1$, exactness gives \begin{align*} R^0F(I) \longrightarrow R^0F(C) \xrightarrow{\delta_A^0} R^1F(A) \longrightarrow 0, \end{align*} so $\delta_A^0$ is also an epimorphism, with kernel equal to the image of $R^0F(I) \to R^0F(C)$. We define $\alpha_A^i: R^iF(A) \to T^i(A)$ as the unique morphism satisfying \begin{align*} \alpha_A^i \circ \delta_A^{i-1} = \partial_A^{i-1} \circ \alpha_C^{i-1}, \end{align*} where $\partial_A^{i-1}:T^{i-1}(C)\to T^i(A)$ is the connecting morphism of $(T^i,\partial^i)_{i\geq0}$ for the chosen short exact sequence. This is well-defined because $\partial_A^{i-1}\circ\alpha_C^{i-1}$ kills $\ker \delta_A^{i-1}$. Indeed, by exactness $\ker\delta_A^{i-1}$ is the image of $R^{i-1}F(I)\to R^{i-1}F(C)$. If $i>1$, this source is zero. If $i=1$, the composite on the image of $R^0F(I)\to R^0F(C)$ equals \begin{align*} T^0(q) \xrightarrow{\partial_A^0} T^1(A) \end{align*} after applying naturality of $\alpha^0$, and this composite is zero by exactness of the long exact sequence for $T$. The morphism $\alpha_A^i$ is independent of the chosen injective embedding. Suppose \begin{align*} 0 \longrightarrow A \xrightarrow{u} I \xrightarrow{q} C \longrightarrow 0, \qquad 0 \longrightarrow A \xrightarrow{v} J \xrightarrow{r} D \longrightarrow 0 \end{align*} are two choices with $I$ and $J$ injective. Since $J$ is injective and $u$ is a monomorphism, there is a morphism $s:I\to J$ such that $s\circ u=v$. Let $h:C\to D$ be the induced morphism of cokernels. Naturality of the long exact sequences gives \begin{align*} \delta_D^{i-1}\circ R^{i-1}F(h)=R^iF(\operatorname{id}_A)\circ\delta_C^{i-1}=\delta_C^{i-1} \end{align*} and \begin{align*} \partial_D^{i-1}\circ T^{i-1}(h)=T^i(\operatorname{id}_A)\circ\partial_C^{i-1}=\partial_C^{i-1}. \end{align*} Therefore the morphisms produced from the two choices agree after precomposition with the epimorphism $\delta_C^{i-1}$: \begin{align*} \alpha_{A,J}^i\circ\delta_C^{i-1} &=\alpha_{A,J}^i\circ\delta_D^{i-1}\circ R^{i-1}F(h) \\ &=\partial_D^{i-1}\circ\alpha_D^{i-1}\circ R^{i-1}F(h) \\ &=\partial_D^{i-1}\circ T^{i-1}(h)\circ\alpha_C^{i-1} \\ &=\partial_C^{i-1}\circ\alpha_C^{i-1} \\ &=\alpha_{A,I}^i\circ\delta_C^{i-1}. \end{align*} Since $\delta_C^{i-1}$ is an epimorphism, $\alpha_{A,J}^i=\alpha_{A,I}^i$. The same comparison proves naturality. Let $f:A\to A'$ be a morphism in $\mathcal A$, choose short exact sequences $0\to A\to I\to C\to0$ and $0\to A'\to I'\to C'\to0$ with $I'$ injective, and extend $A\xrightarrow{f}A'\to I'$ across $A\to I$ to a morphism $s:I\to I'$. Let $h:C\to C'$ be the induced morphism of cokernels. Naturality of the connecting morphisms and the inductive naturality of $\alpha^{i-1}$ give \begin{align*} T^i(f)\circ\alpha_A^i\circ\delta_A^{i-1} &=T^i(f)\circ\partial_A^{i-1}\circ\alpha_C^{i-1} \\ &=\partial_{A'}^{i-1}\circ T^{i-1}(h)\circ\alpha_C^{i-1} \\ &=\partial_{A'}^{i-1}\circ\alpha_{C'}^{i-1}\circ R^{i-1}F(h) \\ &=\alpha_{A'}^i\circ\delta_{A'}^{i-1}\circ R^{i-1}F(h) \\ &=\alpha_{A'}^i\circ R^iF(f)\circ\delta_A^{i-1}. \end{align*} Since $\delta_A^{i-1}$ is an epimorphism, $T^i(f)\circ\alpha_A^i=\alpha_{A'}^i\circ R^iF(f)$. Thus $\alpha^i$ is natural. The defining equation gives compatibility with the connecting morphism for the chosen effacement sequence. To obtain compatibility for an arbitrary short exact sequence \begin{align*} E: 0 \longrightarrow A' \longrightarrow A \longrightarrow A'' \longrightarrow 0, \end{align*} we use the effaceability criterion for cohomological delta functors: if a cohomological delta functor $S^*$ has $S^i$ effaceable for every $i>0$, then the induction above produces a unique extension of every degree-zero natural transformation, and that extension is automatically a morphism of delta functors, meaning it commutes with the connecting morphisms attached to every short exact sequence $E$. Its hypotheses have been verified here: $(R^iF,\delta^i)_{i\geq0}$ is a cohomological delta functor, $R^0F\cong F$, and each $R^iF$ for $i>0$ is effaceable by the injective embedding argument. Therefore, for every such $E$ and every $i\geq0$, the square \begin{align*} \alpha_{A'}^{i+1}\circ \delta_E^i = \partial_E^i\circ \alpha_{A''}^i \end{align*} commutes, where $\delta_E^i:R^iF(A'')\to R^{i+1}F(A')$ and $\partial_E^i:T^i(A'')\to T^{i+1}(A')$ denote the connecting morphisms associated to $E$. Uniqueness follows from the same epimorphism argument on the effacement sequences used in the criterion. If $\beta^i:R^iF\to T^i$ is another degree $i$ extension compatible with connecting morphisms and with $\beta^{i-1}=\alpha^{i-1}$, then for every $A$ and every chosen short exact sequence $0\to A\to I\to C\to0$ with $I$ injective, \begin{align*} \beta_A^i\circ\delta_A^{i-1} =\partial_A^{i-1}\circ\beta_C^{i-1} =\partial_A^{i-1}\circ\alpha_C^{i-1} =\alpha_A^i\circ\delta_A^{i-1}. \end{align*} Because $\delta_A^{i-1}$ is an epimorphism, $\beta_A^i=\alpha_A^i$. Induction constructs and uniquely determines the morphism of cohomological delta functors extending $\alpha^0$.[/step]

[guided]We now prove the universal property directly. Let $(T^i,\partial^i)_{i\geq0}$ be a cohomological delta functor and let \begin{align*} \alpha^0:R^0F\longrightarrow T^0 \end{align*} be a natural transformation. Universality means that there is exactly one family of natural transformations $\alpha^i:R^iF\to T^i$ extending $\alpha^0$ and commuting with every connecting morphism. We construct $\alpha^i$ by induction on $i$. Suppose $i>0$ and that the transformations $\alpha^j$ have already been constructed for all $j<i$, naturally in the object and compatibly with connecting morphisms in lower degrees. Fix an object $A\in\mathcal A$. Since $\mathcal A$ has enough injectives, choose a short exact sequence \begin{align*} 0\longrightarrow A\xrightarrow{u}I\xrightarrow{q}C\longrightarrow0 \end{align*} with $I$ injective. The long exact sequence for the derived functors contains \begin{align*} R^{i-1}F(I)\longrightarrow R^{i-1}F(C)\xrightarrow{\delta_A^{i-1}}R^iF(A)\longrightarrow R^iF(I). \end{align*} Because $I$ is injective, the higher derived functors of $F$ vanish on $I$: $R^mF(I)=0$ for every $m>0$. Hence, if $i>1$, both $R^{i-1}F(I)$ and $R^iF(I)$ vanish, so exactness says that $\delta_A^{i-1}$ is an epimorphism. If $i=1$, the relevant part is \begin{align*} R^0F(I)\longrightarrow R^0F(C)\xrightarrow{\delta_A^0}R^1F(A)\longrightarrow0, \end{align*} so $\delta_A^0$ is still an epimorphism; its kernel is exactly the image of $R^0F(I)\to R^0F(C)$. We want $\alpha^i$ to commute with connecting morphisms, so the only possible definition is forced by \begin{align*} \alpha_A^i\circ\delta_A^{i-1}=\partial_A^{i-1}\circ\alpha_C^{i-1}, \end{align*} where $\partial_A^{i-1}:T^{i-1}(C)\to T^i(A)$ is the connecting morphism of $T$ for the chosen short exact sequence. Since $\delta_A^{i-1}$ is an epimorphism, this equation determines $\alpha_A^i$ uniquely if it exists. It exists because the right-hand side kills $\ker\delta_A^{i-1}$. For $i>1$ this kernel is zero. For $i=1$, an element of the kernel comes from $R^0F(I)\to R^0F(C)$; after applying $\alpha^0$, naturality gives the corresponding map $T^0(I)\to T^0(C)$, and exactness of the long exact sequence for $T$ gives \begin{align*} T^0(I)\longrightarrow T^0(C)\xrightarrow{\partial_A^0}T^1(A) \end{align*} with composite zero. Thus the definition is well-defined in both cases. We must prove that the chosen injective embedding of $A$ does not matter. Take two choices \begin{align*} 0\longrightarrow A\xrightarrow{u}I\xrightarrow{q}C\longrightarrow0, \qquad 0\longrightarrow A\xrightarrow{v}J\xrightarrow{r}D\longrightarrow0 \end{align*} with $I$ and $J$ injective. Since $J$ is injective and $u$ is a monomorphism, the map $v:A\to J$ extends to a morphism $s:I\to J$ satisfying $s\circ u=v$. This induces a morphism $h:C\to D$ and hence a morphism of short exact sequences whose left vertical map is $\operatorname{id}_A$. Naturality of the long exact sequences gives \begin{align*} \delta_D^{i-1}\circ R^{i-1}F(h)=\delta_C^{i-1} \end{align*} and \begin{align*} \partial_D^{i-1}\circ T^{i-1}(h)=\partial_C^{i-1}. \end{align*} Using the inductive naturality of $\alpha^{i-1}$, we compute \begin{align*} \alpha_{A,J}^i\circ\delta_C^{i-1} &=\alpha_{A,J}^i\circ\delta_D^{i-1}\circ R^{i-1}F(h)\\ &=\partial_D^{i-1}\circ\alpha_D^{i-1}\circ R^{i-1}F(h)\\ &=\partial_D^{i-1}\circ T^{i-1}(h)\circ\alpha_C^{i-1}\\ &=\partial_C^{i-1}\circ\alpha_C^{i-1}\\ &=\alpha_{A,I}^i\circ\delta_C^{i-1}. \end{align*} Since $\delta_C^{i-1}$ is an epimorphism, the two definitions agree. Now we prove naturality. Let $f:A\to A'$ be a morphism in $\mathcal A$. Choose short exact sequences \begin{align*} 0\longrightarrow A\longrightarrow I\longrightarrow C\longrightarrow0, \qquad 0\longrightarrow A'\longrightarrow I'\longrightarrow C'\longrightarrow0 \end{align*} with $I'$ injective. The composite $A\xrightarrow{f}A'\to I'$ extends across the monomorphism $A\to I$ to a morphism $s:I\to I'$, and this induces a morphism $h:C\to C'$. Naturality of connecting morphisms gives a morphism of the two long exact sequences. Therefore \begin{align*} T^i(f)\circ\alpha_A^i\circ\delta_A^{i-1} &=T^i(f)\circ\partial_A^{i-1}\circ\alpha_C^{i-1}\\ &=\partial_{A'}^{i-1}\circ T^{i-1}(h)\circ\alpha_C^{i-1}\\ &=\partial_{A'}^{i-1}\circ\alpha_{C'}^{i-1}\circ R^{i-1}F(h)\\ &=\alpha_{A'}^i\circ\delta_{A'}^{i-1}\circ R^{i-1}F(h)\\ &=\alpha_{A'}^i\circ R^iF(f)\circ\delta_A^{i-1}. \end{align*} The map $\delta_A^{i-1}$ is an epimorphism, so \begin{align*} T^i(f)\circ\alpha_A^i=\alpha_{A'}^i\circ R^iF(f). \end{align*} Thus $\alpha^i$ is natural. The defining equation gives compatibility with the connecting morphism for the chosen effacement sequence. The point that still has to be checked is stronger: a morphism of cohomological delta functors must commute with the connecting morphism for every short exact sequence, not just the special sequence $0\to A\to I\to C\to0$ used to define $\alpha_A^i$. This is exactly what the effaceability criterion for cohomological delta functors supplies. The criterion says: if $S^*=(S^i,\delta_S^i)_{i\geq0}$ is a cohomological delta functor and each $S^i$ with $i>0$ is effaceable, then every natural transformation $S^0\to T^0$ into the degree-zero part of any cohomological delta functor $T^*$ extends uniquely to a morphism of cohomological delta functors. We have verified the hypotheses for $S^*=R^*F$: the first step gave the cohomological delta functor structure, and the injective embedding argument proved effaceability of $R^iF$ for all $i>0$. Hence, for every short exact sequence \begin{align*} E:0\longrightarrow A'\longrightarrow A\longrightarrow A''\longrightarrow0 \end{align*} and every $i\geq0$, the connecting morphisms \begin{align*} \delta_E^i:R^iF(A'')\longrightarrow R^{i+1}F(A'), \qquad \partial_E^i:T^i(A'')\longrightarrow T^{i+1}(A') \end{align*} satisfy \begin{align*} \alpha_{A'}^{i+1}\circ\delta_E^i=\partial_E^i\circ\alpha_{A''}^i. \end{align*} This gives the missing compatibility for arbitrary short exact sequences. Finally, uniqueness is forced by the same epimorphism used in the effaceability criterion. If $\beta^i:R^iF\to T^i$ is another compatible extension and $\beta^{i-1}=\alpha^{i-1}$ by induction, then for every chosen effacement sequence $0\to A\to I\to C\to0$, \begin{align*} \beta_A^i\circ\delta_A^{i-1} =\partial_A^{i-1}\circ\beta_C^{i-1} =\partial_A^{i-1}\circ\alpha_C^{i-1} =\alpha_A^i\circ\delta_A^{i-1}. \end{align*} Since $\delta_A^{i-1}$ is an epimorphism, $\beta_A^i=\alpha_A^i$. This completes the induction and proves the universal property.[/guided]

[step:Conclude universality and dualize the argument] The previous step shows that every natural transformation out of $R^0F$ into the degree zero part of any cohomological delta functor extends uniquely to a morphism of cohomological delta functors. Hence $(R^iF)_{i \geq 0}$ is universal. For the dual statement, let $G: \mathcal A \to \mathcal B$ be additive and right exact, and suppose $\mathcal A$ has enough projectives. For every object $A \in \mathcal A$, choose an epimorphism \begin{align*} P \longrightarrow A \end{align*} with $P$ projective. The standard dual vanishing property for left derived functors on projective objects gives \begin{align*} L_iG(P)=0 \qquad \text{for all } i>0. \end{align*} Thus each $L_iG$ with $i>0$ is co-effaceable. Applying the dual effaceability criterion to the long exact homology sequences associated to short exact sequences proves that $(L_iG)_{i \geq 0}$ is a universal homological delta functor. This completes the proof. [/step]