Talk:Middle-dimensional surgery kernel (Ex)
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− | Choose a chain homotopy $T$ between the chain map $f$ from the propsition above and the identity on $C$. We can choose $T$ such that $T^2=0$. Now for $(T+d)_{even}:\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}$ and $(T+d)_{odd}:\bigoplus_{i\in\Zz}C_{n+2i+1}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i}$ the composition $(T+d)_{even}\circ(T+d)_{odd}$ is the identity on $\bigoplus_{i\in\Zz}C_{n+2i+1}$ because $f_{n+2i+1}=0$ for all $i\in\Zz$. And the composition $(T+d)_{odd}\circ(T+d)_{even}$ is the identity on $C_{n+2i}$ for $i\neq 0$ and the projection onto a complement of $V$ in degree $n$. Let $p:C_n\rightarrow V$ be the projection given by $f$ and $g:V\rightarrow H_n(C)$ an isomorphism then the map $$((T+d)_{even},g\circ p):\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}\oplus H_n(C)$$ | + | Choose a chain homotopy $T$ between the chain map $f$ from the propsition above and the identity on $C$. We can choose $T$ such that $T^2=0$ (???). Now for $(T+d)_{even}:\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}$ and $(T+d)_{odd}:\bigoplus_{i\in\Zz}C_{n+2i+1}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i}$ the composition $(T+d)_{even}\circ(T+d)_{odd}$ is the identity on $\bigoplus_{i\in\Zz}C_{n+2i+1}$ because $f_{n+2i+1}=0$ for all $i\in\Zz$. And the composition $(T+d)_{odd}\circ(T+d)_{even}$ is the identity on $C_{n+2i}$ for $i\neq 0$ and the projection onto a complement of $V$ in degree $n$. Let $p:C_n\rightarrow V$ be the projection given by $f$ and $g:V\rightarrow H_n(C)$ an isomorphism then the map $$((T+d)_{even},g\circ p):\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}\oplus H_n(C)$$ |
is an isomorphism. Dualizing gives the statement for cohomology. | is an isomorphism. Dualizing gives the statement for cohomology. | ||
</wikitex> | </wikitex> |
Revision as of 12:08, 4 April 2012
First a little lemma:
Let be a chain complex with projective, and a direct summand. Then also is a direct summand.
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.
Before addressing (2) we will prove the following:
Proposition:
Under the assumptions of (2) there exist a chain map with for and the projection onto a direct summand of isomorphic to and such that is chain homotopy equivalent to the identity on .
In (1) it was proved that is a direct summand. The chain complex defined as
is again finitely generated and degreewise projective. The inclusion induces an isomorphism on homology and since all occurring modules are projective is a chain homotopy equivalence. We can choose a projection with the identity on . Now extend this to a chain map by identity maps on the one side and zero maps on the other and note that . Since is a chain homotopy equivalence this implies that is a chain homotopy inverse of .
By assumption for and since is finitely generated and degreewise projective so is . Thus as before is a direct summand. Hence for the chain complex concentrated in degree with , the inclusion is a chain homotopy equivalence. We can again choose a projection such that is an inverse of and is the identity on .
Putting this together we have chain homotopy equivalences and with a chain homotopy inverse of and the identity on .
Since for we have , also in dimensions and because the map is a projection onto a direct summand of . Since induces an isomorphism in homology has to be isomorphic to . Upon identification of with the proposition is proved.
ad(2):
Choose a chain homotopy between the chain map from the propsition above and the identity on . We can choose such that (???). Now for and the composition is the identity on because for all . And the composition is the identity on for and the projection onto a complement of in degree . Let be the projection given by and an isomorphism then the mapis an isomorphism. Dualizing gives the statement for cohomology.