Homology module

The nth homology H_n(C) = ker(d_n)/im(d_{n+1}) of a chain complex of modules.

Homology module

Let $R$ be a ring and let

\cdots \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \xrightarrow{d_{n-1}} \cdots

be a chain complex of R-modules , i.e. $d_n\circ d_{n+1}=0$ for all $n$ .

Definition

The $n$ th cycles and $n$ th boundaries of $C_\bullet$ are the submodules

Z_n(C_\bullet) := \ker(d_n)\subseteq C_n, \qquad B_n(C_\bullet) := \operatorname{im}(d_{n+1})\subseteq C_n.

Since $d_n\circ d_{n+1}=0$ , one has $B_n(C_\bullet)\subseteq Z_n(C_\bullet)$ . The $n$ th homology module is

H_n(C_\bullet) := Z_n(C_\bullet) / B_n(C_\bullet).

Equivalently, $H_n(C_\bullet)=0$ for all $n$ iff $C_\bullet$ is an exact complex .

Basic properties

A chain map $f:C_\bullet\to D_\bullet$ induces $R$ -linear maps $H_n(f):H_n(C_\bullet)\to H_n(D_\bullet)$ for all $n$ .
chain-homotopic chain maps induce the same maps on homology.

Examples

Example 1: Two-term complex over $\mathbb Z$

Consider the chain complex $C_\bullet$ with $C_1=\mathbb Z$ , $C_0=\mathbb Z$ , and $d_1:\mathbb Z\to\mathbb Z$ given by multiplication by $n$ , with all other $C_i=0$ . Then

H_1(C_\bullet)=\ker(\times n)=0,\qquad H_0(C_\bullet)=\operatorname{coker}(\times n)\cong \mathbb Z/n\mathbb Z.

Example 2: Detecting exactness

Let $0\to A\xrightarrow{u} B\xrightarrow{v} C\to 0$ be a short exact sequence of $R$ -modules, viewed as a chain complex

0\to A \xrightarrow{u} B \xrightarrow{v} C \to 0

concentrated in degrees $2,1,0$ . Then the sequence is exact iff $H_2=H_1=H_0=0$ , i.e. iff the complex is exact .

Example 3: $\mathrm{Tor}$ as homology (concrete computation)

Let $M=\mathbb Z/n\mathbb Z$ . A projective resolution of $M$ over $\mathbb Z$ is

0\to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n\mathbb Z \to 0.

Tensoring with $\mathbb Z/m\mathbb Z$ gives the chain complex

0\to \mathbb Z/m\mathbb Z \xrightarrow{\times n} \mathbb Z/m\mathbb Z \to 0.

Its homology is

H_1 \cong \ker(\times n:\mathbb Z/m\to \mathbb Z/m)\cong \mathbb Z/\gcd(m,n)\mathbb Z,

Tor_1 _1^{\mathbb Z}(\mathbb Z/n,\mathbb Z/m)\cong \mathbb Z/\gcd(m,n)\mathbb Z.

(Here we are using the identification $H_1(P_\bullet\otimes -)\cong Tor_1$ .)