Acyclic bounded-below projective complexes are contractible

Proof. We construct a contracting homotopy \(s_n\colon P_n \to P_{n+1}\) satisfying \(d_{n+1} s_n + s_{n-1} d_n = \id _{P_n}\) by induction. Say \(P_{\bullet }\) is concentrated in degrees \(\geq m\); we set \(s_n = 0\) for \(n < m\).

For the base case \(n = m\): Since \(P_{m-1} = 0\), the differential \(d_m\colon P_m \to P_{m-1}\) is zero. The condition \(H_m(P) = 0\) then says that \(\ker (d_m)/\im (d_{m+1}) = P_m/\im (d_{m+1}) = 0\), meaning \(d_{m+1}\colon P_{m+1} \to P_m\) is an epimorphism. Since \(P_m\) is projective, the identity \(\id _{P_m}\) lifts along this epimorphism to give \(s_m\colon P_m \to P_{m+1}\) with \(d_{m+1} s_m = \id _{P_m}\). This satisfies our requirement since \(s_{m-1} = 0\).

For the inductive step, suppose we have constructed \(s_k\) for all \(k < n\) satisfying the homotopy relation. Define \[ \alpha _n \quad := \quad \id _{P_n} - s_{n-1} d_n \colon P_n \to P_n. \] We claim that \(\im (\alpha _n) \subseteq \im (d_{n+1})\). Indeed, using the induction hypothesis \(d_n s_{n-1} + s_{n-2} d_{n-1} = \id _{P_{n-1}}\), we compute \[ d_n \circ \alpha _n \,=\, d_n - d_n s_{n-1} d_n \,=\, d_n - (\id _{P_{n-1}} - s_{n-2} d_{n-1}) d_n \,=\, s_{n-2} d_{n-1} d_n \,=\, 0. \] Thus \(\im (\alpha _n) \subseteq \ker (d_n) = \im (d_{n+1})\), where the last equality uses \(H_n(P) = 0\). Since \(P_n\) is projective and \(d_{n+1}\colon P_{n+1} \to \im (d_{n+1})\) is an epimorphism, there exists \(s_n\colon P_n \to P_{n+1}\) with \(d_{n+1} s_n = \alpha _n = \id _{P_n} - s_{n-1} d_n\), as required. □