Talk:Middle-dimensional surgery kernel (Ex)
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− | First a little lemma: \ | + | First a little lemma: \newline |
− | 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. \ | + | 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. \newline |
Proof: | Proof: | ||
Note that $\ker(d_n) \subseteq C_n$ is a direct summand iff the sequence | Note that $\ker(d_n) \subseteq C_n$ is a direct summand iff the sequence |
Revision as of 12:21, 3 April 2012
First a little lemma: \newline
Let be a chain complex with projective, and a direct summand. Then also is a direct summand. \newline
Proof:
Note that is a direct summand iff the sequence
splits. By exactness at however, we have which is projective being a direct summand of a projective module.
ad(1): Iterating the lemma we find that is a direct summand of if the same statement holds for some lower . However eventually both terms are zero, since the complex is finite. Being a direct summand in a finitely generated module is then itself finitely generated, and hence also . The second assertion follows immediately from the universal coefficient theorem.