research28 min studyLesson 542 in Mathematics

Exact sequences and connecting homomorphisms

Construct the boundary map by lifting a cycle and identify its obstruction to a closed lift.

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What you’ll understand

  • Build the cohomological connecting map with an explicit sign convention.
  • Check independence of lift and representative.
  • Use exactness to determine lifting obstructions and quotient groups.

A short exact sequence of complexes is exact in every degree, but its cohomology groups usually do not form short exact sequences. The missing information is a connecting homomorphism that moves upward one degree. It records the obstruction to lifting a quotient cycle to a cycle upstairs. Constructing that map once, with the elements visible, makes long exact sequences a usable argument rather than an arrangement of arrows to memorize.

A differential detects the obstruction

Consider a degreewise short exact sequence 0→A→iB→pC→00\to A\xrightarrow{i}B\xrightarrow{p}C\to0 of cochain complexes of modules. Choose a cycle cc in degree nn of the quotient and lift it to bb in BnB^n. Although its image is closed, the lift itself may not be closed. The image of dBbd_Bb under pp is zero, so exactness uniquely writes that differential as i(a)i(a) for an element of An+1A^{n+1}.

The element aa is closed: applying ii to its differential gives the square of the differential of bb, hence zero, and ii is injective. We define the connecting map to send the class of cc to the class of aa, with this positive lift-and-differentiate convention. For chain complexes the same construction lowers degree instead. Merely switching the location of an index without changing the surrounding convention is a common source of an incorrect boundary formula.

p(b)=c,dBb=i(a),δn([c])=[a]∈Hn+1(A)p(b)=c,\qquad d_Bb=i(a),\qquad \delta^n([c])=[a]\in H^{n+1}(A)
The quotient cycle lifts, but its differential may survive in the subcomplex.↔ Scroll wide equations horizontally.

Why choices disappear in cohomology

A second lift differs from the first by i(u)i(u) for some uu in AnA^n. Its differential changes aa by dAud_Au, leaving the cohomology class unchanged. If the representative cc is changed by a boundary, lift the element whose differential supplies that boundary. Adding its differential to bb changes the representative downstairs but does not change dBbd_Bb. Combining these two observations proves that the construction is well defined on cohomology classes.

Addition of lifts proves additivity, and a commutative diagram of short exact sequences carries every step of the construction to the corresponding step below it. This is naturality of the connecting map. For arbitrary abelian categories, the element argument is replaced by kernels and pullbacks or a diagram chase. Here modules permit an explicit calculation, but the proof depends on exactness and compatible differentials rather than on a chosen basis or finite generation.

⋯⟶Hn(A)⟶Hn(B)⟶Hn(C)→δnHn+1(A)⟶⋯\cdots\longrightarrow H^n(A)\longrightarrow H^n(B)\longrightarrow H^n(C)\xrightarrow{\delta^n}H^{n+1}(A)\longrightarrow\cdots
The extra map joins successive degrees into one exact sequence.↔ Scroll wide equations horizontally.

Exactness is a lifting statement

At the quotient term, exactness says that a quotient cohomology class comes from a closed element of BB precisely when its connecting class vanishes. Indeed, if a=dAua=d_Au, replacing bb by b−i(u)b-i(u) kills its differential. The converse follows by choosing a closed lift. This proves one key junction of the long exact sequence directly and gives the connecting map its practical meaning: it tests whether a desired lift can satisfy the extra equation of being closed.

Degreewise split exactness does not make every connecting map zero: chosen splittings might fail to respect differentials. A splitting by cochain maps does make the connecting map vanish, because it supplies closed lifts of all cycles. Even then, a vanishing connecting map alone need not establish a specified chain-level splitting. When using a long exact sequence, identify both the group and the actual homomorphism; an unknown multiplication map can change a cokernel even if all surrounding ranks are known.

A boundary map equal to multiplication by six

Let AA have Z\mathbb Z only in degree one, let BB be Z→6Z\mathbb Z\xrightarrow{6}\mathbb Z in degrees zero and one, and let CC have Z\mathbb Z only in degree zero. Use the evident inclusion and projection to calculate the connecting map.

  1. The sequence is exact degreewise: in degree zero the map B0→C0B^0\to C^0 is the identity, while in degree one the map A1→B1A^1\to B^1 is the identity. Each row splits as a sequence of groups.

  2. An integer rr represents a class of H0(C)H^0(C). Its unique degree-zero lift is the same integer in B0B^0, whose differential is 6r6r. Through the inclusion, this is the element 6r6r in A1A^1.

  3. Thus the relevant exact sequence is 0→H0(B)→Z→6Z→H1(B)→00\to H^0(B)\to\mathbb Z\xrightarrow{6}\mathbb Z\to H^1(B)\to0. Multiplication by six has zero kernel and cokernel the integers modulo six.

The result

The connecting map is multiplication by six, with H0(B)=0H^0(B)=0 and H1(B)=Z/6ZH^1(B)=\mathbb Z/6\mathbb Z. Degreewise splittings did not commute with the differential.

Explore the model

Interactive lab

Interactive lab · modular

See the subgroup behind a multiplication map

Pause/Reset, then set both modulus to 12 and additive step to 6. Predict the image of multiplication by six modulo twelve to be {0,6}. Use Step twice to observe the orbit returning to zero. Reset and set both modulus to 12 and step to 5; predict an image containing all twelve residues.

See the subgroup behind a multiplication mapAddition 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 orbit enumerates the image of a cyclic multiplication map, helping interpret kernels and cokernels; it does not build a long exact sequence.

Check your understanding

For a degreewise short exact sequence of cochain complexes, what does a zero connecting class mean?

Try it yourself

In the worked construction replace the differential by multiplication by negative four, leaving the inclusion and projection unchanged. Determine the connecting map, both cohomology groups of BB, and whether changing the sign changes their isomorphism types.

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

Give me a hint

Follow the chosen lift instead of trying to infer the sign from the resulting quotient group.

Compare with the solution

The lift of rr has differential −4r-4r, so the connecting map is multiplication by negative four. Its kernel is zero and its image is 4Z4\mathbb Z, giving H0(B)=0H^0(B)=0 and H1(B)=Z/4ZH^1(B)=\mathbb Z/4\mathbb Z. Replacing four by negative four changes the homomorphism but not its image or these group isomorphism types. The specified connecting-map sign still matters in a commutative diagram.

Take it with you

  • Lift, differentiate, and identify an element of the subcomplex.
  • A connecting class vanishes exactly when the quotient class has a closed lift.
  • Degreewise splitting and splitting by cochain maps are different conditions.

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