Talk:Middle-dimensional surgery kernel (Ex)

Revision as of 12:20, 3 April 2012

First a little lemma: \\ Let $<wikitex> First a little lemma: \ Let $(C,d)$ be a chain complex with $C_{n-1}$ projective, $H_{n-1}(C) = 0$ and $\ker(d_{n-1}) \subseteq C_{n-1}$ a direct summand. Then also $\ker(d_n) \subseteq C_n$ is a direct summand. \ Proof: Note that $\ker(d_n) \subseteq C_n$ is a direct summand iff the sequence $$ 0 \longrightarrow \ker(d_n) \longrightarrow C_n \stackrel{d_n}{\longrightarrow} \text{im}(d_n) \longrightarrow 0$$ splits. By exactness at $n-1$ however, we have $\text{im}(d_n) = \ker(d_{n-1})$ which is projective being a direct summand of a projective module. ad(1): Iterating the lemma we find that $\ker(d_n)$ is a direct summand of $C_n$ if the same statement holds for some lower $n$. However eventually both terms are zero, since the complex is finite. Being a direct summand in a finitely generated module $\ker(d_n)$ is then itself finitely generated, and hence also $H_n(C)$. The second assertion follows immediately from the universal coefficient theorem. </wikitex>(C,d)$ be a chain complex with $C_{n-1}$ projective, $H_{n-1}(C) = 0$ and $\ker(d_{n-1}) \subseteq C_{n-1}$ a direct summand. Then also $\ker(d_n) \subseteq C_n$ is a direct summand. \\ Proof: Note that $\ker(d_n) \subseteq C_n$ is a direct summand iff the sequence

$$\displaystyle 0 \longrightarrow \ker(d_n) \longrightarrow C_n \stackrel{d_n}{\longrightarrow} \text{im}(d_n) \longrightarrow 0$$

splits. By exactness at $n-1$ however, we have $\text{im}(d_n) = \ker(d_{n-1})$ which is projective being a direct summand of a projective module.

ad(1): Iterating the lemma we find that $\ker(d_n)$ is a direct summand of $C_n$ if the same statement holds for some lower $n$. However eventually both terms are zero, since the complex is finite. Being a direct summand in a finitely generated module $\ker(d_n)$ is then itself finitely generated, and hence also $H_n(C)$. The second assertion follows immediately from the universal coefficient theorem.