Consider the sequence of abelian groups 0 → ℤ →^(×2) ℤ →^(mod 2) ℤ/2ℤ → 0. To verify exactness at the middle ℤ, which condition must be checked?
AThat the map ×2 is surjective onto ℤ
BThat ℤ has no zero divisors
CThat the image of ×2 (which is 2ℤ) equals the kernel of (mod 2) (which is also 2ℤ)
DThat the sequence is exact at 0 first, since exactness propagates from left to right
Exactness at the middle ℤ means: im(×2) = ker(mod 2). The image of multiplication by 2 is 2ℤ — all even integers. The kernel of reduction mod 2 is also all even integers. Since these subgroups are equal, the sequence is exact at the middle ℤ. Exactness is a local condition: you check it at each object independently using only the maps immediately entering and leaving that object. Option D reflects a common misunderstanding — exactness at one position does not imply or enable checking at another; each position is verified independently.
Question 2 Multiple Choice
A student is given the sequence A →^f B →^g C →^h D and checks that im(f) = ker(g). They conclude the sequence is exact. Why is this conclusion premature?
AExactness requires f to be a monomorphism, which the student has not verified
BExactness is a local condition: the student has only verified it at B; they must also verify im(g) = ker(h) at C (and any other intermediate objects) separately
CExactness requires both f and g to be epimorphisms, which is a stronger condition than image-kernel equality
DThe check is complete if A, B, C, D are all finitely generated abelian groups
This is the central misconception the topic's Common Misconceptions section flags: exactness is a local condition that must be verified at every intermediate object independently. Checking im(f) = ker(g) tells you about the 'flow' at B, but says nothing about whether im(g) = ker(h) at C. A sequence can be exact at some objects and fail at others. The student must check each position: exactness at B, exactness at C, and so on for every intermediate object in the sequence.
Question 3 True / False
In a short exact sequence 0 → A →^f B →^g C → 0, the map g: B → C is necessarily an epimorphism (surjective, in the case of modules and abelian groups).
TTrue
FFalse
Answer: True
Exactness at C means im(g) = ker(C → 0). The kernel of the zero map C → 0 is all of C (every element maps to 0). So im(g) = C — meaning g surjects onto C. This is automatic from the definition: the '0' on the right of a short exact sequence is not decorative; it forces the last non-trivial map to be surjective. Similarly, exactness at A forces f to be injective. The short exact sequence encodes precisely: A embeds into B as a subobject, and C is the quotient B/A.
Question 4 True / False
A sequence of morphisms with im(f) ⊆ ker(g) at nearly every position (i.e., the composition of any two consecutive morphisms is zero) is an exact sequence.
TTrue
FFalse
Answer: False
A sequence where every consecutive composition is zero is called a chain complex — it satisfies g ∘ f = 0, which means im(f) ⊆ ker(g). But exactness requires the stronger condition im(f) = ker(g): the image must equal the kernel, not merely be contained in it. The gap between a chain complex and an exact sequence is precisely what homology measures: the homology at B is ker(g)/im(f), which is trivial if and only if the sequence is exact at B. Exact sequences are chain complexes with zero homology. The distinction is fundamental — much of algebraic topology is the study of how far chain complexes deviate from exactness.
Question 5 Short Answer
What does a short exact sequence 0 → A →^f B →^g C → 0 say about the relationship between A, B, and C?
Think about your answer, then reveal below.
Model answer: It says that A embeds into B as a subobject (f is a monomorphism), B maps onto C (g is an epimorphism), and C is isomorphic to the quotient B/im(A). Equivalently, B is an 'extension' of C by A — it contains a copy of A, and the remainder is C.
Unpacking the four exactness conditions: exactness at A (with 0 entering) forces f to be injective — A sits inside B. Exactness at C (with 0 following) forces g to be surjective — every element of C has a preimage. Exactness at B says ker(g) = im(f) — everything that maps to 0 in C is exactly the image of A. Together these say B is 'built from' A and C in the sense that A is a subobject and C is the quotient. The short exact sequence is the categorical language for 'B is an extension of C by A,' and it recurs throughout algebra, topology, and geometry wherever objects are constructed by gluing simpler pieces together.