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Snake lemma corollary: long exact sequence in homology

A short exact sequence of chain complexes induces a natural long exact sequence in homology.

Let

0⟶A∙⟶B∙⟶C∙⟶00 \longrightarrow A_\bullet \longrightarrow B_\bullet \longrightarrow C_\bullet \longrightarrow 0

be a short exact sequence of , meaning that for each nn the sequence

0→An→Bn→Cn→00 \to A_n \to B_n \to C_n \to 0

is exact in the sense of , and all differentials commute with the structure maps.

Then there are connecting homomorphisms

δn:Hn(C∙)⟶Hn−1(A∙)\delta_n : H_n(C_\bullet) \longrightarrow H_{n-1}(A_\bullet)

such that the sequence of

⋯→Hn(A∙)→Hn(B∙)→Hn(C∙)→ δn Hn−1(A∙)→Hn−1(B∙)→⋯\cdots \to H_n(A_\bullet) \to H_n(B_\bullet) \to H_n(C_\bullet) \xrightarrow{\ \delta_n\ } H_{n-1}(A_\bullet) \to H_{n-1}(B_\bullet) \to \cdots

is exact.

This long exact sequence is natural in morphisms of short exact sequences of complexes. The maps δn\delta_n are constructed by a standard diagram chase and can be viewed as arising from the ; see also .

Examples

Example 1: Complexes concentrated in degree 00

If A∙,B∙,C∙A_\bullet,B_\bullet,C_\bullet are concentrated in degree 00, then H0(A∙)=A0H_0(A_\bullet)=A_0, H0(B∙)=B0H_0(B_\bullet)=B_0, H0(C∙)=C0H_0(C_\bullet)=C_0 and Hn(−)=0H_n(-)=0 for n≥1n\ge 1. The long exact sequence collapses to

0→A0→B0→C0→0,0 \to A_0 \to B_0 \to C_0 \to 0,

i.e. it recovers the original short exact sequence of modules.

Example 2: A nontrivial connecting map δ1\delta_1 detecting reduction mod nn

Fix n≥2n\ge 2. Define complexes (nonzero only in degrees 1,01,0):

  • A∙A_\bullet: A1=Z→d1=⋅nA0=ZA_1=\mathbb Z \xrightarrow{d_1=\cdot n} A_0=\mathbb Z.
  • B∙B_\bullet: B1=Z⊕Z→d1(x,y)=nx+yB0=ZB_1=\mathbb Z\oplus \mathbb Z \xrightarrow{d_1(x,y)=nx+y} B_0=\mathbb Z.
  • C∙C_\bullet: C1=Z→C0=0C_1=\mathbb Z \to C_0=0 (so d1=0d_1=0).

Define maps A∙→B∙A_\bullet \to B_\bullet by

A1→B1, x↦(x,0),A0→B0, x↦x,A_1\to B_1,\ x\mapsto (x,0), \qquad A_0\to B_0,\ x\mapsto x,

and B∙→C∙B_\bullet\to C_\bullet by

B1→C1, (x,y)↦y,B0→C0, z↦0.B_1\to C_1,\ (x,y)\mapsto y,\qquad B_0\to C_0,\ z\mapsto 0.

Then 0→A∙→B∙→C∙→00\to A_\bullet\to B_\bullet\to C_\bullet\to 0 is short exact degreewise.

Compute homology:

  • H1(A∙)=0H_1(A_\bullet)=0, H0(A∙)≅Z/nZH_0(A_\bullet)\cong \mathbb Z/n\mathbb Z.
  • H1(C∙)≅ZH_1(C_\bullet)\cong \mathbb Z, H0(C∙)=0H_0(C_\bullet)=0.
  • H1(B∙)=ker⁡(d1)={(x,−nx)}≅ZH_1(B_\bullet)=\ker(d_1)=\{(x,-nx)\}\cong\mathbb Z, and H0(B∙)=0H_0(B_\bullet)=0 (since d1(0,1)=1d_1(0,1)=1).

The relevant part of the long exact sequence is

H1(B∙)→H1(C∙)→δ1H0(A∙)→H0(B∙)=0.H_1(B_\bullet)\to H_1(C_\bullet)\xrightarrow{\delta_1} H_0(A_\bullet)\to H_0(B_\bullet)=0.

Under the identifications above, the map H1(B∙)→H1(C∙)H_1(B_\bullet)\to H_1(C_\bullet) is multiplication by nn, hence its image is nZn\mathbb Z. Exactness forces

ker⁡(δ1)=nZ,\ker(\delta_1)=n\mathbb Z,

so δ1:Z→Z/nZ\delta_1:\mathbb Z\to \mathbb Z/n\mathbb Z is precisely reduction mod nn.

Example 3: Split short exact sequences give δn=0\delta_n=0

If the short exact sequence of complexes splits degreewise (e.g. B∙≅A∙⊕C∙B_\bullet \cong A_\bullet \oplus C_\bullet as complexes), then the induced sequence in homology splits and all connecting maps δn\delta_n are zero.