# Middle-dimensional surgery kernel (Ex)

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


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

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

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