Draft lesson.

Injective cogenerators and faithful module realizations

Original exposition: CC0 1.0.

A generator tests maps by arrows into their source. A cogenerator tests them by arrows out of their target. An injective cogenerator combines this separation with extension across every monomorphism. It gives canonical embeddings into products, and, after reversing the arrows of an abelian completion, provides the projective object used to realize a small abelian category inside a module category.

Fix a Grothendieck universe $\mathcal U$, with choice in an ambient enlargement. Hom sets, rings, indexing categories and families called small belong to $\mathcal U$. Categories may have an ambient set of objects larger than $\mathcal U$; choices over that set are made in the enlargement. A Grothendieck category is a locally small abelian category with all small colimits, exact small filtered colimits and a generator. A generator means an object whose covariant Hom functor is faithful.

We use the full proofs of small subobjects and quotient families, solution sets and adjoints, enough injectives, and abelian Ind completion. Their exact interfaces will be specified where they enter.

1. An injective object that separates all maps

An object $K$ of an abelian category $\mathcal A$ is a cogenerator if $\operatorname{Hom}_{\mathcal A}(-,K)$ is faithful: distinct arrows $f,g: X\to Y$ can be distinguished by some $t: Y\to K$, with $tf\ne tg$. This concerns all pairs of arrows, rather than only nonzero objects.

Theorem 1.1. Every Grothendieck category has an injective cogenerator. For such a cogenerator, a morphism $f$ is invertible if and only if $\operatorname{Hom}(f,K)$ is bijective.

Proof. Choose a generator $G$. Subobjects of $G$ form a small set, by the cited small-subobject theorem. Every quotient is the cokernel of a subobject, so choose a small representative family $(Q_j)_{j\in J}$ of the quotients of $G$, including zero. Form $S=\bigoplus_{j\in J}Q_j. \tag{1.1}$ Each summand inclusion has the retraction that is the identity on that summand and zero on all others. By the full enough-injectives theorem, embed $S$ into an injective $K$. Every quotient of $G$ consequently embeds into $K$.

First, $\operatorname{Hom}(X,K)=0$ implies $X=0$. An arrow $a: G\to X$ has image $Q$, a quotient of $G$. Embed $Q$ into $K$ and extend across $Q\hookrightarrow X$. The extension is zero by the hypothesis. Its restriction is a monomorphism, so $Q=0$ and $a=0$. All maps $G\to X$ vanish. Faithfulness of $\operatorname{Hom}(G,-)$, applied to $1_X$ and $0_X$, gives $1_X=0_X$, hence $X=0$.

For full separation, take a nonzero $d: X\to Y$, with nonzero image $Q$. The preceding result supplies a nonzero $b: Q\to K$. Extend it across $i: Q\hookrightarrow Y$ to $t: Y\to K$. Write $d=ip$, with $p: X\twoheadrightarrow Q$. Then $td=bp\ne0$, since cancellation of the epimorphism would otherwise force $b=0$. Apply this to $d=f-g$. Thus $K$ cogenerates.

Hom into an injective object is contravariantly exact: restriction across a monomorphism is surjective, and its kernel consists precisely of maps factoring through the quotient. Factor $f: X\to Y$ through its image to obtain $\begin{gathered} 0\longrightarrow\operatorname{Hom}(\operatorname{coker}f,K) \longrightarrow\operatorname{Hom}(Y,K) \xrightarrow{f^*}\operatorname{Hom}(X,K)\\ \longrightarrow\operatorname{Hom}(\ker f,K)\longrightarrow0. \end{gathered} \tag{1.2}$ The last map restricts to the kernel. Its kernel consists of maps from the image, each of which extends to $Y$. If $f^*$ is bijective, (1.2) and the zero-object test make both kernel and cokernel of $f$ vanish. The abelian image–coimage comparison makes $f$ invertible. The converse follows by composing with its inverse. $\square$

The construction uses a small set of quotients of one object and one injective extension. It does not use a product indexed by all objects.

2. Representations, limits and left adjoints

First account for the required limits. Subobjects of every Grothendieck object form a small set. Ordinary quotient classes are small as well: an epimorphism is the cokernel of its kernel and is determined, up to its specified quotient isomorphism, by that subobject. The category is thus co-wellpowered, cocomplete and has a separating generator. The complete solution-set construction in Theorem 3.1 represents every contravariant functor sending colimits to limits. Theorem 4.2 then constructs every small limit: its functor of cones is a small limit of represented Hom functors and sends colimits to limits by interchange of compatible tuples. In particular all small products exist. Projectivity of the generator is not assumed.

Theorem 2.1. Let $\mathcal A$ be Grothendieck.

A functor $H: \mathcal A\to\operatorname{Set}_{\mathcal U}$ is covariantly representable if and only if it preserves all small limits.
For any locally small $\mathcal B$, a functor $R: \mathcal A\to\mathcal B$ has a left adjoint if and only if it preserves all small limits.

In the second assertion, preservation means that the image of each limit cone is a limit cone in $\mathcal B$; completeness of all of $\mathcal B$ is not assumed.

Proof. By the limit existence just established, $\mathcal A^{\mathrm{op}}$ is cocomplete. It is co-wellpowered because its quotient classes are subobject classes in $\mathcal A$. Theorem 1.1 supplies the separating generator $K$ there. Apply the separating-family solution-set theorem to this opposite category and its contravariant functor $H$. It gives $T$ and natural bijections $H(X)\simeq\operatorname{Hom}_{\mathcal A}(T,X). \tag{2.1}$ Conversely this Hom functor preserves limits by their defining property. No additivity of $H$ is needed.

If $R$ preserves limits, set $H_B(X)=\operatorname{Hom}_{\mathcal B}(B,RX) \qquad(B\in\mathcal B). \tag{2.2}$ These sets are small. The functor preserves limits, because maps from $B$ into a limit are exactly compatible families. Choose representatives $L(B)$. A map $b: B\to B'$ gives $H_{B'}\to H_B$ by precomposition. Covariant Yoneda gives a unique $L(b): L(B)\to L(B')$ implementing it. Uniqueness proves identities and composition. The bijections $\operatorname{Hom}_{\mathcal A}(L(B),X) \simeq\operatorname{Hom}_{\mathcal B}(B,RX) \tag{2.3}$ are natural in $X$ by representation and in $B$ by construction. They give $L\dashv R$. The unit corresponds to $1_{L(B)}$, and the counit at $X$ corresponds to $1_{RX}$.

Conversely an adjunction identifies maps $B\to R(\lim_iX_i)$ with maps $L(B)\to\lim_iX_i$, and hence with compatible families of maps $B\to RX_i$, naturally in $B$. Therefore the image cone is a limit cone, including for the empty diagram. $\square$

Corollary 2.2. If $\mathcal C$ is a small abelian category, $\operatorname{Ind}_{\mathcal U}(\mathcal C)$ has an injective cogenerator.

Proof. The full Ind-completion theorem cited above proves local smallness, abelianity, small colimits, AB5 and a generator for this small base. Theorem 1.1 applies to this entire Grothendieck interface. The same completion lesson proves that the constant embedding is exact and fully faithful. $\square$

3. Evaluation into products and entire diagrams

Theorem 3.1. Choose an injective cogenerator $K$ in a Grothendieck category $\mathcal A$. There is an ordinary functor $\Psi$ to its full subcategory of injective objects and a natural monomorphism $\delta: 1_{\mathcal A}\to\Psi$, given by $\begin{gathered} \Psi(X)=\prod_{t\in\operatorname{Hom}(X,K)}K,\\ \pi_t\delta_X=t. \end{gathered} \tag{3.1}$

Proof. The Hom index is small and the products exist by §2. A small product of injectives is injective: extend each component of a map from a subobject, choose the small family of extensions, and combine them by the product property. The empty product is zero, which is injective.

Product universality determines $\delta_X$. If $a,b: Y\to X$ agree after it, every $t: X\to K$ has $ta=tb$, so separation gives $a=b$. Thus $\delta_X$ is monic. Define $\Psi(f): \Psi(X)\to\Psi(Y)$ for $f: X\to Y$ by $\pi_t\Psi(f)=\pi_{tf} \qquad(t: Y\to K). \tag{3.2}$ An identity leaves the indices unchanged. For $g: Y\to Z$, the $t$-component of $\Psi(g)\Psi(f)$ is $\pi_{tg}\Psi(f)=\pi_{tgf}$. Product uniqueness proves both functor laws. It also gives $\Psi(f)\delta_X=\delta_Yf, \tag{3.3}$ since every component is $tf$. Thus the monomorphisms are natural, and every value is injective. No additivity is asserted. $\square$

Corollary 3.2. For a small category $I$ and diagram $\alpha: I\to\mathcal A$, the diagram $\beta=\Psi\alpha$ takes values in injectives, and $\delta\alpha: \alpha\to\beta$ is a monomorphism in $\operatorname{Fun}(I,\mathcal A)$.

Proof. Equation (3.3) supplies all naturality squares. Each component is monic. If two natural transformations into $\alpha$ become equal after it, componentwise cancellation makes them equal at every object, so they are equal transformations. This is monicity in the functor category. Equivalently its pointwise kernel is zero; kernels are computed pointwise by their unique universal factors. $\square$

This proves injectivity of every value in $\mathcal A$. Injectivity of the entire diagram as an object of its functor category is a separate assertion and is not used.

4. One projective object for a small family

Let $\mathcal D$ be locally small and abelian, with small coproducts and a projective generator $G$. For every $X$, the evaluation map $e_X: \bigoplus_{a\in\operatorname{Hom}(G,X)}G\longrightarrow X \tag{4.1}$ is epic. Two arrows out of $X$ agreeing after it agree after every $G\to X$, hence agree by the faithful generator test. No preservation of arbitrary coproducts by Hom from $G$ is needed.

Lemma 4.1. For a small family $(X_j)_{j\in J}$, there is a projective generator $P$ admitting an epimorphism onto each $X_j$.

Proof. Take $P=G\oplus\bigoplus_{\substack{j\in J\\a\in\operatorname{Hom}(G,X_j)}}G. \tag{4.2}$ The total family is small, because $J$ and every Hom set are small. A coproduct of projectives is projective: lift separately on each summand across a given epimorphism, choose the small family of lifts, and use the coproduct property. The displayed $G$-summand retract makes $P$ a generator. In detail, a map $G\to X$ distinguishing two arrows extends to a map $P\to X$ by composing with that retraction.

For a specified $j$, use evaluation on its block of summands and zero on the rest. The resulting $P\to X_j$ is epic because its restriction to the block is (4.1). For $J=\varnothing$, the formula gives $P=G$; it does not produce a zero object that might fail to generate. $\square$

Only small coproducts are required; the hypothesis of all small colimits would be stronger.

5. The module comparison on finite quotients

For this section, assume only a locally small abelian category with a projective generator $G$. Small coproducts are not required; finite biproducts already exist by abelianity.

Set $S=\operatorname{End}_{\mathcal D}(G)$, with multiplication $ab=a\circ b$, and $R=S^{\mathrm{op}}$. Define $\begin{gathered} \Phi(X)=\operatorname{Hom}_{\mathcal D}(G,X),\\ a^{\mathrm{op}}\cdot x=x\circ a. \end{gathered} \tag{5.1}$ It is a left $R$-module: $a^{\mathrm{op}}b^{\mathrm{op}}=(ba)^{\mathrm{op}}$, and successive actions give $(x\circ b)\circ a=x\circ(ba)$. Postcomposition makes $\Phi$ an additive functor. It is faithful because $G$ generates. It is exact because Hom from any object preserves kernels and projectivity supplies surjectivity on epimorphisms.

Lemma 5.1. If $X$ is a quotient of $G^{\oplus m}$ for a finite $m\ge0$, then for every $Y$ the comparison $\begin{gathered} \operatorname{Hom}_{\mathcal D}(X,Y)\longrightarrow \operatorname{Hom}_R(\Phi(X),\Phi(Y)),\\ f\longmapsto\Phi(f) \end{gathered} \tag{5.2}$ is bijective.

Proof. The module $\Phi(G)=S$ is the regular left $R$-module under $b\mapsto b^{\mathrm{op}}$. A linear map out of it is determined by its value at $1_G$; a map $y: G\to Y$ gives $b\mapsto y\circ b$. Thus the comparison is bijective for $G$, and finite biproducts make it bijective for $G^{\oplus m}$, including $m=0$.

Choose $p: G^{\oplus m}\twoheadrightarrow X$, with kernel inclusion $n: N\hookrightarrow G^{\oplus m}$. For a linear $v: \Phi(X)\to\Phi(Y)$, the finite comparison gives $h: G^{\oplus m}\to Y$ with $\Phi(h)=v\Phi(p)$. Then $\Phi(hn)=v\Phi(pn)=0. \tag{5.3}$ Faithfulness gives $hn=0$. Since $p$ is the cokernel of $n$, a unique $f: X\to Y$ satisfies $fp=h$. Exactness makes $\Phi(p)$ surjective, so $\Phi(f)\Phi(p)=v\Phi(p)$ gives $\Phi(f)=v$. Faithfulness also gives injectivity of the comparison. $\square$

The kernel $N$ need not itself be a finite quotient of $G$. Only faithfulness on the arrow $hn$ was used. Infinite free-module identifications are unnecessary.

6. Realizing a small abelian category as modules

Theorem 6.1 (Freyd–Mitchell). For a small abelian category $\mathcal C$, there are a small unital ring $R$ and an additive, exact, fully faithful functor $F: \mathcal C\longrightarrow R\text{-}\operatorname{Mod}_{\mathcal U}. \tag{6.1}$ Its essential image is a full abelian subcategory with ambient kernels and cokernels for arrows between its objects.

Proof. The opposite category is small and abelian. By the full completion theorem and Corollary 2.2, $\mathcal E=\operatorname{Ind}_{\mathcal U}(\mathcal C^{\mathrm{op}}) \tag{6.2}$ is locally small and Grothendieck, with an injective cogenerator $K$. Section 2 gives all small products. Hence $\mathcal D=\mathcal E^{\mathrm{op}}$ is abelian, has small coproducts, and $K$, regarded there, is a projective generator: lifting across epimorphisms reverses extension across monomorphisms, and its faithful covariant Hom reverses the original faithful contravariant Hom.

Reversing the exact fully faithful constant embedding gives $j: \mathcal C\to\mathcal D$, exact and fully faithful. The objects $jX$ for $X\in\mathcal C$ form a small family. Lemma 4.1 supplies a projective generator $P$ admitting an epimorphism onto each. Its endomorphism set is small by local smallness, so $R=\operatorname{End}_{\mathcal D}(P)^{\mathrm{op}} \tag{6.3}$ is a small ring. Use $P$ in place of $G$ in §5 and set $F=\Phi j$. Both functors are additive and exact. Every $jX$ is a quotient of a single copy of $P$, so Lemma 5.1 and fullness of $j$ give bijective comparisons on every pair of original objects.

An arrow between objects in the essential image, after conjugation by the defining isomorphisms, is $F(f)$ for a unique $f$ in $\mathcal C$. Exactness identifies its ambient kernel and cokernel with $F(\ker f)$ and $F(\operatorname{coker}f)$. Additivity preserves finite biproducts. Thus this full essential image has the same abelian structures and is equivalent to $\mathcal C$. $\square$

There is no claim of equivalence with the entire module category or of a projective generator inside the original category. Nor is closure under all ambient submodules asserted.

7. Two explicit calculations

Example 7.1 (evaluation can fail additivity). In vector spaces over $k=\mathbb F_2$, take $K=k$. Every map from a subspace to $k$ extends by extending a basis, so $k$ is injective; linear functionals distinguish maps, so it cogenerates. For $X=k$, the Hom labels are $0,1$, and $\begin{gathered} \Psi(k)=k^2,\qquad\delta_k(x)=(0,x),\\ \Psi(0)(u_0,u_1)=(u_0,u_0). \end{gathered} \tag{7.1}$ The last arrow is nonzero: it sends $(1,0)$ to $(1,1)$. Thus $\Psi$ is not generally additive. Nevertheless $\Psi(0)\delta_k=0=\delta_k0$, exactly as naturality requires. The zero label names an independent product coordinate; arbitrary vectors in $\Psi(k)$ need not have zero there.

Example 7.2 (the opposite ring gives the ordinary action). For any small unital ring $D$, let $\mathcal D=D\text{-}\operatorname{Mod}$, with $G={}_DD$. Every endomorphism is $r\mapsto rd$. Composing the maps labelled $d,e$ gives the label $ed$, so $\operatorname{End}(G)=D^{\mathrm{op}}$ and the ring in (5.1) is $D$. Evaluation at $1$ identifies $\Phi(X)$ with $X$, with inverse $x\mapsto(r\mapsto rx)$. Precomposition with right multiplication by $d$ evaluates to $f(d)=df(1)$, the original left action. Lemma 5.1 therefore recovers the usual comparison on finitely generated left modules, without commutativity or finite relations.

8. Four graded exercises with complete solutions

Exercise 1 (standard: the two halves of the injective test). For an injective cogenerator $K$ in an abelian category and $f: X\to Y$, prove that $f^*: \operatorname{Hom}(Y,K)\to\operatorname{Hom}(X,K)$ is injective exactly when $f$ is epic, and surjective exactly when $f$ is monic. Identify the use of injectivity.

Solution. The kernel of $f^*$ is $\operatorname{Hom}(\operatorname{coker}f,K)$, by the cokernel property. Cogeneration makes it zero exactly when $\operatorname{coker}f=0$, which is epimorphy. Sequence (1.2) identifies its cokernel with $\operatorname{Hom}(\ker f,K)$, which is zero exactly when $\ker f=0$, or monicity. That sequence uses extension of maps on the image to $Y$ and maps on the kernel to $X$, both consequences of injectivity. Directly, a monomorphism lets every $X\to K$ extend to $Y$. Cogeneration alone does not supply that extension.

Exercise 2 (advanced: the enlarged generator need not be compact). In $k$-vector spaces take $G=k$, $P=k^{(\mathbb N)}$. Prove that $P$ is a projective generator but its Hom functor does not preserve the coproduct of countably many copies of $k$. Explain why Lemma 5.1 still applies.

Solution. A vector-space map across a surjection lifts by choosing lifts of basis vectors and extending linearly. Thus $P$ is projective. Its first basis vector gives a split copy of $k$, so maps from $P$ distinguish any nonzero linear map by choosing a distinguishing vector on that summand and zero on the others.

A member of $\bigoplus_n\operatorname{Hom}(P,k)$ has only finitely many nonzero component maps. Its image under the coproduct comparison has range in a fixed finite-dimensional coordinate subspace of $k^{(\mathbb N)}=P$. The identity of $P$ is outside this image, since its range contains every basis vector. The comparison is not surjective. Lemma 5.1 uses only finite biproducts of $P$, for which every additive Hom functor has the required identification. Compactness of $P$ is not required.

Exercise 3 (expert: coherent modules from finite projective presentations). Let $\mathcal D$ be locally small and abelian with a projective object $G$. Suppose every object is a quotient of a finite biproduct of copies of $G$. Prove that $R=\operatorname{End}(G)^{\mathrm{op}}$ is left coherent and that $\Phi$ gives an exact equivalence with coherent left $R$-modules. A coherent module is finitely generated and has every finitely generated submodule finitely presented; a ring is left coherent if its finitely generated left ideals are finitely presented.

Solution. The finite covers make $G$ a generator: maps out of $X$ agreeing after all maps from $G$ agree after a chosen epic finite cover. Thus $\Phi$ is faithful and exact, and Lemma 5.1 makes it fully faithful on every object.

Choose $G^{\oplus m}\twoheadrightarrow X$. Its kernel $N$ also has a finite cover $G^{\oplus n}\twoheadrightarrow N$. Applying the exact functor gives $R^n\longrightarrow R^m\longrightarrow\Phi(X)\longrightarrow0. \tag{8.1}$ Conversely a finite presentation $R^n\to R^m\to M\to0$ lifts its first arrow to $G^{\oplus n}\to G^{\oplus m}$ by the finite comparison. Its cokernel $X$ has $\Phi(X)\simeq M$. Thus the essential image is exactly the finitely presented modules.

Let $L\subseteq R^m$ be finitely generated. A surjection $R^n\to L$, followed by inclusion, lifts to $u: G^{\oplus n}\to G^{\oplus m}$. Exactness gives $\Phi(\operatorname{im}u)\simeq L$, so (8.1) makes $L$ finitely presented. Taking $m=1$ proves left coherence, and the argument gives the stronger property for every finitely generated submodule of a finite free module.

That stronger property makes every finitely presented $M$ coherent. Write $M=R^m/N$ with $N$ finitely generated. For a finitely generated $L\subseteq M$, its inverse image $E\subseteq R^m$ is generated by generators of $N$ and chosen lifts of generators of $L$. It is therefore finitely presented. Its quotient $L=E/N$ is finitely presented: take a finite presentation of $E$ and add lifts of the finite generators of $N$ to its finite relation list. Conversely a coherent module is one of its own finitely generated submodules, hence is finitely presented by definition. The two classes coincide here, giving the stated exact equivalence. This includes the zero-category case $G=0$ and the zero ring with its sole unital module.

Exercise 4 (expert: all products in a small category force thinness). Prove that a small category with every small product has at most one arrow between each pair of objects and is equivalent to a partially ordered set with an infimum for every subset, including the empty one.

Solution. Let $M$ be its small set of all arrows. If $a,b: X\to Y$ were distinct, the product $Y^M$ would give an injection $\begin{gathered} \mathcal P(M)\longrightarrow\operatorname{Hom}(X,Y^M),\\ A\longmapsto\text{the arrow with component }a\text{ on }A \text{ and }b\text{ elsewhere}. \end{gathered} \tag{8.2}$ Different subsets differ in a component, so their arrows differ. The target is a subset of $M$, contradicting Cantor's theorem. Indeed an injection $\mathcal P(M)\to M$ would give a surjection back by inversion on its image and the empty set elsewhere, contrary to the diagonal subset argument.

Choose one object per isomorphism class and order the classes by the existence of an arrow. It is well-defined, reflexive and transitive. Arrows in both directions are inverse because each endomorphism is the unique identity; this gives antisymmetry. The representative inclusion is fully faithful and essentially surjective, hence an equivalence with this ordered set.

The product of representatives of any subset exists by smallness. A class maps to its product exactly when it maps to every member, so the product is their greatest lower bound. The empty product is terminal and gives the greatest element, the empty infimum. A category with an empty product is necessarily nonempty.

9. Freely accessible further reading

Emily Riehl, Category Theory in Context, Chapters 2–4, for represented functors, limits and adjunctions. The special adjoint functor theorem explains the complete solution-set interface used in §2.
The Stacks Project, Injectives in Grothendieck categories, for injective existence, whose full construction is provided by the linked course lesson.
The Stacks Project, Embedding abelian categories, for a further account of the embedding theorem. Its route through sheaves differs from the opposite-Ind proof in §6.

Referenced works and linked prerequisite lessons retain their stated licences.