Middle-dimensional surgery kernel (Ex)

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The goal of this exercise is to prove the following statement which will be indispensible in defining the surgery obstruction.

Proposition 0.1. Let f:M\rightarrow X be a degree 1 normal map from a 2k-dimensional (resp. (2k+1)-dimensional) manifold to a geometric Poincaré complex, inducing the isomorphism f_\ast:\pi_1(M)\cong\pi_1(X)=:\pi. Denote by K_i(M)=K_i(\widetilde{M}) the homology surgery kernel \mathbb{Z}[\pi]-module. If f is k-connected the kernel module K_k(M) is finitely generated and stably free.

The statement essentially follows from the technical lemma which you are asked to prove.

Lemma 0.2 [Ranicki2002, Lemma 10.26]. Let R be a ring with involution and C=C_\ast a finite chain complex of finitely generated projective (left) R-modules.

1) If for i<n, H_i(C)=0 for some integer n then the R-module H_n(C) is finitely generated and
\displaystyle H^n(C)\rightarrow H_n(C)^\ast, \quad  f\mapsto (x\mapsto f(x)).
is an isomorphism.

2) If in addition for j>n, H^j(C)=0 for the same integer n then there are isomorphisms
\displaystyle H_n(C)\oplus\sum_{i\in\mathbb{Z}}C_{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C_{n+2j},
\displaystyle H^n(C)\oplus\sum_{i\in\mathbb{Z}}C^{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C^{n+2j},
where under further assumption that C is a chain complex of free modules, the latter isomorphism implies that H_n(C) and H^n(C) are stably free and hence H_n(C) and H^n(C) are dual.

The proposition is given as lemma 4.19 in [Lück2001], however the proof is incomplete. Alternatively a good proof can be found in [Wall1999] and a more detailed one in [Ranicki2002].

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