Questions: Exact Sequences in Homological Algebra

Question 1

In the short exact sequence 0 → Z →^{×2} Z → Z/2Z → 0, what does exactness at the middle Z tell us?

• A. The map Z → Z/2Z is injective
• B. The image of the multiplication-by-2 map (the even integers) equals the kernel of the quotient map Z → Z/2Z (also the even integers)
• C. Z is isomorphic to Z ⊕ Z/2Z
• D. The sequence splits

Answer: (B)

Exactness at the middle term means im(×2) = ker(Z → Z/2Z). The image of ×2 is 2Z (the even integers). The kernel of the quotient Z → Z/2Z is also 2Z. So im = ker = 2Z, confirming exactness. This sequence does NOT split: Z is not isomorphic to Z ⊕ Z/2Z (the former is torsion-free, the latter has 2-torsion). The non-splitting shows that Z is a 'non-trivial extension' of Z/2Z by Z.

Question 2

A short exact sequence 0 → A →^i B →^p C → 0 encodes three exactness conditions. Exactness at A means i is injective, and exactness at C means p is surjective.

• T. True
• F. False

Answer: True

Exactness at A: im(0 → A) = ker(i), so ker(i) = {0}, meaning i is injective. Exactness at C: im(p) = ker(C → 0) = C, so p is surjective. Exactness at B: im(i) = ker(p), meaning the image of the injection equals the kernel of the surjection. Together: A injects into B, the image of A is exactly the 'part of B that maps to zero in C,' and every element of C is hit by p. The sequence encodes that C ≅ B/i(A).

Question 3

Every short exact sequence 0 → A → B → C → 0 of abelian groups with C free (e.g., C ≅ Z^n) splits: B ≅ A ⊕ C.

• T. True
• F. False

Answer: True

When C is free abelian, we can choose a section s: C → B (a homomorphism with p ∘ s = id_C) by sending each generator of C to a preimage under p. Then B = i(A) ⊕ s(C) ≅ A ⊕ C. This is a fundamental fact: free abelian groups are projective modules over Z, meaning every surjection onto them has a section. When C is not free (e.g., C = Z/2Z), the sequence may or may not split — the extension is classified by Ext^1(C, A).

Question 4

Explain why the exactness condition 'im = ker' at each term is the correct algebraic formulation of 'no information is lost or gained.'

Model answer: At each term B in the sequence A → B → C: the image of A → B is the information 'coming in,' and the kernel of B → C is the information 'filtered out' (sent to zero in C). Exactness im(A → B) = ker(B → C) means: everything coming in from A is exactly what gets killed going to C. There is no 'extra' kernel (information destroyed without being accounted for by the incoming map) and no 'missing' image (information from A that survives into C). The sequence is 'tight' — each group is determined up to extension by its neighbors.

Contrast with a chain complex where im ⊆ ker but equality may fail. The quotient ker/im is the homology, measuring the 'gap.' An exact sequence has homology zero at every term — the sequence is 'acyclic' as a chain complex. Exact sequences are the most structured chain complexes: they carry maximum algebraic information about the relationships between the groups.