Talk:Chain duality II (Ex)

Solution

By definition $M\otimes_{\mathbb{A}} N = \Hom_{\mathbb{A}}(T(M),N)$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}M\otimes_{\mathbb{A}} N = \Hom_{\mathbb{A}}(T(M),N)$, where $\Hom$$\Hom$ is the total complex of the Hom complex. We claim that for $C,D$$C,D$ applying $T$$T$ gives a natural map $T\colon \Hom_{\mathbb{A}}(C,D) \rightarrow \Hom_{\mathbb{A}}(T(D), T(C))$$T\colon \Hom_{\mathbb{A}}(C,D) \rightarrow \Hom_{\mathbb{A}}(T(D), T(C))$ of chain complexes. We will leave the subscript $\mathbb{A}$$\mathbb{A}$ understood in the following. On double complexes $T$$T$ induces $T\colon \Hom(C_{-p}, D_q) \rightarrow \Hom(T(D_{q}), T(C_{-p}))$$T\colon \Hom(C_{-p}, D_q) \rightarrow \Hom(T(D_{q}), T(C_{-p}))$ on groups and the contravariance of $T$$T$ assures that this is still a double complex with grading $p,q$$p,q$ as well as the naturality of that map. It remains to indentify $\mathrm{Tot}_{p,q} \Hom(T(D_{q}), T(C_{-p}))$$\mathrm{Tot}_{p,q} \Hom(T(D_{q}), T(C_{-p}))$ with $\mathrm{Tot}_{r,s} \Hom(T(D)_{-r}, T(C)_{s})$$\mathrm{Tot}_{r,s} \Hom(T(D)_{-r}, T(C)_{s})$. But this holds as the complexes are finite, so we can pull all sums out of the $\Hom$$\Hom$ before any totalization and get the same quadruple complexes. Then note that it does not matter in which order one takes totalization.

Consider the diagram

$\displaystyle \xymatrix{ \Hom(TM,N) \ar[d]_{T} & \\ \Hom(TN, T^2 M) \ar[r]^T \ar[d]_{(e_M)_*} & \Hom(T^3 M , T^2 N) \ar[d]_{(T e_M)^*} \\ \Hom(TN, M) \ar[r]^-T & \Hom(TM, T^2N) \ar[r]^{(e_N)_*} & \Hom(TM, N) }$

It commutes because of the naturality of $T$$T$. The vertical left composition induces the map $T_{M,N}\colon M \otimes N \rightarrow N \otimes M$$T_{M,N}\colon M \otimes N \rightarrow N \otimes M$, the lower horizontal composition the map $T_{N,M}$$T_{N,M}$. We show the composition is the identity. An element in $\Hom(TM,N)$$\Hom(TM,N)$ is represented by a map $f$$f$ of chain complexes. It gets mapped to $e_N \circ T^2(f) \circ T(e_M)$$e_N \circ T^2(f) \circ T(e_M)$. As $e\colon T^2 \rightarrow \mathrm{Id}$$e\colon T^2 \rightarrow \mathrm{Id}$ is a natural transformation $e_N \circ T^2 (f) = f \circ e_{TM}$$e_N \circ T^2 (f) = f \circ e_{TM}$ by the chain duality condition. But $e_{TM} \circ T(e_M) = \mathrm{Id}_{TM}$$e_{TM} \circ T(e_M) = \mathrm{Id}_{TM}$, so $f$$f$ gets mapped to $f$$f$.