research30 min studyLesson 544 in Mathematics

Projective and injective resolutions

Replace a module by an exact augmented complex and track the choices through homotopy.

On this page

What you’ll understand

  • Construct projective resolutions from successive kernels.
  • Construct injective resolutions from successive cokernels.
  • Verify a periodic resolution and interpret comparison up to homotopy.

A resolution replaces a difficult module by a complex of modules on which a chosen functor behaves well. Exactness ensures that the replacement retains the original module as its only cohomology. The replacement is rarely unique as a written complex, yet comparison maps make the resulting invariants independent of the construction. This lesson makes the augmentation and the degree convention explicit, because confusing an augmented exact sequence with its unaugmented resolution changes which homology group should survive.

Successive kernels build projective resolutions

For a left module MM, choose a projective module P0P_0 surjecting onto it, then a projective module P1P_1 surjecting onto the kernel, and continue. The resulting augmented chain sequence is exact. Every module over a unital ring permits such a construction with free modules. An abelian category needs enough projectives for this argument. The unaugmented chain complex has zeroth homology MM and no positive homology; including the augmentation makes the entire displayed sequence exact.

To use our cochain convention, place PiP_i in degree −i-i. Thus a module resolution is bounded above by degree zero, even when infinitely many negative degrees are needed. A finite resolution is an additional property, not a consequence of finite generation. Nor does a finite presentation guarantee that all higher kernels disappear. In computations, identify the kernel at each stage before declaring a sequence a resolution; simply repeating a matrix with square zero proves only that it is a complex.

⋯⟶P2⟶P1⟶P0⟶M⟶0,P−i=Pi\cdots\longrightarrow P_2\longrightarrow P_1\longrightarrow P_0\longrightarrow M\longrightarrow0,\qquad P^{-i}=P_i
Exactness includes the augmentation; the unaugmented cochain replacement has cohomology only in degree zero.↔ Scroll wide equations horizontally.

Cokernels build the dual replacement

An injective resolution starts with an embedding N→I0N\to I^0. Embed its cokernel into an injective I1I^1, then repeat. This produces an exact augmented cochain sequence with nonnegative degrees. Left-module categories have enough injectives; in an arbitrary abelian category this must be assumed. Projective and injective resolutions point in opposite directions because one repeatedly covers kernels and the other repeatedly embeds cokernels. Their augmentations therefore have opposite orientations as maps of complexes.

The original object maps into an injective resolution, whereas a projective resolution maps onto the original object. Each map is a quasi-isomorphism when the object is considered in degree zero. A functor applied to a resolution can destroy its exactness; that destruction is precisely the information a derived functor will measure. Applying an exact functor would preserve the exact augmented sequence and produce no positive-degree defect. Resolution machinery is useful because common tensor and Hom functors are only one-sided exact.

0⟶N⟶I0⟶I1⟶⋯ ,H0(I)=N,Hi(I)=0 (i>0)0\longrightarrow N\longrightarrow I^0\longrightarrow I^1\longrightarrow\cdots,\qquad H^0(I)=N,\quad H^i(I)=0\ (i>0)
Injective resolutions are bounded below, with the original object embedded as the degree-zero cycles.↔ Scroll wide equations horizontally.

Comparison explains independence

A module map lifts to a map between projective resolutions by lifting degree zero and then lifting successive maps on kernels. Two such lifts are homotopic through an inductive construction. Applying any additive functor preserves the homotopy identity, so the induced homology maps agree. In particular, two projective resolutions of the same module are homotopy equivalent through comparison maps lifting its identity. The dual statement holds for injective resolutions and extensions across the successive inclusions.

These comparison arguments justify independence, but not careless use of arbitrary unbounded complexes of projectives or injectives. The standard cochain argument uses projectives bounded above or injectives bounded below, allowing induction to start. Unbounded derived constructions require stronger replacement properties. For module-level derived functors the ordinary resolutions already have the needed direction of boundedness, even if their total length is infinite. Later examples will exploit infinite resolutions to detect a genuine failure of finite homological dimension.

An infinite resolution over the dual numbers

Let kk be a field, R=k[ε]/(ε2)R=k[\varepsilon]/(\varepsilon^2), and M=R/(ε)M=R/(\varepsilon). Check the augmented chain sequence with one copy of RR in every nonnegative degree, all positive differentials multiplication by ε\varepsilon, and the usual augmentation to MM.

  1. Every term is a free left RR-module. Consecutive differentials compose to multiplication by ε2=0\varepsilon^2=0, so the proposed arrows form a complex.

  2. Write an arbitrary element as a+bεa+b\varepsilon. Multiplying it by ε\varepsilon gives aεa\varepsilon. Its kernel is therefore the ideal (ε)(\varepsilon), which is also the image of the preceding multiplication map. This verifies exactness at every positive degree.

  3. The kernel of the augmentation R→MR\to M is the same ideal (ε)(\varepsilon), so exactness also holds at degree zero. The augmentation is surjective. Consequently the augmented sequence is a free resolution, not merely a repeating square-zero diagram.

The result

Removing the augmentation leaves homology M≅kM\cong k in degree zero and zero in positive chain degrees. The infinite repetition is permitted; its usefulness will become visible after tensoring.

Explore the model

Interactive lab

Interactive lab · modular

Square zero is not yet exactness

Pause/Reset, then set both modulus to 4 and additive step to 2. Predict image {0,2} and verify it with Step. By hand compare that image with the kernel of multiplication by two modulo four; they agree. Reset and set both modulus to 8 and step to 4. The new image has two elements but the kernel has four, although composing multiplication by four twice is still zero.

Square zero is not yet exactnessAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA residue orbit reveals the image; comparing its size with a separately computed kernel distinguishes exactness from the square-zero condition.

Check your understanding

For a projective resolution written as an unaugmented cochain complex, which degree statement is correct?

Try it yourself

Construct a free resolution of Z/12Z\mathbb Z/12\mathbb Z of length one. Verify exactness at both free terms, and translate the unaugmented resolution into cochain degrees. Why does replacing twelve by zero fail?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the quotient map from the integers and determine its kernel before choosing the preceding differential.

Compare with the solution

The exact sequence is 0→Z→12Z→Z/12Z→00\to\mathbb Z\xrightarrow{12}\mathbb Z\to\mathbb Z/12\mathbb Z\to0. Multiplication by twelve is injective, and its image is exactly the kernel of reduction modulo twelve. Put the first free group in degree minus one and the second in degree zero. The unaugmented cohomology is the quotient in degree zero and zero elsewhere. A zero differential would have the wrong image and a nonzero kernel, failing both exactness checks.

Take it with you

  • Verify the augmentation as well as the internal differentials.
  • Projective resolutions are bounded above in cochain degrees; injective resolutions are bounded below.
  • Comparison maps are unique up to homotopy, which gives derived invariants their independence.

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