Show that chain homotopy of chain maps is an equivalence relation.
Solution
1. Definition of chain homotopy:
1.1. Let \((C_*, \partial^C)\) and \((D_*, \partial^D)\) be chain complexes. Two chain maps \(f, g: C_* \to D_*\) are chain homotopic (denoted \(f \simeq g\)) if there exists a sequence of homomorphisms \(P_n: C_n \to D_{n+1}\) such that:\[ f_n - g_n = \partial^D_{n+1} P_n + P_{n-1} \partial^C_n \quad \text{for all } n \in \mathbb{Z}. \]
standard definition of chain homotopy.
2. Reflexivity:
2.1. For any chain map \(f: C_* \to D_*\), define \(P_n = 0: C_n \to D_{n+1}\) for all \(n\).Then:\[ \partial^D_{n+1} P_n + P_{n-1} \partial^C_n = 0 = f_n - f_n. \]Thus \(f \simeq f\).
zero homomorphism satisfies the chain homotopy identity for \(f - f = 0\).
3. Symmetry:
3.1. Suppose \(f \simeq g\) via chain homotopy \(P = \{P_n\}\).Then \(f_n - g_n = \partial^D_{n+1} P_n + P_{n-1} \partial^C_n\).Define \(Q_n = -P_n: C_n \to D_{n+1}\).Then:\[ g_n - f_n = -(f_n - g_n) = -\big(\partial^D_{n+1} P_n + P_{n-1} \partial^C_n\big) = \partial^D_{n+1} (-P_n) + (-P_{n-1}) \partial^C_n = \partial^D_{n+1} Q_n + Q_{n-1} \partial^C_n. \]Thus \(g \simeq f\).
linearity of boundary operators and negation.
4. Transitivity:
4.1. Suppose \(f \simeq g\) via chain homotopy \(P = \{P_n\}\) and \(g \simeq h\) via chain homotopy \(Q = \{Q_n\}\).Then:\[ f_n - g_n = \partial^D_{n+1} P_n + P_{n-1} \partial^C_n \quad \text{and} \quad g_n - h_n = \partial^D_{n+1} Q_n + Q_{n-1} \partial^C_n. \]
definitions of \(P\) and \(Q\).
4.2. Define \(R_n = P_n + Q_n: C_n \to D_{n+1}\).Adding the two equations yields:\[ f_n - h_n = (f_n - g_n) + (g_n - h_n) = \partial^D_{n+1} (P_n + Q_n) + (P_{n-1} + Q_{n-1}) \partial^C_n = \partial^D_{n+1} R_n + R_{n-1} \partial^C_n. \]Thus \(f \simeq h\).
linearity of boundary operators.
5. Conclusion:Chain homotopy is reflexive, symmetric, and transitive; hence it is an equivalence relation on chain maps. Q.E.D.
2, 3, and 4.