From a generator to module presentations
Written and self-checked with GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original text: CC0.
A generator gives a way to describe an object by maps into it. Composition with endomorphisms of the generator turns those maps into a module. The resulting module remembers the object and every arrow between such objects, but arbitrary modules can contain information that disappears on returning to the original category. The Gabriel–Popescu theorem makes this distinction exact.
We work in a fixed universe. Let \(\mathcal C\) be a Grothendieck abelian category: it is locally small and abelian, admits all small colimits, has exact filtered colimits, and has a generator \(U\). Here generator means that \(\operatorname{Hom}_{\mathcal C}(U,-)\) is faithful. We do not assume that \(U\) is projective or that its Hom functor preserves coproducts. An object and an endomorphism ring below may be large relative to a smaller universe; all presentations use sets in the chosen one.
The prerequisite ring-action lesson constructs tensor products with objects of an abelian category. The generator lesson supplies evaluation epimorphisms. The Serre-quotient lesson supplies the quotient and its local Hom formula.
1. The ring and the two module conventions
Set \(S=\operatorname{End}_{\mathcal C}(U)\), with multiplication \(st=s\circ t\). The object \(U\) carries its tautological internal left \(S\)-action. Define \[ \Gamma(A)=\operatorname{Hom}_{\mathcal C}(U,A), \qquad h\cdot s=h\circ s. \] This is a right \(S\)-module: \((h\cdot s)\cdot t=h\cdot(st)\). For \(a: A\to B\), postcomposition defines the right-linear map \(\Gamma(a)\). Thus \[ \Gamma: \mathcal C\longrightarrow\mathsf{Mod}\text{-}S. \] It is additive, faithful and left exact. Left exactness follows by applying the universal property of a kernel to maps out of \(U\); it does not assert that a map out of \(U\) lifts across every epimorphism.
The tensor construction gives an additive functor \[ F: \mathsf{Mod}\text{-}S\longrightarrow\mathcal C, \qquad F(M)=M\otimes_S U, \] and natural bijections \[ \begin{gathered} \operatorname{Hom}_{\mathcal C}(F(M),A) \\ \simeq\operatorname{Hom}_{\mathsf{Mod}\text{-}S}(M,\Gamma(A)). \end{gathered} \tag{1.1} \] In particular \(F\) is left adjoint to \(\Gamma\), preserves all small colimits, and is right exact. To see the colimit assertion directly, maps from \(F(\operatorname{colim}M_i)\) to \(A\) correspond by (1.1) to compatible module maps \(M_i\to\Gamma(A)\), hence to compatible maps \(F(M_i)\to A\). This is the required colimit universal property, natural in \(A\).
For comparison with a left-module convention, put \(R=S^{\mathrm{op}}\). A right \(S\)-module \(M\) is the same left \(R\)-module, by \(s^{\mathrm{op}}m=ms\). The left \(S\)-action on \(U\) is an internal right \(R\)-action. Consequently the tensor in that convention is \[ M\otimes_{R^{\mathrm{op}}}U. \] One may write \(U\otimes_R M\) if this is declared to be the transposed notation for the same balanced construction. Writing \(M\otimes_R U\) with these specified actions gives the wrong sides. No commutativity of \(S\) is available to repair that expression.
2. Presentations, arrows and the actual unit
Choose a free presentation of a right \(S\)-module: \[ S^{(J)}\xrightarrow{d}S^{(I)} \xrightarrow{p}M\longrightarrow0. \tag{2.1} \] If the \(j\)-th column of \(d\) has entries \(s_{ij}\), there are only finitely many nonzero entries in that column. They give a map \(U\to U^{(I)}\) with components \(s_{ij}: U\to U\). The coproduct of these maps is \[ d_U: U^{(J)}\longrightarrow U^{(I)}. \] The composite of two such maps has the same coefficient order as composition of the right-module maps: the coefficient of a composite is a sum of products \(s_{ij}t_{jk}\), and its operator on \(U\) is the sum of \(s_{ij}\circ t_{jk}\). Finite column support makes every sum defining a column meaningful.
Take \(F(M)=\operatorname{coker}d_U\), with quotient \(\pi\). A map \(a: U^{(I)}\to A\) is a family \(a_i: U\to A\). The condition \(ad_U=0\) says \[ \sum_i a_i s_{ij}=0\quad\text{for each }j. \] These are exactly the relations for the map \(S^{(I)}\to\Gamma(A)\) sending its \(i\)-th basis element to \(a_i\) to factor through \(M\). This proves (1.1), including naturality in \(A\).
Here are the choice and arrow details. If \(b: M\to N\) is a module map, precomposition gives, naturally in \(A\), \[ \operatorname{Hom}_S(N,\Gamma(A)) \longrightarrow\operatorname{Hom}_S(M,\Gamma(A)). \] Through (1.1), this is precomposition by a unique map \(F(b): F(M)\to F(N)\), by the covariant representable version of Yoneda. The equality of precomposition maps for identities and composites gives \(F(1)=1\) and \(F(cb)=F(c)F(b)\). Addition of precomposition maps gives \(F(b+b')=F(b)+F(b')\).
Two presentations of \(M\) represent the same functor \(\operatorname{Hom}_S(M,\Gamma(-))\), with its specified identification. Their unique comparison isomorphism respects that identification. Comparisons for three presentations compose to the comparison for the first and third, since both induce the identity on this functor. The same uniqueness makes every comparison commute with every module arrow. This proves presentation independence, functoriality and all their compatibility, without choosing lifts that might depend on a presentation.
Write \(\eta_M: M\to\Gamma F(M)\) and \(\epsilon_A: F\Gamma(A)\to A\) for the unit and counit of (1.1). The unit corresponds to \(1_{F(M)}\); the counit corresponds to \(1_{\Gamma(A)}\). In (2.1), \[ \eta_M(p(e_i))=\pi\iota_i: U\longrightarrow F(M), \tag{2.2} \] where \(\iota_i\) is the \(i\)-th coproduct injection. This determines the unit on the whole module. The two triangles \[ \begin{gathered} \epsilon_{F(M)}F(\eta_M)=1_{F(M)},\\ \Gamma(\epsilon_A)\eta_{\Gamma(A)}=1_{\Gamma(A)} \end{gathered} \tag{2.3} \] follow by transposing the identity maps in (1.1).
For the free module \(S\), the natural identification \(F(S)=U\) makes \(\eta_S\) the identity of \(S=\Gamma(U)\). On finite free modules it gives \[ \begin{gathered} F(S^n)=U^n,\qquad \Gamma(U^n)=S^n,\\ \eta_{S^n}=1_{S^n}. \end{gathered} \tag{2.4} \] The second equality uses finite biproducts. For an infinite set \(I\), we still have \(F(S^{(I)})=U^{(I)}\), but generally \[ \Gamma(U^{(I)})\ne S^{(I)}. \tag{2.5} \] Replacing (2.4) by an infinite version would silently impose an extra hypothesis on the generator.
3. The retained Gabriel–Popescu theorem
We retain Stacks, Theorem 19.14.3 and its specified functors: for the chosen \(U\), the functors above satisfy
- \(F\dashv\Gamma\);
- \(\epsilon: F\Gamma\to\operatorname{id}_{\mathcal C}\) is an isomorphism;
- \(\Gamma\) is fully faithful;
- \(F\) is exact.
The same section's Lemma 19.14.2 states that a module injection \(M\hookrightarrow\Gamma(A)\) has a monic adjoint \(F(M)\to A\). The ring is allowed to be noncommutative, and these statements hold for every generator.
Sections 1–2 identify the canonical functors, their actions and their natural adjunction with the tensor construction used here. In particular this import is about the same \(F\), rather than a second unspecified left adjoint.
For clarity, the counit and full-faithfulness assertions are equivalent in this adjunction. If \(\epsilon\) is invertible, a module map \(b: \Gamma(A)\to\Gamma(B)\) corresponds to \(c: F\Gamma(A)\to B\), and then to \(a=c\epsilon_A^{-1}: A\to B\). Naturality of \(\epsilon\) and (2.3) show \(\Gamma(a)=b\). The same adjunction shows uniqueness.
Conversely, if \(\Gamma\) is fully faithful, for every \(B\) the map \[ \begin{gathered} \operatorname{Hom}_{\mathcal C}(A,B) \\ \longrightarrow\operatorname{Hom}_{\mathcal C}(F\Gamma(A),B), \\ a\longmapsto a\epsilon_A \end{gathered} \] becomes under (1.1) the bijection \(a\mapsto\Gamma(a)\). It is therefore bijective. Yoneda implies that \(\epsilon_A\) is an isomorphism. Both arguments use the actual counit, so their conclusions are natural in \(A\).
Full faithfulness identifies \(\mathcal C\) with a full subcategory of \(\mathsf{Mod}\text{-}S\). Exactness belongs to the functor returning from modules. The fully faithful functor \(\Gamma\) can fail to be right exact, as Exercise 1 shows.
4. An ordinary exactness criterion
The following elementary criterion supplies the ordinary exactness step when a proof has reduced to submodules of free modules.
Proposition 4.1. Let \(H: \mathsf{Mod}\text{-}S\to\mathcal B\) be an additive right exact functor, with \(\mathcal B\) abelian. Suppose \(H(K)\to H(P)\) is monic for every submodule \(K\subseteq P\) of a free module. Then \(H\) is exact.
Proof. Given \(M'\subseteq M\), choose a free epimorphism \(P\to M\), and let \(L\) be its kernel and \(K\) the inverse image of \(M'\). There are compatible exact sequences \[ \begin{gathered} 0\longrightarrow L\longrightarrow K\longrightarrow M'\longrightarrow0,\\ 0\longrightarrow L\longrightarrow P\longrightarrow M\longrightarrow0. \end{gathered} \tag{4.1} \] By hypothesis \(H(L)\) and \(H(K)\) embed into \(H(P)\). Their factorization \(H(L)\to H(K)\to H(P)\) shows that \(H(L)\to H(K)\) is also monic. We can therefore regard them as nested subobjects \[ H(L)\subseteq H(K)\subseteq H(P). \] Right exactness identifies the images of the last two quotients in (4.1): \[ \begin{gathered} H(M')=H(K)/H(L),\\ H(M)=H(P)/H(L). \end{gathered} \] The induced map between these quotients is monic. Indeed, the kernel of \(H(K)\to H(P)/H(L)\) is the pullback of \(H(L)\hookrightarrow H(P)\) to \(H(K)\), which is precisely \(H(L)\). The coimage–image isomorphism in an abelian category identifies the induced quotient with a subobject of \(H(P)/H(L)\). This induced arrow is \(H(M'\hookrightarrow M)\), by the compatible cokernel diagrams.
Thus \(H\) preserves every monomorphism. Applied to a short exact sequence, right exactness supplies its last two terms and their kernel-image equality, while preservation of the first monomorphism supplies its initial zero. Hence \(H\) is exact. \(\square\)
Only free covers and abelian quotients were used. There is no need to construct a derived functor for this criterion.
5. Correcting the finite kernel argument
There is also a direct way to check the delicate finite step in the free-cover argument. Let \(i: M\hookrightarrow S^{(I)}\) be an inclusion with \(M\) finitely generated. Choose an epimorphism \(p: S^n\to M\), and set \(h=ip\).
Since the images of finitely many basis elements have finite support, \(h\) factors through some finite-coordinate summand: \[ \begin{gathered} S^n\xrightarrow{h_0}S^J \hookrightarrow S^{(I)},\\ J\subseteq I\text{ finite}. \end{gathered} \tag{5.1} \] The second arrow is split. Under (2.4), \(\Gamma F(h_0)\) is \(h_0\) itself: this follows either from the coefficients or from naturality of \(\eta\). Applying \(F\), then \(\Gamma\), to the split inclusion still gives a split monomorphism. Consequently, as submodules of \(\Gamma F(S^n)=S^n\), \[ \begin{aligned} \ker\Gamma F(h)&=\ker h_0\\ &=\ker h=\ker p. \end{aligned} \tag{5.2} \] This calculation has not replaced \(\Gamma(U^{(I)})\) by \(S^{(I)}\).
Let \(\kappa: K\hookrightarrow F(S^n)\) be the kernel of \(F(h)\). Left exactness of \(\Gamma\) identifies \(\Gamma(\kappa)\) with the kernel in (5.2). Naturality of the unit says \[ \Gamma F(p)\eta_{S^n}=\eta_M p. \] As \(\eta_{S^n}\) is the identity under (2.4), it follows that \[ \Gamma F(p)\Gamma(\kappa)=0. \] Faithfulness of \(\Gamma\) gives \(F(p)\kappa=0\). Thus \(\ker F(h)\) is contained in \(\ker F(p)\). The reverse containment follows from \(F(h)=F(i)F(p)\), so these kernels agree.
Right exactness makes \(F(p)\) epic. In an abelian category, the kernel of \(F(i)\) pulls back along this epimorphism to \(\ker F(h)\); the projection of that pullback onto \(\ker F(i)\) is epic. The equality just proved says that this pullback is \(\ker F(p)\), whose map to \(\ker F(i)\) is zero. A zero epimorphism has zero target. Hence \(\ker F(i)=0\), proving that \(F(i)\) is monic.
The required vanishing in this argument is \[ F(p)\kappa=0: \ker F(h)\longrightarrow F(M). \tag{5.3} \] The source instead prints a map from \(\ker F(h)\) to \(F(S^{(I)})\). That map vanishes by the definition of a kernel and gives no information about \(F(M)\). Its corresponding Hom display needs the same target correction. Equation (5.3) states the missing assertion and proves it.
For arbitrary \(M\subseteq S^{(I)}\), take its directed family of finitely generated submodules \(M_\lambda\). Their union is \(M\): each finite set of elements generates one, and the sum of two such submodules is again finitely generated. Each \(F(M_\lambda)\to F(S^{(I)})\) is monic by the finite argument. Since \(F\) preserves colimits and filtered colimits are exact in \(\mathcal C\), their colimit \(F(M)\to F(S^{(I)})\) is monic. More explicitly, apply filtered exactness to the sequences \[ 0\longrightarrow F(M_\lambda)\longrightarrow F(S^{(I)}) \longrightarrow C_\lambda\longrightarrow0, \] where \(C_\lambda\) is the cokernel. The constant middle diagram has colimit \(F(S^{(I)})\). Proposition 4.1 now supplies exactness for arbitrary module monomorphisms.
This checks the printed proof interface using the faithful Hom functor, finite free identifications, right exactness, and AB5. In particular neither an infinite free Hom identification nor a projective generator is hidden in the correction.
6. Four graded exercises with full solutions
Exercise 1 (warm-up: a generator that is not projective). In abelian groups take \(U=\mathbf Z\oplus\mathbf Z/2\). Compute \(S=\operatorname{End}(U)\) and the right action on \(\Gamma(X)\). Prove that \(U\) is a generator and that \(S\) is noncommutative. For the quotient \(q: \mathbf Z\to\mathbf Z/2\), decide whether \(\Gamma(q)\) is epic. Explain how your answer fits the retained theorem.
Solution. The four Hom groups between the two summands give \[ S=\left\{ \begin{pmatrix}r&0\\ b&c\end{pmatrix} \ \middle|\ \substack{r\in\mathbf Z\\ b,c\in\mathbf F_2} \right\}, \] where a product has entries \[ \begin{gathered} \begin{pmatrix}r&0\\ b&c\end{pmatrix} \begin{pmatrix}r'&0\\ b'&c'\end{pmatrix} \\ {}= \begin{pmatrix}rr'&0\\ b\overline{r'}+cb'&cc'\end{pmatrix}. \end{gathered} \] The bar means reduction modulo \(2\). The missing upper-right entry is zero because an element of order dividing \(2\) cannot have a nonzero image in \(\mathbf Z\).
A map \(U\to X\) is a pair \((x,y)\in X\oplus X[2]\), where \(X[2]=\{y: 2y=0\}\). Precomposition gives \[ (x,y)\begin{pmatrix}r&0\\ b&c\end{pmatrix} =(rx+by,cy). \tag{6.1} \] The scalar \(b\) acts on \(y\) through \(\mathbf F_2\), so this expression is well defined. Applying it twice gives the displayed matrix product in exactly its specified order.
If \(a: X\to Y\) is a nonzero homomorphism, some \(x\in X\) has \(a(x)\ne0\). Send \(1\) in the \(\mathbf Z\)-summand of \(U\) to \(x\), and send the other summand to zero. Its composite with \(a\) is nonzero. Subtracting two different maps proves faithfulness, hence the generator assertion.
Let \(e\) be the matrix with diagonal \((1,0)\), and \(n\) the matrix with lower-left entry \(1\) and other entries zero. Then \(ne=n\) and \(en=0\). Thus \(S\) is noncommutative.
Here \(\Gamma(\mathbf Z)=\mathbf Z\oplus0\), whereas \(\Gamma(\mathbf Z/2)=\mathbf F_2\oplus\mathbf F_2\). The map is \[ \Gamma(q)(x,0)=(\overline x,0). \] It misses \((0,1)\) and is not epic in right \(S\)-modules, where epimorphisms are surjections. Equivalently the map \(U\to\mathbf Z/2\) that is identity on its second summand does not lift to \(\mathbf Z\); all maps from that summand to \(\mathbf Z\) vanish. Thus \(U\) is not projective.
Nevertheless \(F\) is exact and \(\Gamma\) is fully faithful by §3. The theorem does not claim that \(\Gamma\) preserves epimorphisms. This example uses precisely the endomorphism-ring action (6.1); forgetting it would lose the full-faithfulness assertion.
Exercise 2 (moderate: a nonzero module erased by \(F\)). Let \(k\) be any field and \(U=k^{(\mathbf N)}\) in \(k\)-vector spaces, with \(S=\operatorname{End}_k(U)\). Let \(D\subseteq S\) consist of the finite-rank operators. Prove that \(D\) is a two-sided ideal and that \(S/D\ne0\), but \(F(S/D)=0\). Determine the image of the canonical coproduct map \[ \bigoplus_{n\in\mathbf N}\Gamma(k) \longrightarrow\Gamma\!\left(\bigoplus_{n\in\mathbf N}k\right)=S. \tag{6.2} \]
Solution. The object \(U\) is a generator: a nonzero map of vector spaces is nonzero on some vector \(v\), and the map from \(U\) sending its first basis vector to \(v\) and the others to zero detects it. A sum of finite-rank operators has image in the sum of their finite-dimensional images. Precomposition and postcomposition of a finite-rank operator remain finite rank. Thus \(D\) is a two-sided ideal. The identity of the infinite-dimensional \(U\) is not finite rank, so its class in \(S/D\) is nonzero.
For a finite-dimensional subspace \(W\subseteq U\), choose a projection \(e_W: U\to W\subseteq U\). Its associated right ideal is \[ e_WS=\{s\in S: \operatorname{im}s\subseteq W\}. \tag{6.3} \] The inclusion from left to right follows from the projection's image. For the converse, \(e_Ws=s\) whenever \(s\) has image in \(W\). The right side of (6.3) is independent of the chosen projection.
The right-linear idempotent \(s\mapsto e_Ws\) on \(S\) has image \(e_WS\). Under \(F(S)=U\), its image under \(F\) is the operator \(e_W\), by the free-module identification. An additive functor preserves the injection and retraction of a split image, so \[ F(e_WS)=W \] with its specified inclusion into \(F(S)=U\). This identification is natural under \(W\subseteq W'\): composing either proposed map into \(U\) gives the inclusion of \(W\), and the inclusion of \(W'\) is monic. It is also independent of the projection, because it identifies the same right ideal's map into \(S\) with the same subspace's map into \(U\).
Finite-dimensional subspaces form a directed set under inclusion, and \[ D=\operatorname{colim}_{W}e_WS. \] Every finite-rank operator belongs to the ideal for its image; the reverse inclusion is immediate. Preservation of colimits by \(F\) now identifies \(F(D)\to F(S)\) with \[ \operatorname{colim}_{W}W\longrightarrow U. \] Every vector lies in a finite-dimensional subspace, so this is an isomorphism. Right exactness on \(D\to S\to S/D\to0\) gives \(F(S/D)=0\).
Finally a member of the source of (6.2) is a finite family of linear functionals \(U\to k\), placed in finitely many output coordinates. It therefore gives an operator with finite-coordinate image, hence finite rank. Conversely a finite-dimensional subspace of \(k^{(\mathbf N)}\) lies in a finite-coordinate subspace: take a finite basis and unite its finite supports. Thus every finite-rank operator has only finitely many nonzero output coordinate functionals and occurs in (6.2). The map is injective, since its coordinates recover each functional, and its image is exactly \(D\). It misses the identity. This gives an explicit failure of coproduct preservation by \(\Gamma\), and explains why the arbitrary-module comparison is a quotient rather than necessarily a module-category equivalence.
Exercise 3 (hard: identify the quotient and all its local objects). In the general setting of this lesson, let \[ \mathcal T=\{M\in\mathsf{Mod}\text{-}S: F(M)=0\}. \] Prove that \(\mathcal T\) is Serre and closed under small colimits. Show that \(F\) induces an equivalence \[ \overline F: (\mathsf{Mod}\text{-}S)/\mathcal T \simeq\mathcal C. \tag{6.4} \] Identify the local modules and the kernel of \(\eta_M\).
Solution. Exactness of \(F\) takes a short exact sequence to a short exact sequence. If its middle term is killed, so are its subobject and quotient; if both ends are killed, so is its middle term. This is Serre closure. Preservation of colimits shows closure under arbitrary small colimits of killed modules. The Serre-quotient theorem from the prerequisite therefore gives an exact quotient functor \(q\), with small Hom sets since a module has a set of submodules. Exact \(F\) descends to the exact functor \(\overline F\) in (6.4).
The first triangle in (2.3), together with invertibility of \(\epsilon\), makes \(F(\eta_M)\) invertible. Exactness then puts \(\ker\eta_M\) and \(\operatorname{coker}\eta_M\) in \(\mathcal T\). Thus \(q\eta_M\) is invertible.
Call a module \(L\) local if \(\operatorname{Hom}_S(-,L)\) takes every map with kernel and cokernel in \(\mathcal T\) to a bijection. For \(L=\Gamma(A)\), adjunction identifies this precomposition map with precomposition by the corresponding \(F(d)\). Exactness makes \(F(d)\) invertible, so \(\Gamma(A)\) is local. The local Hom formula of the Serre-quotient lesson gives \[ \begin{gathered} \operatorname{Hom}_{(\mathsf{Mod}\text{-}S)/\mathcal T}(qX,q\Gamma(A)) \\ \simeq\operatorname{Hom}_S(X,\Gamma(A)). \end{gathered} \tag{6.5} \] In this formula the map induced by \(q\) is the specified bijection.
Replacing \(qM,qN\) by the isomorphic \(q\Gamma F(M),q\Gamma F(N)\), (6.5) and full faithfulness of \(\Gamma\) identify their Hom set with \(\operatorname{Hom}_{\mathcal C}(F(M),F(N))\). This bijection is the one induced by \(\overline F\): for a map between the \(\Gamma\)-values, the counit identifies its image under \(F\) with the unique corresponding map in \(\mathcal C\); naturality of \(\eta\) and \(\epsilon\) gives the same identification after the replacements. Hence \(\overline F\) is fully faithful. Every \(A\) is isomorphic to \(F\Gamma(A)\), so it is essentially surjective. This proves (6.4).
For a local \(M\), apply locality to the denominator \(\eta_M: M\to\Gamma F(M)\). There is a unique map \(a: \Gamma F(M)\to M\) with \(a\eta_M=1_M\). The target \(\Gamma F(M)\) is also local. The maps \(\eta_Ma\) and \(1_{\Gamma F(M)}\) agree after precomposition with \(\eta_M\), so locality gives \(\eta_Ma=1\). Thus \(\eta_M\) is invertible and \(M\) is a \(\Gamma\)-value up to isomorphism. Conversely all such values were proved local. They are exactly the local modules.
If \(T'\subseteq M\) is killed by \(F\), adjunction gives \[ \begin{gathered} \operatorname{Hom}_S(T',\Gamma F(M)) \\ \simeq\operatorname{Hom}_{\mathcal C}(F(T'),F(M))=0. \end{gathered} \] Therefore \(T'\subseteq\ker\eta_M\). The kernel itself is killed, so it is the largest \(\mathcal T\)-submodule of \(M\). The unit's cokernel can still be killed and nonzero. Removing that largest submodule need not make \(M\) local; the full saturation is \(\Gamma F(M)\).
Exercise 4 (expert: when the quotient disappears). Prove that the adjunction \(F\dashv\Gamma\) is an equivalence of categories if and only if \(U\) is projective and the canonical map \[ \begin{gathered} \bigoplus_i\operatorname{Hom}_{\mathcal C}(U,A_i) \\ \longrightarrow \operatorname{Hom}_{\mathcal C}\!\left(U,\bigoplus_i A_i\right) \end{gathered} \tag{6.6} \] is bijective for every small family. State exactly where each hypothesis is used, and check the failures in Exercises 1–2.
Solution. Suppose the two properties hold. Projectivity means that every map \(U\to B\) lifts across every epimorphism \(A\to B\). Thus \(\Gamma\) takes epimorphisms to surjective module maps. Together with its left exactness, this makes \(\Gamma\) exact. The bijections (6.6) are right-linear and compatible with the injections, so \(\Gamma\) preserves coproducts as module coproducts.
For every free module \(S^{(I)}\), (6.6), \(F(S)=U\), and (2.2) identify \[ \eta_{S^{(I)}}: S^{(I)} \xrightarrow{\ \sim\ }\Gamma(U^{(I)}). \] This is where the coproduct hypothesis is used. For a free presentation (2.1), both the identity functor and \(\Gamma F\) take its cokernel to the cokernel of its first map: \(F\) is right exact, and \(\Gamma\) is exact by projectivity. Naturality of \(\eta\) gives a diagram of these cokernel presentations, whose two free-module components are isomorphisms. The induced map on cokernels is therefore an isomorphism. It is \(\eta_M\). This is where projectivity is used, and no finite-presentation restriction has been imposed.
The unit and the already invertible counit are now isomorphisms on all objects, so \(F,\Gamma\) are inverse equivalences.
Conversely suppose they are equivalences. An equivalence preserves epimorphisms and coproducts by their categorical universal properties. Thus \(\Gamma(A)\to\Gamma(B)\) is surjective for every epimorphism \(A\to B\), because epimorphisms of modules are surjective. Elements of these modules are maps from \(U\), proving projectivity. Preservation of coproducts gives (6.6), since the underlying abelian group of a module coproduct is the direct sum of its groups. The preserved injection maps give exactly the canonical map, rather than merely some abstract group isomorphism.
Exercise 1 fails projectivity: the specified map from the \(\mathbf Z/2\)-summand cannot lift across \(q\). Its \(U\) does satisfy (6.6), since a map from each of the two finitely generated summands has finite support in a coproduct, and their two finite supports have finite union. Exercise 2 has projective \(U\): maps from a free vector space lift across surjections by lifting each basis vector. It fails (6.6), since the identity is outside the image of (6.2). These failures occur separately. Neither property follows from being a generator alone.
7. References and scope
The ordinary Gabriel–Popescu theorem and the injection-to-adjoint-monomorphism lemma are retained through Stacks, Section 19.14. Generators and Grothendieck categories are also treated in Pierre Schapira, An Introduction to Categories and Homological Algebra, lecture notes, version of 1 March 2026, Section 5.4.
The tensor construction is the previously proved internal-object construction, including all presentation comparisons. The quotient uses the previously retained Serre-quotient interface. For the ordinary abelian category of sheaves of modules, see the existing Sheaves of modules on a ringed space, Section 2, Theorem 2.1. This lesson uses ordinary categories, modules and exact sequences throughout.