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
F(A^{\oplus n}) = K^{\oplus n}
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
F(\varphi ) : F(A^{\oplus m}) = K^{\oplus m} \longrightarrow K^{\oplus n} = F(A^{\oplus n})
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
A^{\oplus m_ M} \xrightarrow {\varphi _ M} A^{\oplus n_ M} \to M \to 0
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
56.2.5.1
\begin{equation} \label{functors-equation-map} \vcenter { \xymatrix{ A^{\oplus m_ M} \ar[r]_{\varphi _ M} \ar[d]_{\psi _ f} & A^{\oplus n_ M} \ar[r] \ar[d]_{\chi _ f} & M \ar[r] \ar[d]_ f & 0 \\ A^{\oplus m_ N} \ar[r]^{\varphi _ N} & A^{\oplus n_ N} \ar[r] & N \ar[r] & 0 } } \end{equation}
Having made these choices we can define: for an object M of \text{Mod}^{fp}_ A we set
F(M) = \mathop{\mathrm{Coker}}(F(\varphi _ M) : F(A^{\oplus m_ M}) \to F(A^{\oplus n_ M}))
and for a morphism f : M \to N of \text{Mod}^{fp}_ A we set
F(f) = \text{the map }F(M) \to F(N)\text{ induced by } F(\psi _ f)\text{ and }F(\chi _ f)\text{ on cokernels}
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
\xymatrix{ A^{\oplus m_ M} \ar[r]_{\varphi _ M} \ar[d]_\psi & A^{\oplus n_ M} \ar[r] \ar[d]_\chi & M \ar[r] \ar[d]_ f & 0 \\ A^{\oplus m_ N} \ar[r]^{\varphi _ N} & A^{\oplus n_ N} \ar[r] & N \ar[r] & 0 }
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
F(K) \to F(L) \to F(M) \to 0
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
\xymatrix{ A^{\oplus m_ M} \ar[r]_{\varphi _ M} \ar[d]_\psi & A^{\oplus n_ M} \ar[r] \ar[d]_\chi & M \ar[r] \ar[d]_1 & 0 \\ K \ar[r] & L \ar[r] & M \ar[r] & 0 }
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
\mathop{\mathrm{Coker}}(F(K) \to F(L)) = F(M) \oplus E
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
Comments (0)