research33 min studyLesson 547 in Mathematics

Derived categories and quasi-isomorphisms

Localize the homotopy category while retaining the replacement data needed for derived functors.

On this page

What you’ll understand

  • Separate chain isomorphism, homotopy equivalence, and quasi-isomorphism.
  • Explain localization through roofs and its universal property.
  • Compute a derived tensor product with its negative cohomological degree intact.

A resolution is usually more convenient than its original module for applying a one-sided exact functor, even though their terms look very different. The derived category organizes these replacements by making quasi-isomorphisms invertible. This identifies objects through their cohomological equivalences without pretending that every such map already has a chain-map inverse. The distinction is essential: localization changes which morphisms are available, not merely the names of the objects already present.

Three increasingly flexible equivalences

A chain isomorphism has a strict inverse in every degree. A homotopy equivalence has an inverse whose composites are homotopic to the identities. A quasi-isomorphism only requires isomorphisms on cohomology. The first implies the second and the second implies the third, but neither converse holds in general. In particular, an acyclic complex can represent a nonzero object of the homotopy category when its exact sequences do not split, yet it becomes isomorphic to zero after quasi-isomorphisms are inverted.

Let K(R)K(R) denote the homotopy category of cochain complexes of left modules, so its morphisms are cochain maps modulo homotopy. The derived category D(R)D(R) is obtained from it by inverting quasi-isomorphisms. One can use a set-theoretic universe convention to manage the size of module categories. The construction does not declare every cohomology isomorphism between graded objects to be a chain map: a quasi-isomorphism is first an actual map of complexes, with its compatibility equation already satisfied.

D(R)=K(R)[Q−1],Q={s:Hn(s) is an isomorphism for every n}D(R)=K(R)[\mathcal Q^{-1}],\qquad \mathcal Q=\{s:H^n(s)\text{ is an isomorphism for every }n\}
Localization formally supplies inverses to the chosen quasi-isomorphisms.↔ Scroll wide equations horizontally.

Roofs encode replacement before mapping

A derived morphism can be represented by a roof with a quasi-isomorphism pointing toward its source and an ordinary homotopy-class map pointing toward its target. The inverse of the left arrow exists in the localized category, so the roof describes their composite there. Different roofs can describe the same morphism after suitable refinements. Roofs are a presentation of morphisms, not unique coordinate expressions; composing them generally requires a common replacement rather than naive concatenation of unrelated chain maps.

The universal property says that a functor from the homotopy category which sends every quasi-isomorphism to an isomorphism factors through the localization, uniquely up to natural isomorphism. This tells us what the construction accomplishes independently of a chosen roof calculus. A one-sided exact functor need not already invert quasi-isomorphisms when applied termwise. Its derived version first chooses replacements on which that failure disappears, then descends to the localized setting.

C← s P→ f D⇝f∘s−1:C⟶D in D(R)C\xleftarrow{\ s\ }P\xrightarrow{\ f\ }D\quad\rightsquigarrow\quad f\circ s^{-1}:C\longrightarrow D\text{ in }D(R)
The left map is a quasi-isomorphism; its formal inverse supplies the missing direction.↔ Scroll wide equations horizontally.

Derived operations retain lost degrees

For modules over a commutative ring, resolve one tensor factor projectively in nonpositive cochain degrees and tensor with the other. The resulting derived tensor product has cohomology in degree −i-i equal to the homological group Tor⁡i\operatorname{Tor}_i. Its degree-zero cohomology is the ordinary tensor product, but retaining only that group would discard the very data the derived operation adds. Derived Hom similarly packages Ext in nonnegative cohomological degrees for modules in degree zero.

Bounded-above complexes of projectives are sufficient replacements for ordinary module resolutions and for bounded-above derived constructions. Arbitrary unbounded complexes require appropriate K-projective, K-flat, or K-injective replacements; being projective or injective in each degree alone is insufficient. Also, equal graded cohomology groups do not generally force two complexes to be isomorphic in the derived category: extension information between degrees can remain. Localization preserves more structure than the list of cohomology modules, even while it forgets acyclic summands.

A quasi-isomorphism without a homotopy inverse

Let PP be Z→4Z\mathbb Z\xrightarrow{4}\mathbb Z in cochain degrees minus one and zero. Map it to M=(Z/4Z)[0]M=(\mathbb Z/4\mathbb Z)[0] by reduction in degree zero. Prove this is a quasi-isomorphism but not a homotopy equivalence.

  1. Reduction kills the image of multiplication by four, so the proposed degreewise maps commute with differentials. The negative-degree kernel is zero and the degree-zero cokernel is Z/4Z\mathbb Z/4\mathbb Z.

  2. The induced map on degree-zero cohomology is the identity of that quotient, and all other cohomology groups on both sides vanish. Thus the map is a quasi-isomorphism.

  3. Every degree-zero homomorphism from the finite group M0M^0 to P0=ZP^0=\mathbb Z is zero. Consequently every cochain map M→PM\to P is zero. Its composite on MM cannot be homotopic to the identity, since MM has only degree zero and therefore no nonzero homotopy components.

The result

The map becomes invertible in the derived category, although no homotopy inverse exists among the original cochain maps. Formal inversion is doing real work.

Explore the model

Interactive lab

Interactive lab · modular

Retain both degrees of derived tensor

Pause/Reset, then set both modulus to 6 and additive step to 4. Predict image {0,2,4} and kernel {0,3}. Use three Step presses to check the image orbit. These are the multiplication map in the practice problem; its cokernel belongs to degree zero and its kernel to degree minus one. Reset and set both modulus to 6 and step to 5 for the invertible comparison.

Retain both degrees of derived tensorAddition 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 forOne multiplication map contributes two different cohomological degrees after a derived tensor calculation; the orbit alone is not a derived-category model.

Check your understanding

What is guaranteed when a quasi-isomorphism is inverted in the derived category?

Try it yourself

Compute (Z/4Z)⊗ZL(Z/6Z)(\mathbb Z/4\mathbb Z)\otimes_{\mathbb Z}^{\mathbf L}(\mathbb Z/6\mathbb Z) using the worked resolution. State its two possible nonzero cohomology groups with their cochain degrees and compare with the ordinary tensor product.

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

Give me a hint

Tensor each of the two free terms with the second factor, leaving the original degree positions unchanged.

Compare with the solution

The tensor complex is Z/6Z→4Z/6Z\mathbb Z/6\mathbb Z\xrightarrow{4}\mathbb Z/6\mathbb Z in degrees minus one and zero. Its kernel is {0,3}, isomorphic to Z/2Z\mathbb Z/2\mathbb Z, in degree minus one. Its cokernel is also Z/2Z\mathbb Z/2\mathbb Z, in degree zero, since the image is {0,2,4}. There is no other cohomology. Ordinary tensor product gives only the degree-zero quotient and omits the nonzero first Tor recorded in degree minus one.

Take it with you

  • Quasi-isomorphisms can lack chain-level homotopy inverses.
  • Roofs encode morphisms after replacing an object by a quasi-isomorphic complex.
  • Derived tensor uses negative cochain degrees for positive Tor.

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