Talk:Chain duality I (Ex)

(i) Suppose we have a contravariant functor

$\displaystyle T_{\mathbb{A}}: \mathbb{A} \to \mathbb{B}(\mathbb{A}).$ $$\displaystyle T_{\mathbb{A}}: \mathbb{A} \to \mathbb{B}(\mathbb{A}).$$

We would like to extend this to a contravariant functor

$\displaystyle T_{\mathbb{B}}: \mathbb{B}(\mathbb{A}) \to \mathbb{B}(\mathbb{A}).$ $$\displaystyle T_{\mathbb{B}}: \mathbb{B}(\mathbb{A}) \to \mathbb{B}(\mathbb{A}).$$

Let $C$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}C$ be a chain complex in $\mathbb{B}(\mathbb{A})$$\mathbb{B}(\mathbb{A})$, applying $T_\mathbb{A}$$T_\mathbb{A}$ we get the double complex $T_\mathbb{A}(C_{-*})_{*}$$T_\mathbb{A}(C_{-*})_{*}$:

$\displaystyle \xymatrix{ & \vdots \ar[d] & \vdots \ar[d] & \vdots \ar[d] & \\ \ldots \ar[r] & T_\mathbb{A}(C_{i-1})_{j+1} \ar[d] \ar[r] & T_\mathbb{A}(C_{i-1})_{j} \ar[d]^{T_{\mathbb{A}}((d_C)_i)_j} \ar[r] & T_\mathbb{A}(C_{i-1})_{j-1} \ar[d] \ar[r] & \\ \ldots \ar[r] & T_\mathbb{A}(C_i)_{j+1} \ar[d] \ar[r]^{(d_{T_{\mathbb{A}}(C_i)})_{j+1}} & T_\mathbb{A}(C_i)_{j} \ar[d]^{T_{\mathbb{A}}((d_C)_{i+1})_j} \ar[r]^{(d_{T_{\mathbb{A}}(C_i)})_{j}} & T_\mathbb{A}(C_i)_{j-1} \ar[d] \ar[r] & \\ \ldots \ar[r] & T_\mathbb{A}(C_{i+1})_{j+1} \ar[d] \ar[r] & T_\mathbb{A}(C_{i+1})_{j} \ar[d] \ar[r] & T_\mathbb{A}(C_{i+1})_{j-1} \ar[d] \ar[r] & \\ & \vdots & \vdots & \vdots & }$

This can be converted into a chain complex in the usual way:

$\displaystyle T_{\mathbb{B}}(C)_k := \sum_{-i+j=k}{T_\mathbb{A}(C_i)_j}$ $$\displaystyle T_{\mathbb{B}}(C)_k := \sum_{-i+j=k}{T_\mathbb{A}(C_i)_j}$$

$\displaystyle (d_{T_{\mathbb{B}}(C)})_k := d_{T_{\mathbb{A}}(C_{i+1})} + (-1)^jT_{\mathbb{A}}((d_C)_j).$ $$\displaystyle (d_{T_{\mathbb{B}}(C)})_k := d_{T_{\mathbb{A}}(C_{i+1})} + (-1)^jT_{\mathbb{A}}((d_C)_j).$$

This defines what $T_\mathbb{B}$$T_\mathbb{B}$ of an object is, so now we specify its behaviour on morphisms:

$\displaystyle T_\mathbb{B}(f:C\to D)_k := \sum_{-i+j=k}T_\mathbb{A}(f_i)_j$ $$\displaystyle T_\mathbb{B}(f:C\to D)_k := \sum_{-i+j=k}T_\mathbb{A}(f_i)_j$$

where $f_i$$f_i$ denotes the component of $f$$f$ from $C_i \to D_i$$C_i \to D_i$. The claim is that this defines a contravariant functor.

Since $f:C\to D$$f:C\to D$ is a chain map we have

$\displaystyle (d_D)_{i-1}f_i = f_{i-1}(d_C)_i,$ $$\displaystyle (d_D)_{i-1}f_i = f_{i-1}(d_C)_i,$$

so by the contravariance of the functor $T_\mathbb{A}$$T_\mathbb{A}$ we observe that

$\displaystyle T_\mathbb{A}(f_i)T_\mathbb{A}((d_D)_{i-1}) = {T_\mathbb{A}}((d_C)_{i}) T_\mathbb{A}(f_{i-1}).$ $$\displaystyle T_\mathbb{A}(f_i)T_\mathbb{A}((d_D)_{i-1}) = {T_\mathbb{A}}((d_C)_{i}) T_\mathbb{A}(f_{i-1}).$$

Also, because $T_\mathbb{A}$$T_\mathbb{A}$ sends morphisms to morphisms, $T_\mathbb{A}(f_i)$$T_\mathbb{A}(f_i)$ is a chain map, thus

$\displaystyle d_{T_\mathbb{A}(C_i)}T_\mathbb{A}(f_i)_j = T_\mathbb{A}(f_i)_{j-1}d_{T_\mathbb{A}(D_i)}.$ $$\displaystyle d_{T_\mathbb{A}(C_i)}T_\mathbb{A}(f_i)_j = T_\mathbb{A}(f_i)_{j-1}d_{T_\mathbb{A}(D_i)}.$$

These formulae prove precisely that $T_\mathbb{B}(f)$$T_\mathbb{B}(f)$ is a chain map by the commutativity of the following diagram:

$\displaystyle \xymatrix{ & T_\mathbb{A}(D_i)_j \ar[dd]^{T_\mathbb{A}(f_i)_j} \ar[dl]_{d_{T_\mathbb{A}(D_i)}} \ar[dr]^{(-1)^jT_\mathbb{A}(d_{D_j})} & \\ T_\mathbb{A}(D_i)_{j-1} \ar[dd]_{T_\mathbb{A}(f_i)_{j-1}} && T_\mathbb{A}(D_{i+1})_j \ar[dd]^{T_\mathbb{A}(f_{i+1})_j} \\ & T_\mathbb{A}(C_i)_j \ar[dl]^{d_{T_\mathbb{A}(C_i)}} \ar[dr]_{(-1)^jT_\mathbb{A}(d_{C_j})} & \\ T_\mathbb{A}(C_i)_{j-1} && T_\mathbb{A}(C_{i+1})_j \\ }$

Thus $T_\mathbb{A}$$T_\mathbb{A}$ sends objects to objects and morphisms to morphisms. Functoriality of $T_\mathbb{B}$$T_\mathbb{B}$ follows from that of $T_\mathbb{A}$$T_\mathbb{A}$:

$\displaystyle \begin{aligned} T_\mathbb{B}(g\circ f: C \to D \to E)_k &= \sum_{-i+j=k}T_\mathbb{A}((g\circ f)_i)_j \\ &= \sum_{-i+j=k}T_\mathbb{A}(f_i)_jT_\mathbb{A}(g_i)_j \\ &= \sum_{-i+j=k}T_\mathbb{A}(f_i)_j\sum_{-i+j=k}T_\mathbb{A}(g_i)_j \\ &= (T_\mathbb{B}(f))_k \circ (T_\mathbb{B}(g))_k \\ &= (T_\mathbb{B}(f) \circ T_\mathbb{B}(g))_k. \end{aligned}$ $$\displaystyle \begin{aligned} T_\mathbb{B}(g\circ f: C \to D \to E)_k &= \sum_{-i+j=k}T_\mathbb{A}((g\circ f)_i)_j \\ &= \sum_{-i+j=k}T_\mathbb{A}(f_i)_jT_\mathbb{A}(g_i)_j \\ &= \sum_{-i+j=k}T_\mathbb{A}(f_i)_j\sum_{-i+j=k}T_\mathbb{A}(g_i)_j \\ &= (T_\mathbb{B}(f))_k \circ (T_\mathbb{B}(g))_k \\ &= (T_\mathbb{B}(f) \circ T_\mathbb{B}(g))_k. \end{aligned}$$

(ii) Now suppose that the chain complex $C\in \mathbb{B}(\mathbb{A})$$C\in \mathbb{B}(\mathbb{A})$ is concentrated between dimensions $0$$0$ and $n$$n$ and that $T_\mathbb{A}: \mathbb{A} \to \mathbb{A}$$T_\mathbb{A}: \mathbb{A} \to \mathbb{A}$. Then looking at

$\displaystyle T_{\mathbb{B}}(C)_k := \sum_{-i+j=k}{T_\mathbb{A}(C_i)_j}$ $$\displaystyle T_{\mathbb{B}}(C)_k := \sum_{-i+j=k}{T_\mathbb{A}(C_i)_j}$$

we observe that the only non-zero contributions occur when $j=0$$j=0$ and when $i$$i$ is between $0$$0$ and $n$$n$, in other words $T_{\mathbb{B}}(C)_k$$T_{\mathbb{B}}(C)_k$ is only non-zero for $k$$k$ between $-n$$-n$ and $0$$0$ as required.