In a chain complex · · · → B →(d_n) C →(d_{n-1}) D → · · ·, the condition d_{n-1} ∘ d_n = 0 implies which relationship between the image of d_n and the kernel of d_{n-1}?
Aim(d_n) = ker(d_{n-1}), so the complex is exact at C
Bim(d_n) ⊆ ker(d_{n-1}), so every boundary is a cycle, but not every cycle need be a boundary
Cim(d_n) and ker(d_{n-1}) are disjoint
Dd_n = 0 or d_{n-1} = 0
d_{n-1} ∘ d_n = 0 means every element in the image of d_n is sent to zero by d_{n-1}, i.e., im(d_n) ⊆ ker(d_{n-1}). This is weaker than exactness: exactness at C would require equality im(d_n) = ker(d_{n-1}). The gap — elements in ker(d_{n-1}) that are not in im(d_n) — is exactly what the homology group H_n = ker(d_n) / im(d_{n+1}) measures.
Question 2 True / False
If 0 → A →f B →g C → 0 is a short exact sequence of abelian groups, then B is necessarily isomorphic to A ⊕ C.
TTrue
FFalse
Answer: False
Exactness forces f to be injective, g to be surjective, and im(f) = ker(g), so C ≅ B/A. But this does not mean the sequence splits. The sequence 0 → Z →(×2) Z → Z/2Z → 0 is exact, yet Z ≇ Z ⊕ Z/2Z — the integers cannot split into a copy of Z plus a finite group. Splitting requires an additional section (a right inverse of g or a retraction of f), which may not exist.
Question 3 Short Answer
The homology group H_n of a chain complex is defined as ker(d_n)/im(d_{n+1}). What does H_n = 0 mean about the complex at position n?
Think about your answer, then reveal below.
Model answer: H_n = 0 means ker(d_n) = im(d_{n+1}), i.e., the complex is exact at position n: every cycle (element killed by d_n) is a boundary (in the image of d_{n+1}), and there are no 'holes' at that position.
Homology measures the failure of exactness. When H_n = ker(d_n)/im(d_{n+1}) = 0, the quotient is trivial, meaning the two subgroups are equal. Exactness at every position means the entire complex is an exact sequence, and all homology groups vanish. Nontrivial H_n indicates topological or algebraic 'holes' that the complex detects.