Chain homotopy - Knowlpedia

Let $f,g:C_\bullet\to D_\bullet$ be chain maps between chain complexes $(C_\bullet,d^C)$ and $(D_\bullet,d^D)$ . A chain homotopy from $f$ to $g$ is a family of $R$ -linear maps

s_n: C_n \longrightarrow D_{n+1}\qquad (n\in\mathbb Z)

such that, for every $n$ ,

f_n - g_n \;=\; d^D_{n+1}\circ s_n \;+\; s_{n-1}\circ d^C_n.

One writes $f\simeq g$ if such an $s$ exists.

Consequence

If $f\simeq g$ , then the induced maps on homology agree:

H_n(f)=H_n(g)\quad\text{for all }n,

where homology module is used.

Remarks

Special case: a contracting homotopy $\mathrm{id}\simeq 0$ shows a complex is “contractible,” hence exact.
Chain homotopy is the basic equivalence relation behind chain-homotopy categories; compare chain map.

Examples

Degree-0 complexes: homotopy forces equality. If $C_\bullet$ and $D_\bullet$ are concentrated in degree $0$ , then any $s_n$ must be $0$ (there is no $D_{1}$ ), so the homotopy identity becomes $f_0-g_0=0$ . Thus $f\simeq g$ implies $f=g$ in this case.

A contractible 2-term complex. Consider the complex $C_\bullet$ with $C_1=R$ , $C_0=R$ , and $d_1=\mathrm{id}_R$ (all other $C_n=0$ ). This is a chain complex since $d_0=0$ . Define $s_0:C_0\to C_1$ to be $\mathrm{id}_R$ and all other $s_n=0$ . Then for $n=0$ , $(\mathrm{id}-0)_{0} = d_1 s_0 + s_{-1} d_0 = \mathrm{id}\circ \mathrm{id} + 0,$ and for $n=1$ , $(\mathrm{id}-0)_{1} = d_2 s_1 + s_0 d_1 = 0 + \mathrm{id}\circ \mathrm{id}.$ Hence $\mathrm{id}_{C_\bullet}\simeq 0$ . In particular $H_n(C_\bullet)=0$ for all $n$ , so $C_\bullet$ is exact.

Split exact complexes admit contracting homotopies. If each short exact sequence $0\to \operatorname{im}(d_{n+1}) \to C_n \to \operatorname{im}(d_n)\to 0$ splits (as in many algebraic settings), then one can choose splittings to build maps $s_n:C_n\to C_{n+1}$ with $\mathrm{id}\simeq 0$ . This provides a conceptual way to recognize contractible (hence exact) complexes.