Talk:Middle-dimensional surgery kernel (Ex)

First a little lemma:

Let $(C,d)$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}(C,d)$ be a chain complex with $C_{n-1}$$C_{n-1}$ projective, $H_{n-1}(C) = 0$$H_{n-1}(C) = 0$ and $\ker(d_{n-1}) \subseteq C_{n-1}$$\ker(d_{n-1}) \subseteq C_{n-1}$ a direct summand. Then also $\ker(d_n) \subseteq C_n$$\ker(d_n) \subseteq C_n$ is a direct summand.

Proof: Note that $\ker(d_n) \subseteq C_n$$\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$$n-1$ however, we have $\text{im}(d_n) = \ker(d_{n-1})$$\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)$$\ker(d_n)$ is a direct summand of $C_n$$C_n$ if the same statement holds for some lower $n$$n$. However, eventually both terms are zero, since the complex is finite. Being a direct summand in a finitely generated module $\ker(d_n)$$\ker(d_n)$ is then itself finitely generated, and hence also $H_n(C)$$H_n(C)$. 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 $f:C\rightarrow C$$f:C\rightarrow C$ with $f_m=0$$f_m=0$ for $m\neq n$$m\neq n$ and $f_n$$f_n$ the projection onto a direct summand $V$$V$ of $C_n$$C_n$ isomorphic to $H_n$$H_n$ and such that $f$$f$ is chain homotopy equivalent to the identity on $C$$C$.

In (1) it was proved that $\ker(d_n)\subseteq C_n$$\ker(d_n)\subseteq C_n$ is a direct summand. The chain complex $C'$$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$$i':C'\rightarrow C$ induces an isomorphism on homology and since all occurring modules are projective is a chain homotopy equivalence. We can choose a projection $p_n':C_n\rightarrow C_n' = \ker(d_n)$$p_n':C_n\rightarrow C_n' = \ker(d_n)$ with $p_n'\circ i_n'$$p_n'\circ i_n'$ the identity on $C'_n$$C'_n$. Now extend this to a chain map $p'$$p'$ by identity maps on the one side and zero maps on the other and note that $p' \circ i' = id_{C'}$$p' \circ i' = id_{C'}$. Since $i'$$i'$ is a chain homotopy equivalence this implies that $p'$$p'$ is a chain homotopy inverse of $i'$$i'$.

By assumption $H^j(C')=H^j(C)=0$$H^j(C')=H^j(C)=0$ for $j>n$$j>n$ and since $C'$$C'$ is finitely generated and degreewise projective so is $(C')^*$$(C')^*$. Thus as before $\ker((d'_{n+1})^*)\subseteq (C'_n)^*$$\ker((d'_{n+1})^*)\subseteq (C'_n)^*$ is a direct summand. Hence for the chain complex $C''$$C''$ concentrated in degree $n$$n$ with $C''_n:=\ker((d'_{n+1})^*)$$C''_n:=\ker((d'_{n+1})^*)$, the inclusion $i'': C'' \hookrightarrow (C')^*$$i'': C'' \hookrightarrow (C')^*$ is a chain homotopy equivalence. We can again choose a projection $p'':(C')^*\rightarrow C''$$p'':(C')^*\rightarrow C''$ such that $p''$$p''$ is an inverse of $i''$$i''$ and $p''\circ i''$$p''\circ i''$ is the identity on $C''$$C''$.

Putting this together we have chain homotopy equivalences $i:=(i')^{**}\circ(p'')^*:(C'')^{*}\rightarrow C^{**}$$i:=(i')^{**}\circ(p'')^*:(C'')^{*}\rightarrow C^{**}$ and $p:=(i'')^{*}\circ(p')^{**}:C^{**}\rightarrow (C'')^*$$p:=(i'')^{*}\circ(p')^{**}:C^{**}\rightarrow (C'')^*$ with $p$$p$ a chain homotopy inverse of $i$$i$ and $p\circ i$$p\circ i$ the identity on $(C'')^*$$(C'')^*$.

Since for $m\neq n$$m\neq n$ we have $C''_m=0$$C''_m=0$, also $(i\circ p)_m=0$$(i\circ p)_m=0$ in dimensions $\neq n$$\neq n$ and because $p\circ i=\id_{(C'')^*}$$p\circ i=\id_{(C'')^*}$ the map $(i\circ p)_n$$(i\circ p)_n$ is a projection onto a direct summand $V$$V$ of $(C^{**})_n$$(C^{**})_n$. Since $i\circ p$$i\circ p$ induces an isomorphism in homology $V$$V$ has to be isomorphic to $H_n(C)$$H_n(C)$. Upon identification of $C^{**}$$C^{**}$ with $C$$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.