Middle-dimensional surgery kernel (Ex)
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− | '''2)''' If in addition for $j>n$, $H^j(C)=0$ for the same integer $n$ then | + | '''2)''' If in addition for $j>n$, $H^j(C)=0$ for the same integer $n$ then there are isomorphisms$$H_n(C)\oplus\sum_{i\in\mathbb{Z}}C_{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C_{n+2j}, $$$$H^n(C)\oplus\sum_{i\in\mathbb{Z}}C^{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C^{n+2j},$$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. |
{{endthm|Lemma}} | {{endthm|Lemma}} | ||
Revision as of 18:31, 29 March 2012
The goal of this exercise is to prove the following statement which will be indispensible in defining the surgery obstruction.
Proposition 0.1. Let be a degree 1 normal map from a -dimensional (resp. -dimensional) manifold to a geometric Poincaré complex, inducing the isomorphism . Denote by the homology surgery kernel -module. If is -connected the kernel module 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 be a ring with involution and a finite chain complex of finitely generated projective (left) -modules.
1) If for , for some integer then the -module is finitely generated and
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].