Chain map

A chain map $f:C\to C^{\prime }$ is a family $f_i:C_i\to C_i^{\prime }$ satisfying $d_i^{\prime }{f}_i=f_{i-1}{d}_i$.

A chain map inducing isomorphisms on all homology groups. The same definition applies to a map of cochain complexes using cohomology groups. A quasi-isomorphism need not be an isomorphism in each degree.

Let $f$ be a chain map between degreewise finitely generated free abelian groups. If its reduction modulo every prime number is a quasi-isomorphism, the mapping cone acyclicity criterion makes $M(f)\otimes {\mathbb{F}}_p$ acyclic. The short exact sequence of chain complexes $0\to M\stackrel{p}{\to }M\to M\otimes {\mathbb{F}}_p\to 0$ implies that multiplication by $p$ is an isomorphism on every $H_i(M)$. These are finitely generated abelian groups. Their free summands would make multiplication by $p$ fail to be surjective, and any nonzero finite summand would make it fail to be injective for a prime dividing its order. Thus $H_i(M)=0$ and $f$ is a quasi-isomorphism. Degreewise finite generation is essential to this argument.

A chain map sends cycles to cycles and boundaries to boundaries, and therefore induces $f_{\ast }:H_i(C)\to H_i(C^{\prime })$ by $[x]\mapsto [f_i(x)]$.

Chain maps $f,g:C\to C^{\prime }$ are chain homotopic when there are maps $h_i:C_i\to C_{i+1}^{\prime }$ such thatChain-homotopic maps induce the same map on homology.

A contracting homotopy on a chain complex is a degree-one map $h$ whose displayed identity makes the identity chain-homotopic to zero. It implies that all homology vanishes. The converse is not valid for arbitrary complexes of modules.

For cochain maps $f,g:C^{\bullet }\to D^{\bullet }$, a cochain homotopy is a family $K:C^q\to D^{q-1}$ satisfyingCochain-homotopic maps induce the same map on cohomology.

For an open cover $\mathcal{U}$ of $X$, singular chains generated by simplices lying inside one member of $\mathcal{U}$ form a subcomplex chain-homotopy equivalent to the full singular chain complex. Repeated barycentric subdivision makes chains small.