Extensions form a group before any resolution is chosen
An extension of C by A is a short exact sequence 0 → A → B → C → 0. Two are equivalent when an isomorphism of the middle terms commutes with both ends. The Baer sum — pull back along the diagonal of C, push out along the addition of A — makes the equivalence classes an abelian group with the split extension as zero. That group is Ext1(C, A), defined here without a single resolution.
Learning objectives
- Define an extension and equivalence of extensions.
- Construct the Baer sum by pullback and pushout.
- Identify the zero element and inverses.
- Compute a small example by hand.
- Explain why equivalence is finer than isomorphism of middle terms.
Section 01Extensions and equivalence
Two extensions are equivalent when there is φ: B → B′ making both squares commute with the identity on A and C. By the short five lemma such a φ is automatically an isomorphism.
Two extensions can have isomorphic middle modules and still be inequivalent, because the isomorphism may fail to respect the maps to and from A and C. Classification of extensions is a classification of sequences, not of modules.
The split extension, with B = A ⊕ C and the canonical maps, is always present. The content of the theory is how many others there are.
| A | C | Extensions up to equivalence | Ext1(C, A) |
|---|---|---|---|
| ℤ | ℤ/2ℤ | Two: split, and 0 → ℤ → ℤ → ℤ/2ℤ → 0 | ℤ/2ℤ |
| ℤ/2ℤ | ℤ/2ℤ | Two: ℤ/2 ⊕ ℤ/2, and ℤ/4ℤ | ℤ/2ℤ |
| ℤ | ℤ | One: split only | 0 |
| ℤ/2ℤ | ℤ | One: split only | 0 |
| ℚ | any C | Split only — ℚ is injective | 0 |
Section 02The Baer sum
- Given extensions E and E′ of C by A, form the direct sum extension of C ⊕ C by A ⊕ A.
- Pull back along the diagonal Δ: C → C ⊕ C. Yields an extension of C by A ⊕ A.
- Push out along the addition map ∇: A ⊕ A → A. Yields an extension of C by A.
- The class of the result is E + E′. It is independent of the representatives chosen.
- The zero element is the split extension; the inverse of E is E with the map on A negated.
It is the unique operation natural in both variables that has the split extension as identity. It also agrees with the addition Ext1 inherits as a derived functor, which is the theorem tying the two definitions together.
Section 03Functoriality
Ext1 is contravariant in C by pullback and covariant in A by pushout. A map C′ → C pulls an extension back; a map A → A′ pushes it forward.
Given γ: C′ → C, form the fibre product B ×C C′. The result is an extension of C′ by the same A.
Given α: A → A′, form the pushout of B and A′ over A. The result is an extension of the same C by A′.
It matches Hom: contravariant in the first argument, covariant in the second. That is not a coincidence — Ext1 is a derived functor of Hom, and derived functors inherit variance from the functor they derive.
ReferenceFrequently asked questions
Why is the split extension the zero element?
Because pulling back the split extension along the diagonal and pushing out along addition returns the original extension unchanged. It is the neutral element for the operation, which is what zero means here.
Can two inequivalent extensions have isomorphic middle terms?
Yes. Over ℤ/p²ℤ there are extensions of ℤ/p by ℤ/p that are inequivalent but whose middle terms are abstractly isomorphic. Equivalence tracks the maps, not just the objects.
Does the Baer sum require modules?
No — only pullbacks, pushouts and a zero object, so it works in any abelian category. That is why Ext1 can be defined for sheaves and for representations without a resolution being available.
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