Chain Complexes and Homology - Key Properties

Chain homotopy provides a notion of "equivalence" between chain maps that is coarser than equality but finer than inducing the same homology maps.

Definition1.4Chain Homotopy

Let $f, g: C_\bullet \to D_\bullet$ be chain maps. A chain homotopy from $f$ to $g$ is a collection of homomorphisms $h_n: C_n \to D_{n+1}$ such that:

$f_n - g_n = \partial_{n+1}^D \circ h_n + h_{n-1} \circ \partial_n^C$

for all $n$ . We write $f \simeq g$ and say $f$ and $g$ are chain homotopic.

TheoremChain Homotopic Maps Induce Same Homology

If $f, g: C_\bullet \to D_\bullet$ are chain homotopic, then they induce the same homomorphism on homology: $H_n(f) = H_n(g): H_n(C_\bullet) \to H_n(D_\bullet)$ for all $n$ .

Proof

Let $h$ be a chain homotopy from $f$ to $g$ , and let $[z] \in H_n(C_\bullet)$ be represented by a cycle $z \in Z_n(C_\bullet)$ . Then:

$f_n(z) - g_n(z) = (\partial_{n+1}^D \circ h_n + h_{n-1} \circ \partial_n^C)(z)$

Since $z$ is a cycle, $\partial_n^C(z) = 0$ , so: $f_n(z) - g_n(z) = \partial_{n+1}^D(h_n(z))$

This shows $f_n(z) - g_n(z) \in B_n(D_\bullet)$ , hence $f_n(z)$ and $g_n(z)$ represent the same homology class.

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Definition1.5Chain Homotopy Equivalence

A chain map $f: C_\bullet \to D_\bullet$ is a chain homotopy equivalence if there exists a chain map $g: D_\bullet \to C_\bullet$ such that:

$g \circ f \simeq \text{id}_{C_\bullet}$
$f \circ g \simeq \text{id}_{D_\bullet}$

We say $C_\bullet$ and $D_\bullet$ are chain homotopy equivalent.

ExampleContractible Chain Complex

A chain complex $C_\bullet$ is called contractible if it is chain homotopy equivalent to the zero complex. This happens if and only if the identity map $\text{id}_{C_\bullet}$ is null-homotopic, i.e., there exists $h$ such that: $\text{id}_n = \partial_{n+1} \circ h_n + h_{n-1} \circ \partial_n$

Contractible complexes have zero homology in all degrees.

Remark

Chain homotopy equivalence is weaker than isomorphism of chain complexes but stronger than having isomorphic homology. Two chain complexes can have the same homology without being chain homotopy equivalent, but chain homotopy equivalence always implies isomorphic homology.

ExampleMapping Cone

Given a chain map $f: C_\bullet \to D_\bullet$ , the mapping cone $\text{Cone}(f)$ is defined by: $\text{Cone}(f)_n = D_n \oplus C_{n-1}$

with differential $\partial(d, c) = (\partial_D(d) + f_{n-1}(c), -\partial_C(c))$ . The cone measures the failure of $f$ to be a quasi-isomorphism.