[proofplan] We choose one projective resolution for each object and use projectivity to lift every morphism in $\mathcal A$ to a chain map between the chosen resolutions. The possible choices of chain lift are harmless because any two such lifts are chain homotopic, and applying the additive functor $F$ preserves chain homotopies, so they induce the same morphism on homology. This gives well-defined morphisms on the homology objects $H_i(F(P_\bullet(A)))$, and the same homotopy argument proves compatibility with identities and composition. Finally, right exactness of $F$ identifies the zeroth homology of $F(P_\bullet(A))$ with $F(A)$, naturally in $A$. [/proofplan] [step:Choose projective resolutions and define the objects $L_iF(A)$] For each object $A \in \mathcal A$, fix a projective resolution \begin{align*} \cdots \longrightarrow P_2(A) \xrightarrow{d_2^A} P_1(A) \xrightarrow{d_1^A} P_0(A) \xrightarrow{\varepsilon^A} A \longrightarrow 0. \end{align*} Thus each $P_n(A)$ is projective, $\varepsilon^A$ is an epimorphism, and the augmented complex is exact. Since $F$ is additive, the sequence \begin{align*} \cdots \longrightarrow F(P_2(A)) \xrightarrow{F(d_2^A)} F(P_1(A)) \xrightarrow{F(d_1^A)} F(P_0(A)) \end{align*} is a chain complex in $\mathcal B$. Indeed, for every $n \geq 2$, \begin{align*} F(d_n^A) \circ F(d_{n+1}^A) = F(d_n^A \circ d_{n+1}^A) = F(0) = 0, \end{align*} because $d_n^A \circ d_{n+1}^A = 0$ and additive functors preserve zero morphisms. For every $i \geq 0$, define \begin{align*} L_iF(A) := H_i\bigl(F(P_\bullet(A))\bigr), \end{align*} where homology is computed in the abelian category $\mathcal B$. [/step] [step:Lift each morphism to a chain map between projective resolutions] Let $f: A \to A'$ be a morphism in $\mathcal A$. We construct a chain map \begin{align*} \tilde f_\bullet: P_\bullet(A) \to P_\bullet(A') \end{align*} lifting $f$, meaning that \begin{align*} \varepsilon^{A'} \circ \tilde f_0 = f \circ \varepsilon^A \end{align*} and, for every $n \geq 1$, \begin{align*} d_n^{A'} \circ \tilde f_n = \tilde f_{n-1} \circ d_n^A. \end{align*} Since $\varepsilon^{A'}: P_0(A') \to A'$ is an epimorphism and $P_0(A)$ is projective, the morphism $f \circ \varepsilon^A: P_0(A) \to A'$ lifts through $\varepsilon^{A'}$. Hence there exists a morphism \begin{align*} \tilde f_0: P_0(A) \to P_0(A') \end{align*} such that \begin{align*} \varepsilon^{A'} \circ \tilde f_0 = f \circ \varepsilon^A. \end{align*} Assume now that $n \geq 1$ and that morphisms $\tilde f_0,\dots,\tilde f_{n-1}$ have been constructed with the required chain identities up to degree $n-1$. Consider the morphism \begin{align*} \tilde f_{n-1} \circ d_n^A: P_n(A) \to P_{n-1}(A'). \end{align*} It lands in $\ker d_{n-1}^{A'}$ when $n \geq 2$, because \begin{align*} d_{n-1}^{A'} \circ \tilde f_{n-1} \circ d_n^A = \tilde f_{n-2} \circ d_{n-1}^A \circ d_n^A = 0. \end{align*} For $n=1$, it lands in $\ker \varepsilon^{A'}$, because \begin{align*} \varepsilon^{A'} \circ \tilde f_0 \circ d_1^A = f \circ \varepsilon^A \circ d_1^A = 0. \end{align*} By exactness of the projective resolution of $A'$, the morphism $d_n^{A'}: P_n(A') \to \ker d_{n-1}^{A'}$ for $n \geq 2$, and the morphism $d_1^{A'}: P_1(A') \to \ker \varepsilon^{A'}$ for $n=1$, are epimorphisms onto the relevant kernels. Since $P_n(A)$ is projective, $\tilde f_{n-1} \circ d_n^A$ lifts through $d_n^{A'}$. Thus there exists \begin{align*} \tilde f_n: P_n(A) \to P_n(A') \end{align*} such that \begin{align*} d_n^{A'} \circ \tilde f_n = \tilde f_{n-1} \circ d_n^A. \end{align*} Induction constructs the desired chain map $\tilde f_\bullet$. [/step] [step:Show that different lifts induce the same maps on homology] Let \begin{align*} \tilde f_\bullet, \hat f_\bullet: P_\bullet(A) \to P_\bullet(A') \end{align*} be two chain maps lifting the same morphism $f: A \to A'$. We show that they are chain homotopic. Define \begin{align*} \delta_n := \tilde f_n - \hat f_n: P_n(A) \to P_n(A') \end{align*} for every $n \geq 0$. Since both lifts cover $f$, we have \begin{align*} \varepsilon^{A'} \circ \delta_0 = 0. \end{align*} Moreover the family $(\delta_n)_{n \geq 0}$ is a chain map lifting the zero morphism $A \to A'$. We construct morphisms \begin{align*} s_n: P_n(A) \to P_{n+1}(A') \end{align*} for $n \geq 0$ satisfying \begin{align*} \delta_0 &= d_1^{A'} \circ s_0, \\ \delta_n &= d_{n+1}^{A'} \circ s_n + s_{n-1} \circ d_n^A \qquad \text{for } n \geq 1. \end{align*} Since $\varepsilon^{A'} \circ \delta_0 = 0$, the morphism $\delta_0$ factors through $\ker \varepsilon^{A'} = \operatorname{im} d_1^{A'}$. Because $P_0(A)$ is projective and $d_1^{A'}: P_1(A') \to \ker \varepsilon^{A'}$ is an epimorphism, there exists $s_0: P_0(A) \to P_1(A')$ with $\delta_0 = d_1^{A'} \circ s_0$. Assume $s_0,\dots,s_{n-1}$ have been constructed. Define \begin{align*} r_n := \delta_n - s_{n-1} \circ d_n^A: P_n(A) \to P_n(A'). \end{align*} Then \begin{align*} d_n^{A'} \circ r_n &= d_n^{A'} \circ \delta_n - d_n^{A'} \circ s_{n-1} \circ d_n^A \\ &= \delta_{n-1} \circ d_n^A - \bigl(\delta_{n-1} - s_{n-2} \circ d_{n-1}^A\bigr)\circ d_n^A \\ &= s_{n-2} \circ d_{n-1}^A \circ d_n^A \\ &= 0 \end{align*} for $n \geq 2$, and the same computation for $n=1$ gives \begin{align*} d_1^{A'} \circ r_1 = \delta_0 \circ d_1^A - d_1^{A'} \circ s_0 \circ d_1^A = 0. \end{align*} Thus $r_n$ factors through $\ker d_n^{A'} = \operatorname{im} d_{n+1}^{A'}$. Since $P_n(A)$ is projective and $d_{n+1}^{A'}: P_{n+1}(A') \to \ker d_n^{A'}$ is an epimorphism, there exists $s_n: P_n(A) \to P_{n+1}(A')$ such that \begin{align*} d_{n+1}^{A'} \circ s_n = r_n. \end{align*} This proves the chain homotopy identity. Applying the additive functor $F$ gives morphisms \begin{align*} F(s_n): F(P_n(A)) \to F(P_{n+1}(A')) \end{align*} and the same identities show that $F(\tilde f_\bullet)$ and $F(\hat f_\bullet)$ are chain homotopic. Chain homotopic maps induce the same maps on homology: if $x$ is a cycle in degree $n$, then \begin{align*} F(\tilde f_n)(x) - F(\hat f_n)(x) = F(d_{n+1}^{A'})(F(s_n)(x)) + F(s_{n-1})(F(d_n^A)(x)) = F(d_{n+1}^{A'})(F(s_n)(x)), \end{align*} which is a boundary. Hence the induced map on $H_n(F(P_\bullet(A)))$ is independent of the chosen lift. [/step] [step:Define the morphism $L_iF(f)$ and prove functoriality] For a morphism $f: A \to A'$, choose any chain lift \begin{align*} \tilde f_\bullet: P_\bullet(A) \to P_\bullet(A') \end{align*} and define \begin{align*} L_iF(f) := H_i\bigl(F(\tilde f_\bullet)\bigr): L_iF(A) \to L_iF(A'). \end{align*} The preceding step proves that this definition is independent of the chosen lift. We verify identities. For an object $A \in \mathcal A$, the identity chain map \begin{align*} \operatorname{id}_{P_\bullet(A)}: P_\bullet(A) \to P_\bullet(A) \end{align*} lifts $\operatorname{id}_A$. Therefore \begin{align*} L_iF(\operatorname{id}_A) = H_i\bigl(F(\operatorname{id}_{P_\bullet(A)})\bigr) = \operatorname{id}_{H_i(F(P_\bullet(A)))} = \operatorname{id}_{L_iF(A)}. \end{align*} We verify composition. Let $f: A \to A'$ and $g: A' \to A''$ be morphisms in $\mathcal A$. Choose chain lifts \begin{align*} \tilde f_\bullet: P_\bullet(A) \to P_\bullet(A'), \qquad \tilde g_\bullet: P_\bullet(A') \to P_\bullet(A''). \end{align*} Then $\tilde g_\bullet \circ \tilde f_\bullet$ is a chain lift of $g \circ f$. Hence \begin{align*} L_iF(g \circ f) &= H_i\bigl(F(\tilde g_\bullet \circ \tilde f_\bullet)\bigr) \\ &= H_i\bigl(F(\tilde g_\bullet) \circ F(\tilde f_\bullet)\bigr) \\ &= H_i\bigl(F(\tilde g_\bullet)\bigr)\circ H_i\bigl(F(\tilde f_\bullet)\bigr) \\ &= L_iF(g) \circ L_iF(f). \end{align*} Thus $L_iF: \mathcal A \to \mathcal B$ is a functor. [/step] [step:Identify $L_0F$ naturally with $F$] Fix $A \in \mathcal A$. The beginning of the projective resolution of $A$ is exact: \begin{align*} P_1(A) \xrightarrow{d_1^A} P_0(A) \xrightarrow{\varepsilon^A} A \longrightarrow 0. \end{align*} Since $F$ is right exact, the sequence \begin{align*} F(P_1(A)) \xrightarrow{F(d_1^A)} F(P_0(A)) \xrightarrow{F(\varepsilon^A)} F(A) \longrightarrow 0 \end{align*} is exact in $\mathcal B$. Therefore $F(\varepsilon^A)$ is the cokernel of $F(d_1^A)$, and hence \begin{align*} H_0(F(P_\bullet(A))) = \operatorname{coker}\bigl(F(d_1^A): F(P_1(A)) \to F(P_0(A))\bigr) \cong F(A). \end{align*} Denote this isomorphism by \begin{align*} \eta_A: L_0F(A) \to F(A). \end{align*} We prove naturality. Let $f: A \to A'$ be a morphism, and let \begin{align*} \tilde f_\bullet: P_\bullet(A) \to P_\bullet(A') \end{align*} be a chain lift of $f$. The augmentation condition gives \begin{align*} \varepsilon^{A'} \circ \tilde f_0 = f \circ \varepsilon^A. \end{align*} Applying $F$ gives \begin{align*} F(\varepsilon^{A'}) \circ F(\tilde f_0) = F(f) \circ F(\varepsilon^A). \end{align*} Passing to cokernels of $F(d_1^A)$ and $F(d_1^{A'})$, this identity is exactly \begin{align*} \eta_{A'} \circ L_0F(f) = F(f) \circ \eta_A. \end{align*} Thus $\eta: L_0F \to F$ is a natural isomorphism. [/step] [step:Compare different choices of projective resolutions] Suppose another projective resolution \begin{align*} \cdots \longrightarrow Q_2(A) \longrightarrow Q_1(A) \longrightarrow Q_0(A) \longrightarrow A \longrightarrow 0 \end{align*} has been chosen for every object $A \in \mathcal A$, and let $L_i^QF$ be the functors constructed from these choices. Applying the lifting construction to the identity morphism $\operatorname{id}_A: A \to A$ gives chain maps \begin{align*} u_\bullet(A): P_\bullet(A) \to Q_\bullet(A), \qquad v_\bullet(A): Q_\bullet(A) \to P_\bullet(A) \end{align*} lifting $\operatorname{id}_A$. The composites \begin{align*} v_\bullet(A)\circ u_\bullet(A): P_\bullet(A) \to P_\bullet(A), \qquad u_\bullet(A)\circ v_\bullet(A): Q_\bullet(A) \to Q_\bullet(A) \end{align*} both lift $\operatorname{id}_A$. By the chain homotopy uniqueness proved above, they are chain homotopic to the corresponding identity chain maps. After applying $F$, they induce mutually inverse isomorphisms on homology: \begin{align*} H_i(F(u_\bullet(A))): L_iF(A) \to L_i^QF(A), \qquad H_i(F(v_\bullet(A))): L_i^QF(A) \to L_iF(A). \end{align*} The same homotopy uniqueness argument applied to morphisms $f: A \to A'$ shows that these isomorphisms commute with the maps induced by $f$. Hence they form a natural isomorphism \begin{align*} L_iF \cong L_i^QF. \end{align*} Therefore the construction is independent, up to canonical natural isomorphism, of the chosen projective resolutions. Together with the previous steps, this completes the construction of the left derived functors and the natural identification $L_0F \cong F$. [/step]