Exercise 5, § 2

Let C be a complex, s a splitting of C, $\delta : C \to C$ a graded homomorphism of degree – 1, such that $(d + \delta) \circ (d + \delta) = 0$. We denote by C’ the complex $(C, d + \delta)$. We suppose that the a-linear map $\theta$ of C into C defined by $\theta = 1_C + \delta s$ is bijective. (a) In order that $s \theta^{-1}$ be a splitting of C’, it is necessary and sufficient that $\delta(Z(C)) \subset \theta(B(C))$. (b) Suppose the conditions of (a) are satisfied. Now prove that there exist direct sum decompositions: $C = B(C) \oplus \mathrm{Ker}\,(ds) = B(C') \oplus \mathrm{Ker}\,(ds)$. If $p$ (resp. $p'$) is the projector with image $\mathrm{Ker}\,(ds)$ and kernel $B(C)$ (resp. $B(C')$), we have $p' = p \theta^{-1}$. (c) Suppose moreover that the a-linear map $\lambda$ of C into C defined by $\lambda = 1_C + s \delta$ is bijective. Now prove that $Z(C') = \lambda^{-1}(Z(C))$ and $C = Z(C) \oplus \mathrm{Im}\,(sd) = Z(C') \oplus \mathrm{Im}\,(sd)$. If $q$ (resp. $q'$) is the projector with image $Z(C)$ (resp. $Z(C')$) and kernel $\mathrm{Im}\,(sd)$, we have $q' = \lambda^{-1} q$.