Draft lesson.

Injective tests and finite homological dimension

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

Level: advanced. Prerequisites: abelian categories, quasi-isomorphism roofs, smart truncations, exact Hom sequences of triangles and compatible right derived functors.

Injectivity can be tested by a single extension degree even in a category without enough injectives. A global vanishing bound also has a test in one degree. The existence of enough injectives enters later, when we turn that bound into finite resolutions. Finally, an example shows why taking the first cohomology of a derived composition requires left exactness of the outer functor.

Conventions and prerequisites

Let \(\mathcal A\) be an abelian category. Objects are regarded as complexes concentrated in degree zero, and shifts satisfy \(H^j(X[m])=H^{j+m}(X)\). We use

Hom groups are taken in an ambient universe where the indicated localization is defined. Cohomological bounds and truncation adjunctions, Sections 1–4 proves the roof descriptions, bounded representatives and the natural identifications

Its Section 5 supplies truncation triangles. Triangles after quasi-isomorphism localization, Sections 1–6 proves the derived triangulation, its exact Hom sequences and the distinguished triangle of a short exact row in the heart. We use these exact sequences, rather than assuming an injective resolution exists.

An object \(I\) is injective if every map \(A\to I\) extends across every monomorphism \(A\hookrightarrow B\). An object \(P\) is projective if every map \(P\to B\) lifts across every epimorphism \(A\twoheadrightarrow B\). Enough injectives means that every object embeds in an injective; enough projectives means that every object is a quotient of a projective.

For an integer \(n\geq-1\), say that \(\mathcal A\) has homological dimension at most n if

Dimension at most minus one means that the category is zero, as we will prove. A category whose groups do not have any finite bound has infinite homological dimension.

1. Testing a single injective or projective object

Theorem 1.1. For an object \(I\), the following conditions are equivalent:

1. \(I\) is injective.
2. \(\operatorname{Ext}^1(A,I)=0\) for every object \(A\).
3. \(\operatorname{Ext}^j(A,I)=0\) for every object \(A\) and every \(j>0\).

No supply of injective objects is assumed.

Proof. Suppose first that \(I\) is injective. Let \(E\) be any acyclic complex and let \(f: E\to I[m]\) be a chain map. Put \(k=-m\), the only degree in which its target has a nonzero term. Its only possible component is \(f^k: E^k\to I\). The chain-map condition gives \(f^k d_E^{k-1}=0\). Since \(E\) is acyclic, \(\operatorname{im}d_E^{k-1}=\ker d_E^k\). Thus \(f^k\) factors through the quotient by \(\ker d_E^k\), or equivalently through the image of \(d_E^k\):

Extend \(\bar f\) across that image inclusion to \(h^{k+1}: E^{k+1}\to I\). Take all other components of a degree-minus-one homotopy to be zero. Its equation in degree \(k\) is \(f^k=h^{k+1}d_E^k\). In other degrees both sides are zero, since the target differential is zero and no other homotopy component can contribute. This is a null-homotopy of \(f\).

For a quasi-isomorphism \(s: U\to A\), its cone and all shifts are acyclic. Apply \(\operatorname{Hom}_K(-,I[j])\) to its cone triangle. The preceding vanishing makes precomposition an isomorphism

In an incoming roof \(A\leftarrow U\to I[j]\), the numerator therefore has a unique homotopy class coming from \(A\). These inverse identifications commute with all denominator refinements: if \(su=s'\), then \(u^*s^*=(s')^*\). The nonempty filtered roof index consequently gives

For \(j\ne0\), there is no common nonzero degree of the two complexes, so every chain map is zero. This proves condition 3, and condition 3 immediately implies condition 2.

Assume condition 2. Given \(u: A\hookrightarrow B\), let \(C=\operatorname{coker}u\). The triangle of \(0\to A\to B\to C\to0\) gives the exact segment

The first arrow is restriction along \(u\), so every map from \(A\) extends. This is injectivity. \(\square\)

Theorem 1.2. Dually, \(P\) is projective if and only if \(\operatorname{Ext}^1(P,A)=0\) for every \(A\), and this is equivalent to \(\operatorname{Ext}^j(P,A)=0\) for every \(A\) and \(j>0\).

Proof. Here is the lifting argument explicitly. For a map \(f: P[0]\to E[m]\) with \(E\) acyclic, its only component has image in \(Z^mE\). Acyclicity gives the epimorphism \(E^{m-1}\twoheadrightarrow Z^mE\). Projectivity lifts \(f^0\) through it. Multiplying the lift by \((-1)^m\), if necessary, supplies \(h^0: P\to E^{m-1}\) with \(d_{E[m]}^{-1}h^0=f^0\); all other homotopy components vanish. Thus maps from \(P\) to shifted acyclic complexes are null-homotopic. The exact Hom sequence of a quasi-isomorphism cone now makes postcomposition on \(\operatorname{Hom}_K(P,-)\) invertible. Outgoing roofs give \(\operatorname{Hom}_K(P,A[j])=\operatorname{Hom}_D(P,A[j])\), which is zero for \(j\ne0\).

Conversely, for \(B\twoheadrightarrow C\) with kernel \(A\), the exact segment \(\operatorname{Hom}(P,B)\to\operatorname{Hom}(P,C)\to\operatorname{Ext}^1(P,A)\) shows that every map to \(C\) lifts when the last group vanishes. \(\square\)

2. A global bound tested in one degree

Lemma 2.1. If \(s\ge1\) and \(\operatorname{Ext}^s(A,B)=0\) for all pairs of objects, then \(\operatorname{Ext}^r(A,B)=0\) for all pairs and all \(r\ge s\).

Proof. Only \(r>s\) needs proof. Represent an arbitrary class by a roof \(A\xleftarrow{q}U\xrightarrow{f}B[r]\). The bounded denominator comparison permits the smart upper truncation at zero. Next smart-truncate below degree \(-r\). The numerator descends across this lower truncation: its component in degree \(-r\) kills the image of \(d_U^{-r-1}\), by its chain-map equation. The denominator to \(A\) also descends. Because \(U\) has cohomology only in degree zero, both truncation comparisons are quasi-isomorphisms. We may therefore take a representative \(P\), supported in degrees \([-r,0]\), with a quasi-isomorphism \(q: P\to A\) and a numerator \(f: P\to B[r]\).

Put \(t=-r+s\), so \(-r<t<0\). Let \(L\) be the brutal upper truncation, namely the quotient complex with \(L^j=P^j\) for \(j\le t\) and zero for \(j>t\). Its last differential is zero. The projection \(\pi: P\to L\) is a chain map. Since the only possible nonzero component of \(f\) is in degree \(-r<t\), it factors as \(g\pi\), where \(g: L\to B[r]\) has this same component.

All cohomology of \(L\) below \(t\) is zero. Its degree-\(t\) cohomology is \(T=\operatorname{coker}(d_P^{t-1})\). The cycle quotient map in its last degree gives a quasi-isomorphism \(L\to T[-t]\): below \(t\) both cohomology groups vanish, and in degree \(t\) the map induces the identity of \(T\). Thus the original derived arrow factors as

After shifting the last arrow back by \(r-s\), it belongs to \(\operatorname{Ext}^s(T,B)\), which is zero by hypothesis. The original arrow is therefore zero. The argument used only kernels, cokernels and bounded roof comparisons. \(\square\)

Theorem 2.2. For every integer \(n\ge-1\), the category has homological dimension at most \(n\) if and only if

Proof. For \(n\ge0\), apply Lemma 2.1 with \(s=n+1\); necessity follows from the definition. When \(n=-1\), (2.2) says that every Hom group in the heart is zero. In particular \(1_A=0\) for every \(A\). Its unique maps to and from the zero object are then inverse, so every object is zero. Every derived complex and every Ext group is zero as well. Conversely a zero category satisfies the asserted bound. \(\square\)

The integer restriction matters. Over a nonzero field, vector spaces form a nonzero abelian category. Their negative Ext groups vanish by (0.2). Consequently vanishing in degree \(n+1<0\) holds for every \(n\le-2\), but does not imply a bound at most \(n\), since identities in degree zero are nonzero. Vanishing in a negative test degree supplies no such dimension criterion.

3. Finite injective resolutions and arbitrary terminal objects

An object \(X\) has injective dimension at most \(n\ge0\) if there is an exact sequence

with all \(I^j\) injective. For \(n=0\) this means that \(X\) itself is injective.

Theorem 3.1. Suppose \(\mathcal A\) has enough injectives and \(n\ge0\). For a fixed \(X\), injective dimension at most \(n\) is equivalent to \(\operatorname{Ext}^{n+1}(A,X)=0\) for all \(A\). Either condition makes \(\operatorname{Ext}^j(A,X)=0\) for every \(j>n\).

Proof. For \(n=0\), this is Theorem 1.1. For \(n>0\), successively embed \(C^0=X\) and its ensuing cokernels into injectives, producing the short exact rows

For every \(r\ge1\), the long exact sequence gives a connecting isomorphism

Indeed the two groups adjacent to this arrow are \(\operatorname{Ext}^r(A,I^j)\) and \(\operatorname{Ext}^{r+1}(A,I^j)\), both zero by Theorem 1.1. Composing these actual connecting maps, in the order \(j=n-1,n-2,\ldots,0\), gives

If the asserted degree \(n+1\) vanishes, (3.4) with \(r=1\) and Theorem 1.1 show that \(C^n\) is injective. The spliced rows (3.2), ending in \(I^n=C^n\), give (3.1). Conversely, any resolution (3.1) supplies the same rows with an injective final cokernel. Equation (3.4) then proves every stated higher-degree vanishing. \(\square\)

Corollary 3.2. Under enough injectives, the following conditions are equivalent for \(n\ge0\):

1. \(\mathcal A\) has homological dimension at most \(n\).
2. Every object has a resolution (3.1).
3. In every right-ended exact string

whose displayed \(I^j\) are injective, the object \(T\) is injective. For \(n=0\), condition 3 means that every \(T\) is injective.

Proof. Conditions 1 and 2 are equivalent by Theorems 2.2 and 3.1. For \(n>0\), set \(X=\ker(I^0\to I^1)\), interpreting the first arrow as \(I^0\to T\) when \(n=1\). Exactness of (3.5) supplies rows (3.2) with final cokernel \(C^n=T\). Condition 1 and (3.4) force \(\operatorname{Ext}^1(A,T)=0\) for every \(A\), so condition 3 follows. Conversely, enough injectives construct the first \(n\) rows (3.2) for any \(X\); their terminal \(C^n\) is injective by condition 3, yielding condition 2. When \(n=0\), conditions 2 and 3 both assert that every object is injective, and Theorem 1.1 plus Theorem 2.2 give condition 1. \(\square\)

There is no initial zero in (3.5). Its first kernel may be any object; adding an initial-zero hypothesis would test fewer strings than the corollary requires.

4. The projective counterpart

An object has projective dimension at most \(n\ge0\) if it has an exact resolution \(0\to P_n\to\cdots\to P_0\to X\to0\) with projective terms.

Theorem 4.1. If \(\mathcal A\) has enough projectives, this condition for \(X\) is equivalent to \(\operatorname{Ext}^{n+1}(X,B)=0\) for all \(B\). The following global conditions are equivalent: homological dimension at most \(n\); such resolutions for all objects; and projectivity of every \(T\) in every left-ended exact string

with projective displayed terms. For \(n=0\) the last condition means that every \(T\) is projective.

Proof. Put \(C_0=X\) and take successive projective covers, meaning epimorphisms from projectives, not minimal covers. Their rows are \(0\to C_{j+1}\to P_j\to C_j\to0\). The contravariant long exact sequence and Theorem 1.2 give

Compose from \(j=n-1\) down to zero. The result identifies \(\operatorname{Ext}^1(C_n,B)\) with \(\operatorname{Ext}^{n+1}(X,B)\), so the latter vanishing makes \(C_n\) projective by Theorem 1.2. Conversely a projective terminal kernel and (4.2) kill every degree above \(n\). The global criterion follows from Theorem 2.2.

For (4.1) with \(n>0\), put \(X=\operatorname{coker}(P_1\to P_0)\), interpreting this as \(\operatorname{coker}(T\to P_0)\) for \(n=1\). Its successive kernels are the images in the string and its last kernel is \(T\). Equation (4.2) proves projectivity of \(T\) under the global bound. In the reverse direction, enough projectives produce the first \(n\) covers, and the stipulated projectivity of their terminal kernel completes the resolution. Theorem 1.2 handles \(n=0\). No terminal zero was required in (4.1). \(\square\)

5. The first degree of a derived composition

Write \(R^+F\) for the universal bounded-below right derived functor of an additive functor, and \(R^jF(X)=H^j(R^+F(X))\). Universal derived functors, Sections 5 and 9 proves computation by an F-injective class and the unit-normalized compatible-composition comparison. First derived cohomology, Theorem 2.1 proves, for left exact \(G\) and \(Y\in D^{\ge n}\), the natural kernel comparison \(G(H^nY)\xrightarrow\sim H^n(R^+G(Y))\).

Proposition 5.1. Let \(F: \mathcal A\to\mathcal B\) and \(G: \mathcal B\to\mathcal C\) be additive, and suppose the actual compatible comparison

is an isomorphism. For example, the prerequisite supplies this when full additive computing classes \(\mathcal J,\mathcal J'\) are respectively F-injective and G-injective, with \(F(\mathcal J)\subseteq\mathcal J'\). If \(G\) is left exact and \(R^jF(X)=0\) for \(j<n\), then

Proof. Set \(Y=R^+F(X)\), which belongs to \(D^{\ge n}(\mathcal B)\) by hypothesis. Compose \(H^n(c_X)\) with the inverse of the kernel comparison just recalled:

Both comparisons are specified natural maps, so their composite is the asserted natural isomorphism. Its naturality holds for all derived maps between objects satisfying the same bound. This argument needs only a lower bound; it does not assume that \(Y\) has just one nonzero cohomology object. \(\square\)

Example 5.2 (the left-exactness hypothesis is necessary). Work in abelian groups. Fix \(m\ge2\), let \(F\) be the identity and let \(G(M)=M\otimes_{\mathbb Z}\mathbb Z/m\). The tensor universal property gives a natural isomorphism \(G(M)=M/mM\): a balanced map out of the tensor is determined by the image of \(x\otimes\overline1\), and that additive map kills \(mM\); conversely \((x,\bar a)\mapsto ax+mM\) is balanced and defines the inverse.

Every injective abelian group \(I\) is divisible. To find an m-th root of \(x\in I\), extend the map \(m\mathbb Z\to I\) taking \(m\) to \(x\) across \(m\mathbb Z\hookrightarrow\mathbb Z\). Thus \(I/mI=0\). Abelian groups have enough injectives: the complete character embedding and Zorn extension proof is Countable Mittag–Leffler systems, Section 4. Applying \(G\) to any bounded-below injective replacement gives the zero complex. The class-computation theorem therefore gives \(R^+G=0\), with its zero unit. Consequently all three derived functors exist, \(R^+F\) is the identity, and \(R^+(GF)=R^+G=0\).

Even the compatible classes can be the class of injective groups on both sides: they compute the identity and \(G\), and \(F\) takes the first class into the second. Thus (5.1) is indeed an isomorphism here. For \(X=\mathbb Z[0]\) and \(n=0\), its lower-bound premise holds but

The functor \(G\) is not left exact: it sends the monomorphism \(m: \mathbb Z\to\mathbb Z\) to the zero endomorphism of the nonzero group \(\mathbb Z/m\). Thus existence of the derived functors, even with compatible classes and an invertible comparison, does not replace left exactness in (5.2).

6. Four graded exercises with complete solutions

Exercise 1 — An extension test over the integers (introductory)

For \(m\ge2\), prove directly that \(\mathbb Z\) is not an injective group and compute \(\operatorname{Ext}^j_{\mathbb Z}(\mathbb Z/m,\mathbb Z)\) for every integer \(j\). Relate the two answers.

Solution. The map \(m\mathbb Z\to\mathbb Z\), \(m\mapsto1\), cannot extend to \(\mathbb Z\): an extension would have \(m f(1)=1\), impossible in integers. The free two-term resolution \(P=(\mathbb Z\xrightarrow{m}\mathbb Z)\), in degrees \(-1,0\), has augmentation to \(\mathbb Z/m\) and zero negative cohomology. Hereditary categories, Lemma 3.1 proves that such a projective complex computes the actual derived Hom. Its Hom complex into \(\mathbb Z[0]\) has a copy of \(\mathbb Z\) in degrees zero and one and differential \(-m\), by \(d(f)=d_{\mathbb Z}f-(-1)^{|f|}f d_P\). The degree-zero kernel is zero and the degree-one cokernel is \(\mathbb Z/m\); all other groups vanish. Hence \(\operatorname{Ext}^1(\mathbb Z/m,\mathbb Z)=\mathbb Z/m\) and every other degree is zero. This nonzero degree-one test also detects noninjectivity by Theorem 1.1.

Exercise 2 — Dimension zero and the negative endpoint (intermediate)

Prove that homological dimension at most zero is equivalent to splitting of every short exact sequence. Deduce that every object is both injective and projective, without assuming either supply. Explain the case of dimension at most minus one.

Solution. Under the bound, let \(0\to A\to B\xrightarrow{v}C\to0\) be exact. The segment \(\operatorname{Hom}(C,B)\to\operatorname{Hom}(C,C)\to\operatorname{Ext}^1(C,A)=0\) lifts \(1_C\) to \(s: C\to B\) with \(vs=1_C\). The map \(A\oplus C\to B\), \((a,c)\mapsto u(a)+s(c)\), is an isomorphism: \(1_B-sv\) factors uniquely through the kernel \(u\), supplying the inverse components, and the two composites are identities.

Conversely, if every short exact row splits, every mono has a retraction and every epi has a section. A map from the source of a mono to any object extends by composing its retraction; a map from any object to the target of an epi lifts by composing its section. Thus every object is both injective and projective. Theorem 1.1 gives all positive Ext vanishing. Dimension at most minus one additionally forces every identity in degree zero to vanish, hence forces the zero category, as in Theorem 2.2. A nonzero vector space category has dimension zero: choosing preimages of a basis gives a section of every epimorphism, but its nonzero identities prevent dimension minus one.

Exercise 3 — An injective resolution which never ends (advanced)

Let \(k\) be a field, \(R=k[\varepsilon]/(\varepsilon^2)\), and \(S=R/(\varepsilon)\). Prove that \(R\) is injective as an R-module. Construct explicit projective and injective resolutions of \(S\), compute \(\operatorname{Ext}^j_R(S,S)\) in every degree, and show that neither dimension of \(S\) is finite.

Solution. Let \(\lambda: R\to k\) extract the coefficient of \(\varepsilon\). The pairing \((a,b)\mapsto\lambda(ab)\) is nondegenerate: in the basis \(1,\varepsilon\) its matrix is \(\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\), which is invertible over any field. Hence \(\phi: R\to R^*=\operatorname{Hom}_k(R,k)\), \(a\mapsto(b\mapsto\lambda(ab))\), is an R-linear isomorphism for the action \((r\ell)(b)=\ell(br)\).

For every R-module \(M\), evaluation at one gives the natural isomorphism \(\operatorname{Hom}_R(M,R^*)\simeq\operatorname{Hom}_k(M,k)\). Its inverse sends \(f\) to the R-linear map \(m\mapsto(b\mapsto f(bm))\): evaluation recovers \(f\), and R-linearity verifies the other composite. A k-linear functional on a subspace extends by extending a basis. Consequently every map from an R-submodule to \(R^*\) extends across its ambient module, by this isomorphism. This proves injectivity of \(R^*\) and therefore of \(R\).

Multiplication by \(\varepsilon\) has image and kernel both equal to \(k\varepsilon\). The following two sequences are thus exact:

The first is a projective resolution, with copies of \(R\) in all degrees at most zero. The second is an injective resolution, with copies of \(R\) in all degrees at least zero. In the first resolution, every differential after applying \(\operatorname{Hom}_R(-,S)\) is zero because \(\varepsilon S=0\), and each nonzero term is \(S\simeq k\). The complete bounded-above projective-complex Hom proof in the hereditary prerequisite applies even though this resolution is infinite to the left. It gives \(\operatorname{Ext}^j_R(S,S)=k\) for every \(j\ge0\), and (0.2) gives zero for \(j<0\). The injective resolution gives the same answer: \(\operatorname{Hom}_R(S,R)=k\varepsilon\), and multiplication by \(\varepsilon\) is zero on this subspace.

If a finite injective resolution of length \(n\) existed, the direction of Theorem 3.1 using an existing resolution would kill all degrees above \(n\), contrary to this calculation. A finite projective resolution would contradict Theorem 4.1 in the same way. These implications use the displayed finite resolution alone, so no further supply assumption is needed to obtain the contradiction.

Exercise 4 — A failing composition and a positive calculation (advanced)

First verify that Example 5.2 satisfies every premise of Proposition 5.1 except left exactness of \(G\). For a positive example, take \(R=k[t]\), \(F=\operatorname{Hom}_R(k,-): \operatorname{Mod}R\to\operatorname{Vect}_k\), with \(t\) acting by zero on \(k\), and \(G=\operatorname{Hom}_k(V,-)\) for an arbitrary vector space \(V\). Compute \(R^j(GF)(R)\), including its first nonzero degree.

Solution. In Example 5.2 all three functors exist by injective computation; \(R^+F\) is the identity, the two classes of injective groups are compatible, and the comparison identifies the zero functor with itself. For \(\mathbb Z[0]\), all negative \(R^jF\) vanish. But \(G\) takes the mono \(m\) to zero, and the degree-zero values in (5.4) differ. Thus this example also excludes a repair which adds only compatible classes while still allowing an arbitrary additive outer functor.

For the positive example, enough injectives in \(\operatorname{Mod}R\) follow from the full coinduction and character-product construction in the countable Mittag–Leffler prerequisite. Every vector space is injective and projective, by basis extension and basis lifting. The injective R-modules compute \(F\); all vector spaces compute the exact functor \(G\), so these computing classes are compatible. Exactness of \(G\) follows directly: a linear map from \(V\) into the target of an epimorphism lifts by selecting preimages of the images of a basis of \(V\), and Hom always preserves kernels. This applies to any basis cardinality.

The free resolution \((R\xrightarrow{t}R)\) of \(k\), in degrees \(-1,0\), has exact augmentation since multiplication by \(t\) is injective and has cokernel \(k\). Its Hom complex into \(R\) has differential \(-t\), zero kernel, and cokernel \(R/tR=k\). The injective/projective comparison for this computation is proved in Derived Hom, Sections 5–6. Thus \(R^jF(R)\) is zero unless \(j=1\), when it is \(k\). Its derived object is canonically \(k[-1]\), by the singleton-cohomology quotient in the first-derived-cohomology prerequisite. Exactness of \(G\) makes \(R^+G\) its termwise functor: it sends acyclic complexes to acyclic complexes and hence already descends through quasi-isomorphisms, with the identity unit. Applying the compatible comparison gives \(R^+(GF)(R)=\operatorname{Hom}_k(V,k)[-1]\). Therefore degree one is \(V^*\), and every other degree is zero. The isomorphism in degree one is exactly (5.2). If \(V=0\), all degrees vanish, as the same formula states.

Freely accessible references

The following sources provide further reading on the definitions and results. The arguments used in this lesson are supplied above or in the linked prerequisite lessons.

• The Stacks Project, Ext groups, Section 13.27, tag 06XP: derived Ext, triangle exact sequences and propagation of global degree bounds.
• The Stacks Project, Injectives, Section 12.27, tag 0134: extension tests defining an injective object.
• The Stacks Project, Injective abelian groups, Section 15.55, tag 01D6: divisible groups and injectivity.

Original text: CC0 1.0 Universal. Written by OpenAI GPT-6.1 Sol.