You have a commutative diagram with two rows and five vertical morphisms α, β, γ, δ, ε. All four outer morphisms (α, β, δ, ε) are isomorphisms. A student concludes that γ must also be an isomorphism based on commutativity alone — without checking the rows. What is the flaw in this reasoning?
AThe student needs to check that α and ε are both surjective, not just isomorphisms
BThe five lemma requires both rows to be exact sequences; commutativity alone is not sufficient
CThe conclusion is only valid if the diagram has at least six columns
DThe student should apply the short five lemma instead, which does not require exactness
Commutativity alone is not enough — exactness of both rows is essential. The entire diagram-chasing proof of the five lemma uses exactness at every position: to lift an element into a kernel, to assert that something in a kernel comes from the previous term, or to conclude that a residual is zero. Without exactness, the chains of logical steps break down and the conclusion can fail. The common misconception is to think that commutativity plus four isomorphisms is sufficient; it is not.
Question 2 Multiple Choice
In algebraic topology, a geometric map f: X → Y induces a map on long exact sequences of a pair. At every position except one, the induced maps on homology groups are known to be isomorphisms. What is the most efficient way to conclude that the remaining map is also an isomorphism?
ACompute the remaining homology group directly using the definition
BApply the five lemma: the map of long exact sequences with four known isomorphisms forces the fifth by diagram chasing
CUse the universal coefficient theorem to relate homology to cohomology at the unknown position
DApply Mayer-Vietoris to decompose the spaces and compute the missing group
This is the canonical application of the five lemma in topology. A map of long exact sequences is a commutative diagram with two exact rows. If four consecutive vertical maps are isomorphisms, the five lemma immediately gives the fifth. This avoids direct computation — which may be unavailable or extremely difficult — and replaces it with a structural argument. Direct computation, universal coefficients, and Mayer-Vietoris are all more expensive tools for what diagram chasing handles automatically.
Question 3 True / False
The short five lemma states: given a commutative diagram with two short exact sequences 0 → A → B → C → 0 as rows, if the vertical maps at A and C are isomorphisms, then the vertical map at B is also an isomorphism.
TTrue
FFalse
Answer: True
The short five lemma is the most frequently applied special case. It is the five lemma applied to a diagram of the form 0 → A → B → C → 0 on both rows (where the '0 → A' and 'C → 0' positions automatically provide isomorphisms at the trivial objects). Knowing isomorphisms at A and C forces an isomorphism at B. This appears constantly when comparing two extensions of C by A: a map of short exact sequences with isomorphisms at both ends implies the middle map is an isomorphism, meaning the extensions are equivalent.
Question 4 True / False
The five lemma can be applied to any commutative diagram with five columns, even when the rows are not exact sequences, as long as the outer four morphisms are isomorphisms.
TTrue
FFalse
Answer: False
Exactness of both rows is a non-negotiable hypothesis. The proof of the five lemma at every step uses exactness to conclude that an element in a kernel arose from a previous term — without this, the diagram chase cannot proceed. In a commutative diagram without exactness, four outer isomorphisms tell you very little about the middle morphism. The five lemma is not simply a 'four isomorphisms imply the fifth' result for arbitrary commutative diagrams.
Question 5 Short Answer
Describe the diagram-chasing argument for why γ must be injective in the five lemma. What specific roles do exactness and commutativity each play in the proof?
Think about your answer, then reveal below.
Model answer: Suppose γ(c) = 0. Commutativity says the map from C to D' via γ then the bottom row equals the map via the top row then δ: so δ(image of c in D) = 0. Since δ is injective (an isomorphism), the image of c in D is 0. Exactness at D says ker(D → E) = im(C → D), so c maps to 0 in D means c is in the kernel of C → D; by exactness at C in the top row, c comes from some b in B. Now apply β: commutativity gives the image of b in B' maps to the image of c = 0 in C'. Since β is injective, c in C' coming from 0... actually β(b) maps to 0 in C'. By exactness at B' in the bottom row, β(b) comes from some a' in A'. Since α is surjective, a' = α(a) for some a. By commutativity, α(a) maps to β(b), but also a maps to b via exactness at B in the top row would give b = 0 if the sequence is correct... Exactness provides the 'kernel = image' conditions that let you track the element back and forward; commutativity ensures the paths around the diagram give consistent results, allowing conclusions about one morphism from information about adjacent ones.
The proof splits into two halves (injectivity and surjectivity), each a chain of 4–5 standard moves: map an element forward or backward, apply commutativity to change paths, use exactness to lift into a kernel or conclude an element is zero, use an isomorphism hypothesis to conclude injectivity or surjectivity. The power of the argument is that it never requires knowing what the objects actually are — it works in any abelian category.