ProofComplete

Proof of the Mayer-Vietoris Exact Sequence

The Mayer-Vietoris sequence is the primary computational tool in de Rham cohomology. Its proof is a careful application of the short exact sequence of differential form complexes together with the snake lemma.


Statement

Theorem5.2Mayer-Vietoris sequence

Let M=UVM = U \cup V with U,VU, V open. There is a long exact sequence

Hk(M)rHk(U)Hk(V)sHk(UV)δHk+1(M)\cdots \to H^k(M) \xrightarrow{r} H^k(U) \oplus H^k(V) \xrightarrow{s} H^k(U \cap V) \xrightarrow{\delta} H^{k+1}(M) \to \cdots

Proof

Proof

Step 1: Short exact sequence of complexes. Define the sequence of cochain complexes

0Ωk(M)rΩk(U)Ωk(V)sΩk(UV)0,0 \to \Omega^k(M) \xrightarrow{r} \Omega^k(U) \oplus \Omega^k(V) \xrightarrow{s} \Omega^k(U \cap V) \to 0,

where r(ω)=(ωU,ωV)r(\omega) = (\omega|_U, \omega|_V) and s(α,β)=αUVβUVs(\alpha, \beta) = \alpha|_{U \cap V} - \beta|_{U \cap V}.

Injectivity of rr: If ωU=0\omega|_U = 0 and ωV=0\omega|_V = 0, then ω=0\omega = 0 on UV=MU \cup V = M.

kers=imr\ker s = \operatorname{im} r: s(α,β)=0s(\alpha, \beta) = 0 means αUV=βUV\alpha|_{U \cap V} = \beta|_{U \cap V}, so α\alpha and β\beta patch to a global form ω\omega with r(ω)=(α,β)r(\omega) = (\alpha, \beta).

Surjectivity of ss: Let γΩk(UV)\gamma \in \Omega^k(U \cap V) and {ρU,ρV}\{\rho_U, \rho_V\} be a partition of unity subordinate to {U,V}\{U, V\}. Set α=ρVγ\alpha = \rho_V \gamma (extended by zero outside UVU \cap V, smooth on UU) and β=ρUγ\beta = -\rho_U \gamma (smooth on VV). Then s(α,β)=ρVγ+ρUγ=γs(\alpha, \beta) = \rho_V \gamma + \rho_U \gamma = \gamma.

Step 2: Long exact sequence. The short exact sequence of cochain complexes induces a long exact sequence in cohomology by the snake lemma (zig-zag lemma). This gives the Mayer-Vietoris sequence.

Step 3: Connecting homomorphism. We describe δ\delta explicitly. Given a closed kk-form γ\gamma on UVU \cap V, lift it: set α=ρVγ\alpha = \rho_V \gamma on UU and β=ρUγ\beta = -\rho_U \gamma on VV. Then dαd\alpha on UU and dβd\beta on VV agree on UVU \cap V:

dαUV=d(ρVγ)UV=dρVγ+ρV0=dρVγ,d\alpha|_{U \cap V} = d(\rho_V \gamma)|_{U \cap V} = d\rho_V \wedge \gamma + \rho_V \cdot 0 = d\rho_V \wedge \gamma,dβUV=d(ρUγ)UV=dρUγ=dρVγ,d\beta|_{U \cap V} = -d(\rho_U \gamma)|_{U \cap V} = -d\rho_U \wedge \gamma = d\rho_V \wedge \gamma,

(since dρU+dρV=0d\rho_U + d\rho_V = 0). So dαd\alpha and dβd\beta patch to a global closed (k+1)(k+1)-form Γ\Gamma on MM, and δ[γ]=[Γ]\delta[\gamma] = [\Gamma].

Step 4: Exactness verification. The exactness at each position follows from standard homological algebra (snake lemma). We verify:

  • At Hk(U)Hk(V)H^k(U) \oplus H^k(V): sr=0s \circ r = 0 (immediate), and kersimr\ker s \subset \operatorname{im} r follows from patching.
  • At Hk(UV)H^k(U \cap V): δs=0\delta \circ s = 0 because if γ=αUVβUV\gamma = \alpha|_{U \cap V} - \beta|_{U \cap V} with dα=0,dβ=0d\alpha = 0, d\beta = 0, then the lift can be chosen with d=0d = 0.
  • At Hk+1(M)H^{k+1}(M): rδ=0r \circ \delta = 0 because δ[γ]U=[dα]\delta[\gamma]|_U = [d\alpha] which is exact. \blacksquare

Applications

RemarkComputational strategy

To compute H(M)H^*(M) via Mayer-Vietoris: choose U,VU, V so that H(U)H^*(U), H(V)H^*(V), and H(UV)H^*(U \cap V) are known (or simpler). Then the long exact sequence, combined with knowledge of the maps, determines H(M)H^*(M). The key difficulty is usually computing the connecting homomorphism δ\delta.

Example$H^*(S^1)$ via Mayer-Vietoris

Cover S1S^1 by two overlapping arcs U,VRU, V \cong \mathbb{R}, with UVRRU \cap V \cong \mathbb{R} \sqcup \mathbb{R}. The sequence gives 0H0(S1)R2R2δH1(S1)00 \to H^0(S^1) \to \mathbb{R}^2 \to \mathbb{R}^2 \xrightarrow{\delta} H^1(S^1) \to 0. The map R2R2\mathbb{R}^2 \to \mathbb{R}^2 sends (a,b)(ab,ab)(a,b) \mapsto (a-b, a-b), which has rank 1. So H0(S1)RH^0(S^1) \cong \mathbb{R} and H1(S1)RH^1(S^1) \cong \mathbb{R}.