Questions: Relative Homology and the Long Exact Sequence of a Pair

Question 1

In the long exact sequence ... → H_n(A) →^{i_*} H_n(X) →^{j_*} H_n(X,A) →^{∂} H_{n-1}(A) → ..., what does the connecting homomorphism ∂ represent geometrically?

• A. It maps a cycle in X to its image in A
• B. It takes a relative cycle (a chain in X whose boundary lies in A) and maps it to that boundary, viewed as a cycle in A
• C. It maps a homology class of X to the homology class of its boundary in A
• D. It computes the intersection number of cycles in X and A

Answer: (B)

A relative n-cycle is a chain c in C_n(X) whose boundary d(c) lies in C_{n-1}(A) (rather than being zero). The connecting homomorphism ∂ sends the class [c] in H_n(X,A) to the class [d(c)] in H_{n-1}(A). Geometrically: a relative cycle is a 'chain in X with its boundary in A,' and ∂ extracts that boundary. The boundary is automatically a cycle in A (since d∘d = 0), and its homology class in H_{n-1}(A) measures the obstruction to extending the chain from X into A.

Question 2

If A is a deformation retract of X, then H_n(X, A) = 0 for all n.

• T. True
• F. False

Answer: True

If A is a deformation retract of X, then the inclusion i: A → X is a homotopy equivalence, so i_*: H_n(A) → H_n(X) is an isomorphism for all n. In the long exact sequence, the maps H_n(A) → H_n(X) being isomorphisms forces H_n(X, A) = 0 for all n (by exactness, the kernel of H_n(X, A) → H_{n-1}(A) is the image of the surjection H_n(X) → H_n(X, A), and the image of H_n(X, A) → H_{n-1}(A) is the kernel of the injection H_{n-1}(A) → H_{n-1}(X); both force H_n(X, A) = 0).

Question 3

The long exact sequence of the pair (D^n, S^{n-1}) gives H_k(D^n, S^{n-1}) ≅ Z for k = n and 0 otherwise. Why?

• A. Because D^n is contractible and S^{n-1} has known homology, so the connecting homomorphisms determine the relative groups
• B. Because D^n/S^{n-1} ≅ S^n, and relative homology equals the homology of the quotient
• C. Both A and B are correct and give the same answer
• D. Because relative homology always equals the reduced homology of the subspace

Answer: (C)

Both methods work. Method A: D^n is contractible (H_k(D^n) = 0 for k > 0), so the long exact sequence gives isomorphisms H_k(D^n, S^{n-1}) ≅ H_{k-1}(S^{n-1}) for k ≥ 2. Inductively, this gives H_n(D^n, S^{n-1}) ≅ Z. Method B: collapsing the boundary to a point gives D^n/S^{n-1} ≅ S^n, and for good pairs (which this is), H_k(X, A) ≅ H̃_k(X/A), so H_k(D^n, S^{n-1}) ≅ H̃_k(S^n). Both are standard and reinforce each other.

Question 4

Explain why the long exact sequence of a pair is 'exact' — what does exactness mean at the term H_n(X)?

Model answer: Exactness at H_n(X) means that the image of i_*: H_n(A) → H_n(X) equals the kernel of j_*: H_n(X) → H_n(X, A). In other words: a homology class in H_n(X) maps to zero in the relative group H_n(X, A) if and only if it comes from a class in H_n(A). Geometrically: a cycle in X is 'relatively trivial' (its homology class in X modulo A is zero) precisely when it is homologous to a cycle lying entirely within A.

Exactness at each term has a similar interpretation. At H_n(A): a class in H_n(A) maps to zero in H_n(X) iff it is in the image of the connecting homomorphism from H_{n+1}(X, A). At H_n(X, A): a relative class maps to zero via the connecting homomorphism iff it comes from H_n(X). These three exactness conditions together encode all the relationships between the homologies of A, X, and X relative to A.