advanced38 min studyLesson 483 in Mathematics

Relative homology and long exact sequences

Quotient chains in a subspace and track how relative cycles connect to ordinary boundaries one degree lower.

On this page

What you’ll understand

  • Recognize a relative cycle whose boundary lies in the subspace.
  • Construct the connecting homomorphism of a pair.
  • Use exactness with actual maps rather than only group names.

A path joining two points is usually not an absolute cycle, but it becomes a relative cycle when both endpoints belong to a subspace whose chains are being ignored. Relative homology formalizes this controlled forgetting through a quotient chain complex. The boundary is not erased without consequence: a connecting homomorphism records it one degree lower in the subspace. The long exact sequence combines this boundary information with ordinary homology and often turns an otherwise difficult calculation into a kernel-and-cokernel problem.

Quotient the chain complex, not the homology group

For a subspace A⊆XA\subseteq X, singular simplices lying in A form a subcomplex of C∗(X;R)C_*(X;R). Define relative chains as C∗(X,A;R)=C∗(X;R)/C∗(A;R)C_*(X,A;R)=C_*(X;R)/C_*(A;R) and then take homology. A chain c represents a relative cycle precisely when its boundary lies in the A-subcomplex. It may have a nonzero absolute boundary. Relative boundaries are images of higher chains after passing to the quotient. These definitions are valid for arbitrary pairs of spaces with fixed coefficients.

Taking homology of a quotient complex is not generally the same as quotienting the two homology groups. A subspace can contain a cycle that becomes a boundary in the whole space, producing additional relative classes one degree higher. For suitable good pairs with A nonempty, such as nonempty CW subcomplex pairs, relative homology also agrees with reduced homology of the quotient space obtained by collapsing A. That identification needs its hypotheses; the relative chain construction itself does not require pretending that every arbitrary quotient has the same behavior.

Cn(X,A;R)=Cn(X;R)/Cn(A;R),[c] relative cycle ⟺ ∂c∈Cn−1(A;R)C_n(X,A;R)=C_n(X;R)/C_n(A;R),\qquad[c]\text{ relative cycle}\ \Longleftrightarrow\ \partial c\in C_{n-1}(A;R)
The condition is imposed on a chain representative before taking homology; it permits boundaries entirely contained in A.↔ Scroll wide equations horizontally.

The boundary of a relative cycle becomes a class in A

For a relative cycle represented by c, its boundary is an A-chain and has zero boundary because ∂2=0\partial^2=0. Send its relative class to the class of that boundary in Hn−1(A;R)H_{n-1}(A;R). Changing c by an A-chain changes its boundary by an A-boundary, and changing c by an absolute boundary changes it by zero. These checks make the connecting homomorphism well-defined. Its degree drop is essential: an n-dimensional relative filling detects an (n−1)-dimensional boundary.

The short exact sequence of chain complexes produces a natural long exact sequence. Exactness at each term says that the incoming image equals the outgoing kernel. It does not say that every map is injective or surjective, and it does not describe the maps solely from knowing adjacent groups. Two maps from the integers to the integers, multiplication by one and multiplication by five, have different cokernels even though their domain and codomain are identical. Keep the inclusion and connecting maps visible in every calculation.

⋯→Hn(A;R)→Hn(X;R)→Hn(X,A;R)→∂Hn−1(A;R)→Hn−1(X;R)→⋯\cdots\to H_n(A;R)\to H_n(X;R)\to H_n(X,A;R)\xrightarrow{\partial}H_{n-1}(A;R)\to H_{n-1}(X;R)\to\cdots
The connecting map takes a relative cycle to the homology class of its boundary inside the subspace.↔ Scroll wide equations horizontally.

Low degrees and orientation choices

An oriented disk relative to its boundary has an integer relative class whose connecting image is the oriented boundary sphere. Ordinary disk homology alone does not contain that top-dimensional class. For a nonempty connected disk and its nonempty boundary, the degree-zero part of the exact sequence also matters; replacing an ordinary H₀ term by zero without using a reduced sequence gives the wrong conclusion. State whether a calculation uses ordinary or reduced groups, and preserve that choice through the whole sequence.

Changing an orientation can change a connecting map's generator by a minus sign while leaving its kernel, image, and resulting isomorphism types unchanged. Such a sign choice is harmless only when applied consistently to all related formulas. In computations with several cells or several pieces, inconsistent orientations can destroy the chain identity and produce false ranks or torsion. Relative homology is therefore both a conceptual localization tool and a useful bookkeeping framework for translating geometric attachments into algebraic maps.

A relative disk records a degree-five attachment

Let X be formed from A=S1 by attaching one two-cell whose boundary wraps five times positively around A. Compute the connecting map H2(X,A;Z)→H1(A;Z), then find H2(X;Z) and H1(X;Z).

  1. As a CW pair, the relative chain complex has one generator in degree two and no relative generators in degrees one or zero. Therefore H2(X,A)=Z and H1(X,A)=0. The oriented relative disk's boundary represents five times the circle generator.

  2. The relevant exact sequence becomes 0→H2(X)→Z→Z→H1(X)→0, where the middle map is multiplication by five. The zero on the left uses H2(A)=0.

  3. Multiplication by five is injective over Z, so its kernel is zero and H2(X)=0. Its cokernel is Z/5Z, giving H1(X)=Z/5Z. Both X and A are connected, so H0(X)=Z and relative H0(X,A)=0.

The result

The relative class is infinite cyclic, while its boundary map creates finite absolute H1. Ignoring the degree-five connecting map would lose the torsion.

Explore the model

Interactive lab

Interactive lab · modular

Inspect the quotient produced by the attachment

Pause and Reset, then set both controls to modulus m=5 and additive step a=1. Predict residues 0,1,2,3,4 and return to zero after five Steps. Reset and set m=5 and a=2; predict the same five-element group in another visiting order. This additive residue model illustrates the absolute quotient Z/5Z in the example, not the relative generator Z or the connecting map itself.

Inspect the quotient produced by the attachmentAddition 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 forThe same relative generator can produce a finite absolute quotient through a nontrivial connecting map.

Check your understanding

How can an n-chain c represent a relative cycle in (X,A) without being an absolute cycle?

Try it yourself

For the pair (D2,S1), compute H2(D2,S1;Z), H1(D2,S1;Z), and H0(D2,S1;Z). Describe its nonzero connecting map.

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

Give me a hint

The disk is contractible and the boundary circle is connected. Examine both the positive-degree segment and the H0 map in the exact sequence.

Compare with the solution

The segment 0→H2(D2,S1)→H1(S1)=Z→H1(D2)=0 makes the connecting map an isomorphism, so H2(D2,S1)=Z. At the bottom, H0(S1)=Z→H0(D2)=Z is an isomorphism because inclusion preserves the single component. Exactness then gives H1(D2,S1)=0 and H0(D2,S1)=0. With compatible orientations the relative disk generator maps to the positively oriented boundary circle, not to zero merely because the disk has trivial positive-degree absolute homology.

Take it with you

  • Relative homology is homology of a quotient chain complex.
  • The connecting map lowers degree and remembers the relative cycle's boundary.
  • Exactness needs the actual maps and consistent ordinary or reduced conventions.

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