Exact sequences and connecting homomorphisms
Construct the boundary map by lifting a cycle and identify its obstruction to a closed lift.
On this page
The question for today
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 of cochain complexes of modules. Choose a cycle in degree of the quotient and lift it to in . Although its image is closed, the lift itself may not be closed. The image of under is zero, so exactness uniquely writes that differential as for an element of .
The element is closed: applying to its differential gives the square of the differential of , hence zero, and is injective. We define the connecting map to send the class of to the class of , 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.
Why choices disappear in cohomology
A second lift differs from the first by for some in . Its differential changes by , leaving the cohomology class unchanged. If the representative is changed by a boundary, lift the element whose differential supplies that boundary. Adding its differential to changes the representative downstairs but does not change . 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.
Exactness is a lifting statement
At the quotient term, exactness says that a quotient cohomology class comes from a closed element of precisely when its connecting class vanishes. Indeed, if , replacing by 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.
Work it through
A boundary map equal to multiplication by six
Let have only in degree one, let be in degrees zero and one, and let have only in degree zero. Use the evident inclusion and projection to calculate the connecting map.
The sequence is exact degreewise: in degree zero the map is the identity, while in degree one the map is the identity. Each row splits as a sequence of groups.
An integer represents a class of . Its unique degree-zero lift is the same integer in , whose differential is . Through the inclusion, this is the element in .
Thus the relevant exact sequence is . Multiplication by six has zero kernel and cokernel the integers modulo six.
The connecting map is multiplication by six, with and . Degreewise splittings did not commute with the differential.
The living notebook
Explore the model
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.
- Residue r0
- 0
- Orbit period · m/gcd(m,a)
- 12
- Distinct residues visited
- 1 of 12
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.
Pause & reason
Check your understanding
Your turn
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 , 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 has differential , so the connecting map is multiplication by negative four. Its kernel is zero and its image is , giving and . 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.
Keep the idea
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.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.