LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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Relative homology Mayer–Vietoris for closed supports

Statement

Let $A, B$ be closed subsets of a space $X$ , and let $R$ be a commutative unital ring. Write $H_{k} (X ∥ K; R) = H_{k} (X, X ∖ K; R)$ . There is an exact sequence $\dots \to H_{k + 1} (X ∥ A \cap B; R) \to H_{k} (X ∥ A \cup B; R) \overset{j}{\to} H_{k} (X ∥ A; R) \oplus H_{k} (X ∥ B; R) \overset{q}{\to} H_{k} (X ∥ A \cap B; R) \to \dots,$ where $j$ is the pair of restrictions and $q (a, b) = a ∣_{A \cap B} - b ∣_{A \cap B}$ . Empty supports and zero coefficients are included. No AC is used.

Facts & Assumptions

[F1]

Relative singular homology defines relative chains as the quotient of singular chains by the subspace chain complex.

[F2]

Relative cup product for an excisive triad proves that for two subspaces $U, V$ open in their union the canonical map $C_{*} (X) / (C_{*} (U) + C_{*} (V)) \to C_{*} (X) / C_{*} (U \cup V)$ is a chain homotopy equivalence, by the explicit $P = 1 - d E - E d$ construction before dualization.

[F3]

The long exact sequence in homology gives the long exact homology sequence of a short exact sequence of chain complexes.

Proof

Given: $X, A, B, R$ as stated. Put $U = X ∖ A$ , $V = X ∖ B$ , $C = C_{*} (X; R)$ , $D = C_{*} (U; R)$ , and $E = C_{*} (V; R)$ .

1.1F1given

The singular simplex generators common to $D$ and $E$ are exactly the maps with image in $U \cap V$ . Hence $D \cap E = C_{*} (U \cap V; R)$ , including for the zero ring. The chain maps $0 ⟶ C / (D \cap E) \overset{j}{\to} (C / D) \oplus (C / E) \overset{q}{\to} C / (D + E) ⟶ 0,$ given by $j [c] = ([c], [c])$ and $q ([a], [b]) = [a - b]$ , are well-defined and commute with boundary because $D, E$ are subcomplexes.

2.1step 1.1algebra

The map $j$ is injective since a representative mapping to zero lies in both $D$ and $E$ . The map $q$ is onto since $q ([a], 0) = [a]$ , and $q j = 0$ . If $q ([a], [b]) = 0$ , write $a - b = d + e$ with $d \in D, e \in E$ . Then $c = a - d = b + e$ has residues $[a]$ in $C / D$ and $[b]$ in $C / E$ , so $j [c] = ([a], [b])$ . This proves exactness in every degree; the argument uses only the existence of a decomposition for one element of $D + E$ .

3.1F1F2F3step 1.1step 2.1

By De Morgan's laws, $U \cap V = X ∖ (A \cup B)$ and $U \cup V = X ∖ (A \cap B)$ . The first two nonzero complexes in step 1.1 therefore have exactly the relative homology groups displayed in the statement. Since $U, V$ are open, [F2] identifies the homology of the final quotient with $H_{*} (X ∥ A \cap B; R)$ . Applying [F3] to step 2.1 yields the asserted sequence. Composition of $q$ with this canonical quotient map is the difference of the two relative quotient maps, so the printed sign is precisely $a ∣_{A \cap B} - b ∣_{A \cap B}$ .

4.1F1F2step 1.1step 2.1step 3.1∎

If $A = \emptyset$ , then $U = X$ , $D = C$ , and the sequence reduces to identity maps on the groups supported in $B$ , with zero groups for empty support; the other empty case is symmetric. If $A = B$ , the diagonal and difference sequence has the stated exactness. For $X = \emptyset$ , $R = 0$ or negative chain degrees all complexes concerned are zero. Degree zero follows from the same degreewise short exact sequence, with no reduced-group substitution. All singular generators, including degenerate ones, were retained in step 1.1. The quotient equivalence in [F2] uses prescribed small-chain operators; no AC or choice of a splitting is required.

Depends on

Dependency tree · 2 levels

Sources

Hatcher, Algebraic Topology, Lemma 3.27 proof, first exact sequence, p.237