Talk:Middle-dimensional surgery kernel (Ex)

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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. Before addressing (2) we will prove the following: Proposition: 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 $C$. 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 $i'$. 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 $C''$. 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 $(C'')^*$. 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. 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>(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

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

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.

Before addressing (2) we will prove the following:

Proposition:

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 $C$ .

In (1) it was proved that $\ker(d_n)\subseteq C_n$ is a direct summand. The chain complex $C'$ defined as

$\displaystyle C'_m:=\left\{\begin{matrix}C_m&\text{ if }m>n\\\ker(d_n)&\text{ if }n=m\\0&\text{ if }m<n\end{matrix}\right.,~~d'_n:=d_n|_{C_n'}$ $\displaystyle C'_m:=\left\{\begin{matrix}C_m&\text{ if }m>n\\\ker(d_n)&\text{ if }n=m\\0&\text{ if }m<n\end{matrix}\right.,~~d'_n:=d_n|_{C_n'}$

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 $i'$ .

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 $C''$ .

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 $(C'')^*$ .

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.

ad(2):

Choose a chain homotopy $T$ $T$ between the chain map $f$ $f$ from the propsition above and the identity on $C$ $C$ . We can choose $T$ $T$ such that $T^2=0$ $T^2=0$ . Now for $(T+d)_{even}:\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}$ $(T+d)_{even}:\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}$ and $(T+d)_{odd}:\bigoplus_{i\in\Zz}C_{n+2i+1}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i}$ $(T+d)_{odd}:\bigoplus_{i\in\Zz}C_{n+2i+1}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i}$ the composition $(T+d)_{even}\circ(T+d)_{odd}$ $(T+d)_{even}\circ(T+d)_{odd}$ is the identity on $\bigoplus_{i\in\Zz}C_{n+2i+1}$ $\bigoplus_{i\in\Zz}C_{n+2i+1}$ because $f_{n+2i+1}=0$ $f_{n+2i+1}=0$ for all $i\in\Zz$ $i\in\Zz$ . And the composition $(T+d)_{odd}\circ(T+d)_{even}$ $(T+d)_{odd}\circ(T+d)_{even}$ is the identity on $C_{n+2i}$ $C_{n+2i}$ for $i\neq 0$ $i\neq 0$ and the projection onto a complement of $V$ $V$ in degree $n$ $n$ . Let $p:C_n\rightarrow V$ $p:C_n\rightarrow V$ be the projection given by $f$ $f$ and $g:V\rightarrow H_n(C)$ $g:V\rightarrow H_n(C)$ an isomorphism then the map

$\displaystyle ((T+d)_{even},g\circ p):\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}\oplus H_n(C)$ $\displaystyle ((T+d)_{even},g\circ p):\bigoplus_{i\in\Zz}C_{n+2i}\rightarrow \bigoplus_{i\in\Zz}C_{n+2i+1}\oplus H_n(C)$

is an isomorphism. Dualizing gives the statement for cohomology.

Talk:Middle-dimensional surgery kernel (Ex)

Revision as of 12:01, 4 April 2012

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@@ Line 10: / Line 10: @@
 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.
+Before addressing (2) we will prove the following:
+Proposition:
+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 $C$.
+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 $i'$.
+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 $C''$.
+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 $(C'')^*$.
+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.
+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>