Lemma 56.2.5. Let $A$ be a ring. Let $\mathcal{B}$ be an additive category with cokernels. There is an equivalence of categories between
the category of functors $F : \text{Mod}^{fp}_ A \to \mathcal{B}$ which are right exact, and
the category of pairs $(K, \kappa )$ where $K \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ and $\kappa : A \to \text{End}_\mathcal {B}(K)$ is a ring homomorphism
given by the rule sending $F$ to $F(A)$ with its natural $A$-action.
Proof. Let $(K, \kappa )$ be as in (2). We will construct a functor $F : \text{Mod}^{fp}_ A \to \mathcal{B}$ such that $F(A) = K$ endowed with the given $A$-action $\kappa $. Namely, given an integer $n \geq 0$ let us set
Given an $A$-linear map $\varphi : A^{\oplus m} \to A^{\oplus n}$ with matrix $(a_{ij}) \in \text{Mat}(n \times m, A)$ we define
to be the map with matrix $(\kappa (a_{ij}))$. This defines an additive functor $F$ from the full subcategory of $\text{Mod}^{fp}_ A$ with objects $0$, $A$, $A^{\oplus 2}$, $\ldots $ to $\mathcal{B}$; we omit the verification.
For each object $M$ of $\text{Mod}^{fp}_ A$ choose a presentation
of $M$ as an $A$-module. Let us use the trivial presentation $0 \to A^{\oplus n} \xrightarrow {1} A^{\oplus n} \to 0$ if $M = A^{\oplus n}$ (this isn't necessary but simplifies the exposition). For each morphism $f : M \to N$ of $\text{Mod}^{fp}_ A$ we can choose a commutative diagram
Having made these choices we can define: for an object $M$ of $\text{Mod}^{fp}_ A$ we set
and for a morphism $f : M \to N$ of $\text{Mod}^{fp}_ A$ we set
Note that this rule extends the given functor $F$ on the full subcategory consisting of the free modules $A^{\oplus n}$. We still have to show that $F$ is a functor, that $F$ is additive, and that $F$ is right exact.
Let $f : M \to N$ be a morphism $\text{Mod}^{fp}_ A$. We claim that the map $F(f)$ defined above is independent of the choices of $\psi _ f$ and $\chi _ f$ in (56.2.5.1). Namely, say
is also commutative. Denote $F(f)' : F(M) \to F(N)$ the map induced by $F(\psi )$ and $F(\chi )$. Looking at the commutative diagrams, by elementary commutative algebra there exists a map $\omega : A^{\oplus n_ M} \to A^{\oplus m_ N}$ such that $\chi = \chi _ f + \varphi _ N \circ \omega $. Applying $F$ we find that $F(\chi ) = F(\chi _ f) + F(\varphi _ N) \circ F(\omega )$. As $F(N)$ is the cokernel of $F(\varphi _ N)$ we find that the map $F(A^{\oplus n_ M}) \to F(M)$ equalizes $F(f)$ and $F(f)'$. Since a cokernel is an epimorphism, we conclude that $F(f) = F(f)'$.
Let us prove $F$ is a functor. First, observe that $F(\text{id}_ M) = \text{id}_{F(M)}$ because we may pick the identities for $\psi _ f$ and $\chi _ f$ in the diagram above in case $f = \text{id}_ M$. Second, suppose we have $f : M \to N$ and $g : L \to M$. Then we see that $\psi = \psi _ f \circ \psi _ g$ and $\chi = \chi _ f \circ \chi _ g$ fit into (56.2.5.1) for $f \circ g$. Hence these induce the correct map which exactly says that $F(f) \circ F(g) = F(f \circ g)$.
Let us prove that $F$ is additive. Namely, suppose we have $f, g : M \to N$. Then we see that $\psi = \psi _ f + \psi _ g$ and $\chi = \chi _ f + \chi _ g$ fit into (56.2.5.1) for $f + g$. Hence these induce the correct map which exactly says that $F(f) + F(g) = F(f + g)$.
Finally, let us prove that $F$ is right exact. It suffices to show that $F$ commutes with coequalizers, see Categories, Lemma 4.23.3. For this, it suffices to prove that $F$ commutes with cokernels. Let $K \to L \to M \to 0$ be an exact sequence of $A$-modules with $K$, $L$, $M$ finitely presented. Since $F$ is an additive functor, this certainly gives a complex
and we have to show that the second arrow is the cokernel of the first in $\mathcal{B}$. In any case, we obtain a map $\mathop{\mathrm{Coker}}(F(K) \to F(L)) \to F(M)$. By elementary commutative algebra there exists a commutative diagram
Applying $F$ to this diagram and using the construction of $F(M)$ as the cokernel of $F(\varphi _ M)$ we find there exists a map $F(M) \to \mathop{\mathrm{Coker}}(F(K) \to F(L))$ which is a right inverse to the map $\mathop{\mathrm{Coker}}(F(K) \to F(L)) \to F(M)$. This first implies that $F(L) \to F(M)$ is an epimorphism always. Next, the above shows we have
where the direct sum decomposition is compatible with both $F(M) \to \mathop{\mathrm{Coker}}(F(K) \to F(L))$ and $\mathop{\mathrm{Coker}}(F(K) \to F(L)) \to F(M)$. However, then the epimorphism $p : F(L) \to E$ becomes zero both after composition with $F(K) \to F(L)$ and after composition with $F(A^{n_ M}) \to F(L)$. However, since $K \oplus A^{n_ M} \to L$ is surjective (algebra argument omitted), we conclude that $F(K \oplus A^{n_ M}) \to F(L)$ is an epimorphism (by the above) whence $E = 0$. This finishes the proof. $\square$
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