Two sequences, opposite directions, same connecting construction
Ext is a bifunctor, so a short exact sequence in either argument yields a long exact sequence. In the second variable the sequence follows the original direction; in the first, contravariance reverses it, so the sub-object contributes on the right. Both are natural, which is what permits comparison arguments, and both begin with the corresponding Hom sequence — making the failure of Hom's exactness visible as the first connecting map.
Learning objectives
- Write both long exact sequences correctly.
- Explain the direction reversal in the first variable.
- Identify where the Hom sequence sits inside each.
- Use naturality to compare two long exact sequences.
Section 01The second variable
From 0 → A′ → A → A″ → 0 and a fixed C:
The direction matches the original sequence throughout, because the functor is covariant. The first connecting map measures exactly which homomorphisms into A″ fail to lift to A.
Section 02The first variable
From 0 → C′ → C → C″ → 0 and a fixed A:
The quotient C″ now appears first and the submodule C′ second — the reverse of the second-variable sequence. Writing the sequence in the wrong order is the most common error in Ext computations, and it produces conclusions that look plausible. Fix the variable in writing before starting.
| Second variable (covariant) | First variable (contravariant) | |
|---|---|---|
| Order of outer terms | A′, A, A″ — unchanged | C″, C, C′ — reversed |
| Connecting map raises degree | Yes | Yes |
| Interpretation of ∂ in degree 0 | A map to A″ that does not lift | A map out of C′ that does not extend |
| Vanishes when | A′ injective | C″ projective |
Section 03Naturality and comparison
- Take a morphism between two short exact sequences in the same variable.
- Both rows produce long exact Ext sequences.
- Naturality of the connecting homomorphism makes every square commute, giving a morphism of long exact sequences.
- Apply the five lemma at each position: if two out of three vertical maps are isomorphisms, so is the third.
- Conclude by induction along the sequence. This is the shape of most proofs in the subject.
Almost no Ext group is computed from a resolution in practice. The normal method is to embed the module in a short exact sequence with known terms and read the unknown group off the long exact sequence. Choosing that sequence well is the whole skill.
ReferenceFrequently asked questions
Do both sequences terminate?
They continue indefinitely to the right. They terminate in practice when the modules have finite projective or injective dimension — over a PID both stop after Ext1, which is why abelian group computations are short.
Can the two sequences be combined?
Not into a single long exact sequence, but a short exact sequence in each variable simultaneously produces a commutative diagram of long exact sequences, and the resulting spectral sequence is one route to the balance theorem.
What does the connecting map actually do?
In degree 0 of the second-variable sequence it takes a homomorphism C → A″ and returns the extension of C by A′ obtained by pulling back the given short exact sequence along it. The obstruction to lifting is realised as an explicit extension class.
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