Talk:Middle-dimensional surgery kernel (Ex)
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First a little lemma: | First a little lemma: | ||
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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 | ||
$$ 0 \longrightarrow \ker(d_n) \longrightarrow C_n \stackrel{d_n}{\longrightarrow} \text{im}(d_n) \longrightarrow 0$$ | $$ 0 \longrightarrow \ker(d_n) \longrightarrow C_n \stackrel{d_n}{\longrightarrow} \text{im}(d_n) \longrightarrow 0$$ | ||
− | 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. | + | 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. | + | 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. |
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+ | '''Comment:''' The second assertion does ''not'' follow from the universal coefficient theorem since the unversal coefficient theorem does not hold in this generality. R is an arbitrary ring with involution! | ||
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+ | Before addressing (2) we will prove the following: | ||
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+ | Proposition: | ||
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+ | Under the assumptions of (2) there exist a chain map $f:C\rightarrow C$ with $f_m=0$ for $m\neq n$ and $f_n$ the projection onto a direct summand $V$ of $C_n$ isomorphic to $H_n$ and such that $f$ is chain homotopy equivalent to the identity on $C$. | ||
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+ | In (1) it was proved that $\ker(d_n)\subseteq C_n$ is a direct summand. The chain complex $C'$ defined as | ||
+ | $$C'_m:=\left\{\begin{matrix}C_m&\text{ if }m>n\\\ker(d_n)&\text{ if }n=m\\0&\text{ if }m<n\end{matrix}\right.,~~d'_n:=d_n|_{C_n'}$$ | ||
+ | is again finitely generated and degreewise projective. The inclusion $i':C'\rightarrow C$ induces an isomorphism on homology and since all occurring modules are projective is a chain homotopy equivalence. We can choose a projection $p_n':C_n\rightarrow C_n' = \ker(d_n)$ with $p_n'\circ i_n'$ the identity on $C'_n$. Now extend this to a chain map $p'$ by identity maps on the one side and zero maps on the other and note that $p' \circ i' = id_{C'}$. Since $i'$ is a chain homotopy equivalence this implies that $p'$ is a chain homotopy inverse of $i'$. | ||
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+ | By assumption $H^j(C')=H^j(C)=0$ for $j>n$ and since $C'$ is finitely generated and degreewise projective so is $(C')^*$. Thus as before $\ker((d'_{n+1})^*)\subseteq (C'_n)^*$ is a direct summand. Hence for the chain complex $C''$ concentrated in degree $n$ with $C''_n:=\ker((d'_{n+1})^*)$, the inclusion $i'': C'' \hookrightarrow (C')^*$ is a chain homotopy equivalence. We can again choose a projection $p'':(C')^*\rightarrow C''$ such that $p''$ is an inverse of $i''$ and $p''\circ i''$ is the identity on $C''$. | ||
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+ | Putting this together we have chain homotopy equivalences $i:=(i')^{**}\circ(p'')^*:(C'')^{*}\rightarrow C^{**}$ and $p:=(i'')^{*}\circ(p')^{**}:C^{**}\rightarrow (C'')^*$ with $p$ a chain homotopy inverse of $i$ and $p\circ i$ the identity on $(C'')^*$. | ||
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+ | Since for $m\neq n$ we have $C''_m=0$, also $(i\circ p)_m=0$ in dimensions $\neq n$ and because $p\circ i=\id_{(C'')^*}$ the map $(i\circ p)_n$ is a projection onto a direct summand $V$ of $(C^{**})_n$. Since $i\circ p$ induces an isomorphism in homology $V$ has to be isomorphic to $H_n(C)$. | ||
+ | Upon identification of $C^{**}$ with $C$ the proposition is proved. | ||
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+ | ad(2): | ||
+ | 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. | ||
</wikitex> | </wikitex> |
Latest revision as of 20:42, 28 May 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.
Comment: The second assertion does not follow from the universal coefficient theorem since the unversal coefficient theorem does not hold in this generality. R is an arbitrary ring with involution!
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.