Talk:Structures on M x I (Ex)

Recall that an element of the simple structure set $S^s(M\times I,M\times\{0,1\})$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}S^s(M\times I,M\times\{0,1\})$ in the topological category is represented by a manifold $W$$W$ with boundary together with a simple homotopy equivalence $(f,\partial f):(W,\partial W)\rightarrow (M\times I,M\times\{0,1\})$$(f,\partial f):(W,\partial W)\rightarrow (M\times I,M\times\{0,1\})$ such that $\partial f:\partial W\rightarrow M\times \{0,1\}$$\partial f:\partial W\rightarrow M\times \{0,1\}$ is a homeomorphism.

Two such objects $(W,\partial W,f,\partial f)$$(W,\partial W,f,\partial f)$ and $(W',\partial W',f',\partial f')$$(W',\partial W',f',\partial f')$ are equivalent in the structure set if there exists a homeomorphism $h:W\rightarrow W'$$h:W\rightarrow W'$ such that the maps commute up to simple homotopy equivalence $f\sim_s f'\circ h$$f\sim_s f'\circ h$ and commute strictly on the boundary $\partial f=\partial f'\circ h|_{\partial W}$$\partial f=\partial f'\circ h|_{\partial W}$.

Take a representative $(W,\partial W,f,\partial f)$$(W,\partial W,f,\partial f)$.

Since $f:W\rightarrow M\times I$$f:W\rightarrow M\times I$ is a simple homotopy equivalence and $M\times I$$M\times I$ is an s-cobordism, the same is true for $W$$W$ and by the s-cobordism theorem $W$$W$ is homeomorphic to a cylinder $M_0\times I$$M_0\times I$ and every element of the structure set is represented by an element of the form $(M_0\times I,M_0\times\{0,1\},f,\partial f)$$(M_0\times I,M_0\times\{0,1\},f,\partial f)$. $\partial f|_{M_0\times\{1\}}$$\partial f|_{M_0\times\{1\}}$ defines a homeomorphism $g:M_0\rightarrow M$$g:M_0\rightarrow M$ and precomposing the map $f:M_0\times I\rightarrow M\times I$$f:M_0\times I\rightarrow M\times I$ with $g^{-1}\times\id_I:M\times I\rightarrow M_0\times I$$g^{-1}\times\id_I:M\times I\rightarrow M_0\times I$ shows that every element of the structure set is even represented by an element of the form $(M\times I,M\times\{0,1\},f,(f_0,id))$$(M\times I,M\times\{0,1\},f,(f_0,id))$ with $f_0$$f_0$ a self homeomorphism $f_0:M\rightarrow M$$f_0:M\rightarrow M$. Using this we will denote these representatives by $(f,f_0)$$(f,f_0)$.

Define the map

$\displaystyle F:S^s(M\times I,M\times\{0,1\})\rightarrow \tilde \pi_0 \textup{SHomeo}_+(M), \quad [(f,f_0)] \mapsto[f_0].$ $$\displaystyle F:S^s(M\times I,M\times\{0,1\})\rightarrow \tilde \pi_0 \textup{SHomeo}_+(M), \quad [(f,f_0)] \mapsto[f_0].$$

To see that $F$$F$ is well defined assume that $(f,f_0)$$(f,f_0)$ and $(f',f_0')$$(f',f_0')$ are equivalent. Then there exists a self homeomorphism $h$$h$ of $M\times I$$M\times I$ such that $h|_{M\times\{1\}}=\id$$h|_{M\times\{1\}}=\id$ and $h|_{M\times\{0\}}=f\circ g^{-1}$$h|_{M\times\{0\}}=f\circ g^{-1}$. So $k:=h\circ(g\times \id)$$k:=h\circ(g\times \id)$ is a self homeomorphism of $M\times I$$M\times I$ which is $f$$f$ on one side and $g$$g$ on the other side. So $f$$f$ and $g$$g$ are pseudo-isotopic and $F$$F$ is well-defined.

Take an element $f\in \tilde \pi_0 \textup{SHomeo}_+(M)$$f\in \tilde \pi_0 \textup{SHomeo}_+(M)$ and choose a homotopy $H$$H$ to the identity. This induces a map $K:M\times I\rightarrow M\times I,(m,t)\mapsto(H(m,t),t)$$K:M\times I\rightarrow M\times I,(m,t)\mapsto(H(m,t),t)$ with $K|_{M\times\{0\}}=f$$K|_{M\times\{0\}}=f$ and $K|_{M\times{1}}=\id$$K|_{M\times{1}}=\id$. So $[(K,f)]\in S^s(M\times I, M\times\{0,1\})$$[(K,f)]\in S^s(M\times I, M\times\{0,1\})$ is a preimage of $f$$f$ under $F$$F$, i.e. $F$$F$ is surjective.

Since $S^s(M\times I, M\times\{0,1\})$$S^s(M\times I, M\times\{0,1\})$ is abelian it remains to check that $F$$F$ is a homomorphism of groups to show that $\tilde \pi_0 \textup{SHomeo}_+(M)$$\tilde \pi_0 \textup{SHomeo}_+(M)$ is abelian.

Let $(f,f_0),(f',f_0')$$(f,f_0),(f',f_0')$ be two representatives as above. $(f',f'_0)$$(f',f'_0)$ is in the structure set equivalent to $(M\times I,M\times\{0,1\},((f_0\times \id))\circ f',(f_0\circ f'_0,f_0))$$(M\times I,M\times\{0,1\},((f_0\times \id))\circ f',(f_0\circ f'_0,f_0))$ and the addition $(g,g_0)$$(g,g_0)$ of $(f,f_0)$$(f,f_0)$ and $(f',f_0')$$(f',f_0')$ is defined by gluing together $(M\times I, M\times\{0,1\},f,(f_0,id))$$(M\times I, M\times\{0,1\},f,(f_0,id))$ and $(M\times I,M\times\{0,1\},((f_0\times \id))\circ f',(f_0\circ f'_0,f_0))$$(M\times I,M\times\{0,1\},((f_0\times \id))\circ f',(f_0\circ f'_0,f_0))$ at the 1- respectively 0-boundary. So $g_0=f_0\circ f_0'$$g_0=f_0\circ f_0'$, which is the product of $f_0,f_0'$$f_0,f_0'$ in $\tilde \pi_0 \textup{SHomeo}_+(M)$$\tilde \pi_0 \textup{SHomeo}_+(M)$. So $F$$F$ is a group homomorphism.

A proof of this exercise can also be found in [Wang1974].