Questions: The Five Lemma

Question 1

In the five lemma diagram with exact rows and vertical maps α, β, γ, δ, ε, the conclusion that γ is an isomorphism requires which conditions?

• A. α, β, δ, ε are all isomorphisms
• B. β and δ are isomorphisms, α is surjective, and ε is injective
• C. All vertical maps are surjective
• D. The rows are split exact

Answer: (B)

The five lemma has two halves. Injectivity of γ requires: β injective and α surjective (these are used in the diagram chase to show γ(x) = 0 implies x = 0). Surjectivity of γ requires: δ surjective and ε injective. If all four outer maps are isomorphisms, both conditions hold and γ is an isomorphism. The weaker conditions (B) are sufficient and are sometimes needed in practice when the outer maps are not full isomorphisms.

Question 2

The five lemma is often used as follows: two long exact sequences are connected by a map, and isomorphisms at four terms force an isomorphism at the fifth. Give an example from algebraic topology.

Model answer: Example: proving that a homotopy equivalence f: X → Y induces isomorphisms on relative homology H_n(X, A) → H_n(Y, B) when f maps A to B. The induced map of long exact sequences of pairs gives a commutative diagram where f_*: H_n(A) → H_n(B) and f_*: H_n(X) → H_n(Y) are isomorphisms (by homotopy invariance). The five lemma then forces f_*: H_n(X, A) → H_n(Y, B) to be an isomorphism at each term.

Another standard application: proving that simplicial homology equals singular homology. One constructs a map from the simplicial chain complex to the singular chain complex, shows it is a chain map, and uses the five lemma on the resulting maps of long exact sequences (after establishing the result for simplices and cones). The five lemma reduces the problem from all spaces to elementary building blocks.

Question 3

The five lemma works only for abelian groups (or more generally, abelian categories).

• T. True
• F. False

Answer: True

The proof of the five lemma uses diagram chasing, which relies on the ability to add, subtract, and manipulate elements through homomorphisms — operations that require the abelian group structure. In non-abelian settings (e.g., groups that are not abelian), the five lemma can fail. There is a version for groups (using normal subgroups and quotients), but it requires additional hypotheses. For algebraic topology, where homology and cohomology are abelian groups, the standard abelian five lemma is sufficient.

Question 4

State and prove the 'short five lemma': if 0 → A → B → C → 0 and 0 → A' → B' → C' → 0 are exact with α: A → A' and γ: C → C' isomorphisms, then β: B → B' is an isomorphism.

Model answer: This is the five lemma applied to the diagram 0 → A → B → C → 0 over 0 → A' → B' → C' → 0, where the maps from 0 to 0 on each end are trivially isomorphisms. Injectivity of β: suppose β(b) = 0. Then γ(p(b)) = p'(β(b)) = 0, and since γ is injective, p(b) = 0. By exactness, b = i(a) for some a. Then i'(α(a)) = β(i(a)) = β(b) = 0, so α(a) = 0 (since i' is injective). Since α is injective, a = 0, so b = 0. Surjectivity is similar, using surjectivity of α and γ.

The short five lemma is the most commonly used special case. It says that in a morphism of short exact sequences, if the 'ends' are isomorphisms, the 'middle' is too. This is used constantly: to show B ≅ B', it suffices to show A ≅ A' and C ≅ C' with compatible maps, plus exactness of both rows.