Talk:Supplement II (Ex)

Definition 0.1. Let $K$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}K$ be a subcomplex of a simplicial complex $L$$L$. The barycentric subdivision $L^\prime$$L^\prime$ is the simplicial complex with $k$$k$-simplices $\lbrace \tau_0 < \cdots < \tau_k \rbrace$$\lbrace \tau_0 < \cdots < \tau_k \rbrace$ for $\tau_i$$\tau_i$ simplices of $L$$L$ (totally-ordered by inclusion), and $K^\prime \subseteq L^\prime$$K^\prime \subseteq L^\prime$ is the subcomplex whose $k$$k$-simplices are $\lbrace \tau_0 < \cdots < \tau_k \rbrace$$\lbrace \tau_0 < \cdots < \tau_k \rbrace$ with $\tau_i \in K$$\tau_i \in K$ for all $i$$i$. The supplement $\overline{K}$$\overline{K}$ of $K$$K$ in $L$$L$ is the subcomplex $\overline{K}\subseteq L^\prime$$\overline{K}\subseteq L^\prime$ whose $k$$k$-simplices are $\lbrace \tau_0 < \cdots < \tau_k \rbrace$$\lbrace \tau_0 < \cdots < \tau_k \rbrace$ with $\tau_i \notin K$$\tau_i \notin K$ for all $i$$i$.

Definition 0.2. For two spaces $X$$X$ and $Y$$Y$, the join $X*Y$$X*Y$ is by definition the quotient of $X\times I\times Y$$X\times I\times Y$ by the equivalence relation given by

$\displaystyle \begin{aligned} (x,0,y) &\sim (x^\prime ,0,y)\\ (x,1,y) &\sim (x,1,y^\prime ); \end{aligned}$ $$\displaystyle \begin{aligned} (x,0,y) &\sim (x^\prime ,0,y)\\ (x,1,y) &\sim (x,1,y^\prime ); \end{aligned}$$

in other words $X\times Y$$X\times Y$ is collapsed to $Y$$Y$ at the $0$$0$ end and is collapsed to $X$$X$ at the $1$$1$ end. So formally its elements are $(x,t,y)$$(x,t,y)$ subject to this equivalence relation, but we will write them as formal sums $tx+(1-t)y$$tx+(1-t)y$. (This notation indicates that when $t=0$$t=0$ it doesn't matter what $x$$x$ is and when $t=1$$t=1$ it doesn't matter what $y$$y$ is.)

Construction 0.3. We will embed $|L^\prime| \hookrightarrow |K^\prime|*|\overline{K}|$$|L^\prime| \hookrightarrow |K^\prime|*|\overline{K}|$. First, we embed the subspaces $|K^\prime|$$|K^\prime|$ and $|\overline{K}|$$|\overline{K}|$ of $|L^\prime|$$|L^\prime|$ in the obvious way, i.e.

$\displaystyle \begin{aligned} x \mapsto 1x+0 \;\colon\; |K^\prime| &\hookrightarrow |K^\prime|*|\overline{K}|\\ y \mapsto 0+1y \;\colon\; |\overline{K}| &\hookrightarrow |K^\prime|*|\overline{K}|. \end{aligned}$ $$\displaystyle \begin{aligned} x \mapsto 1x+0 \;\colon\; |K^\prime| &\hookrightarrow |K^\prime|*|\overline{K}|\\ y \mapsto 0+1y \;\colon\; |\overline{K}| &\hookrightarrow |K^\prime|*|\overline{K}|. \end{aligned}$$

Let $\sigma$$\sigma$ be a simplex of $L^\prime$$L^\prime$ which is not in $K^\prime$$K^\prime$ or $\overline{K}$$\overline{K}$. So

$\displaystyle \sigma \;=\; \lbrace \tau_0 < \cdots < \tau_j < \tau_{j+1} < \cdots < \tau_k \rbrace$ $$\displaystyle \sigma \;=\; \lbrace \tau_0 < \cdots < \tau_j < \tau_{j+1} < \cdots < \tau_k \rbrace$$

with $\tau_0, \ldots, \tau_j$$\tau_0, \ldots, \tau_j$ in $K$$K$ and $\tau_{j+1}, \ldots, \tau_k$$\tau_{j+1}, \ldots, \tau_k$ not in $K$$K$ ($k>j\geq 0$$k>j\geq 0$). A point in $|\sigma|$$|\sigma|$ is given in barycentric coordinates by

$\displaystyle \sum_{i=0}^k s_i \tau_i \quad\text{with}\quad \sum_{i=0}^k s_i = 1.$ $$\displaystyle \sum_{i=0}^k s_i \tau_i \quad\text{with}\quad \sum_{i=0}^k s_i = 1.$$

We define the embedding $|\sigma| \hookrightarrow |K^\prime|*|\overline{K}|$$|\sigma| \hookrightarrow |K^\prime|*|\overline{K}|$ by the formula

$\displaystyle \sum_{i=0}^k s_i \tau_i \;\mapsto\; \left( \sum_{i=0}^j s_i \right) \lbrace \tau_0 < \cdots < \tau_j \rbrace + \left( \sum_{i=j+1}^k s_i \right) \lbrace \tau_{j+1} < \cdots < \tau_k \rbrace .$ $$\displaystyle \sum_{i=0}^k s_i \tau_i \;\mapsto\; \left( \sum_{i=0}^j s_i \right) \lbrace \tau_0 < \cdots < \tau_j \rbrace + \left( \sum_{i=j+1}^k s_i \right) \lbrace \tau_{j+1} < \cdots < \tau_k \rbrace .$$

These embeddings of $|\sigma|$$|\sigma|$, for $\sigma\in L^\prime \setminus (K^\prime \cup \overline{K})$$\sigma\in L^\prime \setminus (K^\prime \cup \overline{K})$, together with the embeddings of $|K^\prime|$$|K^\prime|$ and $|\overline{K}|$$|\overline{K}|$ above, glue to give a well-defined embedding of $|L^\prime|$$|L^\prime|$ in $|K^\prime|*|\overline{K}|$$|K^\prime|*|\overline{K}|$. (Since the definitions agree on overlaps.)

Deformation retracts 0.4. Now define

$\displaystyle \begin{aligned} N &:= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t\geq \tfrac12 \rbrace \cap |L^\prime|\\ \overline{N} &:= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t\leq \tfrac12 \rbrace \cap |L^\prime|, \end{aligned}$ $$\displaystyle \begin{aligned} N &:= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t\geq \tfrac12 \rbrace \cap |L^\prime|\\ \overline{N} &:= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t\leq \tfrac12 \rbrace \cap |L^\prime|, \end{aligned}$$

and note that

$\displaystyle \begin{aligned} |K^\prime| &= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t=1 \rbrace \\ |\overline{K}| &= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t=0 \rbrace . \end{aligned}$ $$\displaystyle \begin{aligned} |K^\prime| &= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t=1 \rbrace \\ |\overline{K}| &= \lbrace tx+(1-t)y \in |K^\prime|*|\overline{K}| \;|\; t=0 \rbrace . \end{aligned}$$

There is then an obvious deformation retraction

$\displaystyle r\colon \quad tx+(1-t)y \mapsto x \qquad N \longrightarrow |K^\prime|,$ $$\displaystyle r\colon \quad tx+(1-t)y \mapsto x \qquad N \longrightarrow |K^\prime|,$$

where $N\xrightarrow{r} |K^\prime| \hookrightarrow N$$N\xrightarrow{r} |K^\prime| \hookrightarrow N$ is homotopic to the identity via

$\displaystyle (tx+(1-t)y,s) \mapsto (t+(1-t)s)x + (1-t)(1-s)y \qquad N\times I \longrightarrow N.$ $$\displaystyle (tx+(1-t)y,s) \mapsto (t+(1-t)s)x + (1-t)(1-s)y \qquad N\times I \longrightarrow N.$$

Similarly there is a deformation retraction

$\displaystyle \overline{r}\colon \quad tx+(1-t)y \mapsto y \qquad \overline{N} \longrightarrow |\overline{K}|,$ $$\displaystyle \overline{r}\colon \quad tx+(1-t)y \mapsto y \qquad \overline{N} \longrightarrow |\overline{K}|,$$

with $\overline{N}\xrightarrow{\overline{r}} |\overline{K}| \hookrightarrow \overline{N}$$\overline{N}\xrightarrow{\overline{r}} |\overline{K}| \hookrightarrow \overline{N}$ homotopic to the identity via

$\displaystyle (tx+(1-t)y,s) \mapsto t(1-s)x + ((1-t)+st)y \qquad \overline{N} \times I \longrightarrow \overline{N}.$ $$\displaystyle (tx+(1-t)y,s) \mapsto t(1-s)x + ((1-t)+st)y \qquad \overline{N} \times I \longrightarrow \overline{N}.$$