Testing injectivity on generators and constructing enough injectives

Written and self-checked with GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original exposition: CC0. Referenced canonical proofs retain their stated licences.

An injective object extends maps from every subobject. In a Grothendieck category, a family of generators reduces that test to subobjects of the generators themselves. The reduction also explains the existence theorem: attach extensions for all these smaller tests, then continue through an ordinal long enough that each map appears before the end.

Work in a fixed universe with choice. Categories are locally small, all rings are unital, and modules are left modules unless a side is specified. A Grothendieck category is abelian, admits all small colimits, has exact filtered colimits and has a generator. A small family \((G_i)_{i\in I}\) generates when its separate Hom functors jointly detect isomorphisms. In an abelian category this agrees with joint faithfulness by Generators and small quotient families, Theorem 2.2; retain that complete equivalence, the presentations in §3 and the small-subobject proof in §4. Retain the complete extension and lifting tests in Extending maps, §3.

1. The set and union needed for a maximal extension

Let \(A\hookrightarrow B\) and \(h:A\to Z\) be fixed. Choose one representative for each subobject of \(B\). The retained small-subobject theorem makes these a small set; each representative has a small Hom set into \(Z\). Thus extensions \((Y,g)\), where \(A\subset Y\subset B\) and \(g:Y\to Z\) restricts to \(h\), form a set up to unique isomorphism over \(B\). Order them by inclusion with compatible restriction. Monicity into \(B\) makes each possible arrow between extensions unique, so this is a partial order on those classes. It is nonempty, since \((A,h)\) is an extension.

Here is the precise use of exact filtered colimits, or AB5. For a nonempty chain of subobjects \(Y_t\subset B\), the compatible rows

\[ 0\longrightarrow Y_t\longrightarrow B \longrightarrow B/Y_t\longrightarrow0 \tag{1.1} \]

give a monomorphism \(Y=\operatorname{colim}_tY_t\to B\). The colimit of the constant diagram \(B\) is \(B\): a cocone over a nonempty chain has all its component maps equal. The subobject \(Y\) contains every \(Y_t\) and has the union universal property by the colimit property. Compatible \(g_t\) give a unique \(g:Y\to Z\). Since each extends \(h\), so does \(g\). Hence \((Y,g)\) is a chain upper bound. The empty chain has the upper bound \((A,h)\). Choice and Zorn's lemma now give a maximal extension. This argument uses a set of extensions and a small filtered colimit, rather than applying Zorn's lemma to a class.

For later use, a filtered increasing union commutes with inverse images. Suppose \(B=\operatorname{colim}_{t}B_t\), with \(B_t\hookrightarrow B\), and \(f:M\to B\). Put \(P_t=M\times_BB_t\). The full signed-kernel construction in Exact squares, §1 gives

\[ 0\longrightarrow P_t\longrightarrow M\oplus B_t \xrightarrow{(f,-j_t)}B. \tag{1.2} \]

AB5 preserves this kernel. Here diagram exactness is the complete pointwise convention in Pointwise abelian structure, §2; an exact additive colimit functor preserves the kernel by the retained exactness tests. The colimit of \(M\oplus B_t\) is \(M\oplus B\), since the coproduct universal property identifies its compatible cocones with a single map out of \(M\) and a cocone on the \(B_t\). The last object is again the constant colimit \(B\). Consequently \(\operatorname{colim}P_t\) is the kernel of \((f,-1_B)\), which is isomorphic to \(M\) via its components \((1_M,f)\). This proves the inverse-image assertion categorically; it uses no elements of a general abelian category.

2. The criterion for an entire generator family

Theorem 2.1. For a Grothendieck category and any small generating family \((G_i)\), an object \(Z\) is injective if and only if every restriction map

\[ \operatorname{Hom}(G_i,Z)\longrightarrow \operatorname{Hom}(W,Z) \tag{2.1} \]

is surjective, for every \(i\) and every subobject \(W\hookrightarrow G_i\).

The single-generator theorem and its full maximal-extension proof are retained in the AI-integrated Stacks baseline, lemma-characterize-injective. The following proves the residual family version and its union and signed-pushout interfaces.

Proof. Necessity is exactly the retained extension test. For sufficiency, start with an arbitrary \(A\hookrightarrow B\) and \(h:A\to Z\), and take a maximal extension \((Y,g)\) from §1. If \(Y\hookrightarrow B\) is not invertible, some \(\phi:G_i\to B\) does not factor through \(Y\). Indeed all the maps \(\operatorname{Hom}(G_i,Y)\to\operatorname{Hom}(G_i,B)\) are injective; if they were all surjective, detection would make the inclusion invertible.

Form \(W=Y\times_BG_i\), with projections \(p:W\to Y\) and \(q:W\to G_i\). The latter is monic because \(Y\to B\) is monic. The assumption extends \(gp:W\to Z\) to \(\chi:G_i\to Z\), with \(\chi q=gp\).

Let \(u:Y\oplus G_i\to B\) have components the inclusion \(m:Y\to B\) and \(\phi\). Its kernel is the signed pair

\[ W\xrightarrow{(p,-q)}Y\oplus G_i. \tag{2.2} \]

To verify this identification, an arrow with components \(a:T\to Y\), \(b:T\to G_i\) is killed by \(u\) precisely when \(ma=\phi(-b)\). The pullback property gives a unique map \(T\to W\) with projections \(a,-b\); its composite with (2.2) is \((a,b)\). Thus the cokernel \(E\) of (2.2) is \(\operatorname{coim}u\), and the abelian comparison identifies it with \(\operatorname{im}u\subset B\). The complete coimage comparison, its universal maps and the abelian axiom are retained in Kernels and cokernels, §§1 and 3–4.

The map \((g,\chi):Y\oplus G_i\to Z\) kills (2.2), since \(gp-\chi q=0\). It therefore factors uniquely through \(E\). Both \(Y\to B\) and \(\phi\) factor through \(E\subset B\), and the new map \(E\to Z\) restricts to \(g\) on \(Y\). The arrow \(Y\to E\) is monic, since its composite with \(E\hookrightarrow B\) is monic. It cannot be invertible, because then \(\phi\) would factor through \(Y\). This is a strictly larger extension, contradicting maximality. Hence \(Y=B\), and \(h\) extends. The full extension test gives injectivity. \(\square\)

This proof tests each generator separately. It does not assume that extension from subobjects of each \(G_i\) immediately gives extension from arbitrary subobjects of their coproduct. The coproduct of the family is nevertheless a generator: its Hom functor is the product of the separate Hom groups, and equality in that product tests each component, so it is faithful. The retained abelian equivalence then gives detection. An empty generating family forces every object to be zero: the map \(0\to X\) passes all empty tests and therefore is invertible. The theorem includes that zero category.

3. Ordinary modules and the field case

For any unital, possibly noncommutative ring \(R\), retain the full arbitrary-left-module abelian and small-colimit construction in Kernels and cokernels, §5, and the full general filtered-colimit exactness proof in Hom, tensor and module limits, §§3–4. Evaluation \(\operatorname{Hom}_R(R,M)\simeq M\) at \(1\), with inverse \(m\mapsto(r\mapsto rm)\), makes \(R\) a generator. These are the complete Grothendieck hypotheses; no Noetherian or commutativity hypothesis enters.

Theorem 2.1 with this generator yields the left-sided Baer test: \(Z\) is injective exactly when every linear map from every left ideal \(J\subset R\) extends to \(R\). Subobjects of the left regular module are exactly left ideals. The canonical elementary (1)↔(3) proof in More on Algebra, lemma-characterize-injective-bis is retained; its separate Ext equivalence is not needed here. The displayed left-ideal interface and Theorem 2.1 supply the arbitrary noncommutative generality.

Free left modules are projective at arbitrary rank over every unital ring, without commutativity. Here is the complete finite-support lift. For a small set \(S\), a map \(R^{(S)}\to N\) is specified by elements \(n_s\). Given a surjection \(M\twoheadrightarrow N\), choose a preimage \(m_s\) for each \(n_s\). The rule

\[ \sum_{s\in S}r_se_s\longmapsto\sum_{s\in S}r_sm_s \tag{3.1} \]

defines a left-linear lift, since each input has finite support. The order of multiplication stays on the left. This also works for the empty basis. The retained projective lifting test proves projectivity. For any \(M\), the free module on its underlying set maps onto it by \(e_m\mapsto m\); each \(m\) is the image of its own basis vector. Thus there are enough projectives, at all ranks.

For enough injectives, retain Injective modules and bounded below derived functors, Lemmas 2.1–2.2, including the full Zorn/Baer, divisible-group and character-separation arguments, and the complete side and coordinate comparison in Character duals, Sections 1–3. In particular \(\mathbb Q/\mathbb Z\) is divisible by dividing rational representatives; the complete group and Baer proofs establish its injectivity. Lemma 2.2 retains the arbitrary associative unital ring action, every separation check and the naturality of the functorial embedding. Only these ordinary module lemmas are used; no sheaf or derived assertion is imported.

Finally, over a field every vector space is both injective and projective. Retain the full basis-extension proof in Locally nilpotent operators, §1 and the full semisimple and opposite lifting interface in Extending maps, §5. They cover all dimensions in the chosen universe, with choice; no natural basis or splitting is asserted.

4. Retain the stronger Grothendieck existence theorem

The AI-integrated Stacks theorem theorem-injective-embedding-grothendieck proves more than existence: every Grothendieck category admits functorial injective embeddings. Retain its complete proof together with proposition-objects-are-small and the complete Sets proposition on large cofinality. Here are the full structural checks that connect those proofs to this course's conventions.

All small coproducts in a Grothendieck category preserve short exact sequences. A coproduct is the filtered colimit of its finite partial biproducts: a compatible cocone on all finite subsets is exactly a choice of one map from each summand. For a finite family of maps \(f_s\), the sum of their kernel inclusions is the kernel of \(\bigoplus f_s\): a map into the finite biproduct is killed precisely when every projected component is killed, and the component kernel factors combine uniquely. Reversing arrows gives the component cokernel assertion. Thus finite biproducts are exact by the complete kernel/cokernel tests in Testing exactness, §3. AB5 gives the asserted exactness of the arbitrary coproduct. This includes the empty coproduct, which gives the zero row.

Choose a generator \(U\). Its subobjects form a small set by the retained small-subobject theorem. Consequently the canonical attachment diagram, indexed by all pairs \((N\subset U,\varphi:N\to M)\), uses small coproducts. The arrow from the coproduct of the \(N\)'s to the corresponding coproduct of copies of \(U\) is monic by the preceding exact-coproduct assertion.

Pushouts of monomorphisms are monomorphisms here with the required signs. For \(i:A\hookrightarrow B\) and \(f:A\to M\), take the cokernel of \((f,-i):A\to M\oplus B\). The latter arrow is monic. If \(k:K\to M\) is the kernel of the induced \(M\to P\), then \((k,0):K\to M\oplus B\) factors through \((f,-i)\), since this monomorphism is the kernel of its cokernel in an abelian category. Write that factor as \(a:K\to A\). Then \(ia=0\), so \(a=0\), and hence \(k=0\). Thus \(M\to P\) is monic. The cokernel universal property is exactly the pushout property, as retained in the signed construction of Exact squares. This proves that each attachment embeds its input.

At a limit stage of the canonical ordinal construction, every earlier term still embeds in the colimit. Fix an earlier stage \(\beta\). Its tail is cofinal: a cocone on the tail extends uniquely to the whole ordinal diagram by the transition maps. On that tail the monomorphisms from the constant term \(M_\beta\) are natural. Exact filtered colimits preserve their monicity, and the constant colimit is \(M_\beta\). This proves the required stage inclusion, rather than assuming it from the word “union.”

The smallness bound is uniform over every \(N\subset U\). Composition of subobjects with \(N\hookrightarrow U\) injects \(\operatorname{Sub}(N)\) into \(\operatorname{Sub}(U)\): an isomorphism over \(U\) is over \(N\) by cancellation. Thus \(|\operatorname{Sub}(N)|\leq\kappa=|\operatorname{Sub}(U)|\). For a chain \(B_\beta\) of monomorphisms indexed by a limit ordinal \(\lambda\) with \(\operatorname{cf}(\lambda)>\kappa\), a map \(N\to B=\operatorname{colim}_{\beta<\lambda}B_\beta\) has inverse-image subobjects whose colimit is \(N\), by (1.2). Choose one index for each distinct inverse-image subobject. There are at most \(\kappa\) choices, so their indices have an upper bound \(\gamma<\lambda\). Every inverse image is then contained in the one at \(\gamma\). Since their union is \(N\), that one is \(N\); the map factors through \(B_\gamma\). Factorizations that become equal in \(B\) already agree after moving to a common stage, since that stage embeds in \(B\). This supplies both directions of the canonical Hom-colimit comparison.

The complete Sets proof provides such a small limit ordinal in the chosen universe. Applied to the retained attachment construction, every map from a subobject of \(U\) therefore appears at some \(\gamma<\lambda\), and the next attachment extends it to \(U\). Since \(\lambda\) is a limit ordinal, \(\gamma+1<\lambda\). Theorem 2.1 proves injectivity of the terminal object. The retained construction is functorial because the attachment index sends \((N,\varphi)\) to \((N,u\varphi)\) for \(u:M\to M'\), and the pushout and limit universal maps preserve identities and composition. Its embedding is natural at each successor and limit stage. The ordinal bound depends on the fixed generator, not on the input object. This checks all the interfaces of the stronger canonical theorem; functoriality alone does not claim an additive embedding functor.

5. Four graded exercises with full solutions

Exercise 1 (introductory: one multiplication test misses an obstruction). Fix a prime \(p\), and let \(T=\mathbb Z[1/p]\) as an abelian group. Show that multiplication by \(p\) is onto, but \(T\) is not injective. Give a map from a subgroup of the generator \(\mathbb Z\) which does not extend.

Solution. Every \(a/p^n\) is \(p\) times \(a/p^{n+1}\). Choose a prime \(\ell\ne p\). Define \(h\colon\ell\mathbb Z\to T\) by \(h(\ell a)=a\). An extension \(\mathbb Z\to T\) would have value \(t\) at \(1\) satisfying \(\ell t=1\). Writing \(t=a/p^n\) would imply \(\ell a=p^n\), impossible since \(\ell\nmid p\). Thus the generator-subobject test fails. Equivalently the retained divisible/injective criterion requires every positive multiplication test, not just repeated powers of \(p\).

Exercise 2 (intermediate: retain both generators). Let \(K\) be a field and \(R\) its ring of upper triangular \(2\times2\) matrices. Put \(P_1=Re_{11}\), \(P_2=Re_{22}\). Prove that \((P_1,P_2)\) generates left modules. Show that \(Z=P_1\) passes every extension test on subobjects of \(P_1\) but fails a test on \(P_2\).

Solution. The decomposition \(R=Re_{11}\oplus Re_{22}\) is a decomposition of left modules, since right multiplication by the two idempotents supplies its projections. The Hom functor of \(R\) is faithful by evaluation, so the two Hom functors are jointly faithful and hence detecting by the retained abelian equivalence. The module \(P_1=Ke_{11}\) is simple: the diagonal scalar matrices act by arbitrary scalars, and a nonzero vector spans it. Its only subobjects are zero and itself, whose restriction tests are automatically onto for every target. Thus \(Z\) passes those tests.

Inside \(P_2=Ke_{12}\oplus Ke_{22}\) the submodule \(W=Ke_{12}\) is isomorphic to \(P_1\), by \(e_{11}\mapsto e_{12}\). Indeed left multiplication by \(\begin{pmatrix}a&b\\0&c\end{pmatrix}\) multiplies both \(e_{11}\) and \(e_{12}\) by \(a\). Let \(h:W\to Z\) send \(e_{12}\) to \(e_{11}\). Every linear \(f:P_2\to Z\) has \(f(e_{22})=e_{22}f(e_{22})=0\), since \(e_{22}\) acts as zero on \(Z\). Also \(f(e_{12})=e_{12}f(e_{22})=0\). Hence \(f=0\), so none extends \(h\). The second generator supplies an essential test, and \(Z\) is not injective.

Exercise 3 (hard: an infinite coproduct of injectives can fail). Let \(R=\prod_{n\geq1}K\) for a field \(K\), and let \(e_n\) be the coordinate idempotents. Give \(E_n=K\) the left action through the \(n\)-th coordinate. Show that each \(E_n\) and their product are injective, but \(\bigoplus_nE_n\) is not.

Solution. For any left module \(M\), restriction identifies \(\operatorname{Hom}_R(M,E_n)\) with \(\operatorname{Hom}_K(e_nM,K)\). The inverse to restriction sends a functional \(\alpha\) to \(m\mapsto\alpha(e_nm)\). It is linear because \(e_nrm=r_ne_nm\). These maps are inverse and natural. The functor \(M\mapsto e_nM\) preserves a short exact row: injectivity and the kernel equation restrict directly, and a lift of \(c=e_nc\) is replaced by \(e_nm\). The full field injectivity proof makes the contravariant Hom functor exact. Hence \(E_n\) is injective. The full product extension proof retained in Character duals, Lemma 2.1, applies on the left by passing to \(R^{op}\), so \(\prod E_n\) is injective as well.

Let \(J\subset R\) be the ideal of finitely supported tuples. It identifies with \(E=\bigoplus_nE_n\). The identity \(h:J\to E\) is linear. If it extended to \(f:R\to E\), put \(z=f(1)\); then \(f(r)=rz\). Since \(z\) has finite support, choose \(n\) outside that support. We would have \(f(e_n)=e_nz=0\), whereas \(h(e_n)=e_n\ne0\). No extension exists, so Baer's test makes \(E\) noninjective. Exactness of coproducts is therefore weaker than preservation of injective objects by coproducts.

Exercise 4 (advanced: a functorial embedding need not be additive). Use the retained functor \(I_R(M)\) of the ordinary module theorem, and write \(D=\operatorname{Hom}_{\mathbb Z}(R,\mathbb Q/\mathbb Z)\). Construct a natural direct summand of \(I_R\) which contains its embedded copy of \(M\), is injective at every \(M\), and takes zero to zero. Show that even this modified functor need not be additive.

Solution. Projection to the coordinate indexed by the zero character defines \(p_M:I_R(M)\to D\). The constant tuple defines \(s_M:D\to I_R(M)\), and \(p_Ms_M=1_D\). For \(u:M\to N\), the defining formula of \(I_R(u)\) selects the coordinate indexed by \(\chi u\); with \(\chi=0\) this is zero, so \(p_NI_R(u)=p_M\). Constant tuples remain constant, so \(I_R(u)s_M=s_N\). Thus these are natural transformations to and from the constant functor with value \(D\).

Put \(J_R(M)=\ker p_M\). Naturality restricts \(I_R(u)\) to \(J_R(u)\), giving a functor. The maps

\[ \begin{gathered} D\oplus J_R(M)\longrightarrow I_R(M),\\ (d,z)\longmapsto s_M(d)+z,\\ y\longmapsto(p_My,\ y-s_Mp_My) \end{gathered} \tag{5.1} \]

are inverse natural linear maps. Hence \(J_R(M)\) is a direct summand of the injective module \(I_R(M)\), and is injective by the complete retained retract proof. The zero-character coordinate of \(\iota_M(m)\) is zero, so the natural monomorphism \(\iota_M\) factors into \(J_R(M)\). At \(M=0\) there is exactly one character, the zero map, so \(I_R(0)=D\) and \(p_0\) is the identity; thus \(J_R(0)=0\).

Take \(R=\mathbb F_2\). Its group character dual \(D\) has two elements, because a character is determined by the image of \(1\), which is either \(0\) or \(1/2+\mathbb Z\). It is a one-dimensional left \(R\)-module. For \(M=R^d\) there are \(2^d\) linear maps to \(D\). Removing the zero coordinate gives \(J_R(R^d)\simeq R^{2^d-1}\). Thus \(J_R(R)\) has dimension one, but \(J_R(R\oplus R)\) has dimension three. An additive functor preserves the specified binary biproduct: applying it to the inclusion/projection equations gives a compatible biproduct isomorphism, as retained in Extending maps, §2. Such an isomorphism here would equate dimensions three and two. Therefore \(J_R\) is not additive, despite being functorial, zero-preserving and equipped with natural injective embeddings.

6. References and exact scope