Talk:Structures on M x I (Ex)

Latest revision as of 14:06, 31 May 2012

Recall that an element of the simple structure set $<wikitex>; Recall that an element of the simple structure set $S^s(M\times I,M\times\{0,1\})$ in the topological category is represented by a manifold $W$ with boundary together with a simple homotopy equivalence $(f,\partial f):(W,\partial W)\rightarrow (M\times I,M\times\{0,1\})$ such that $\partial f:\partial W\rightarrow M\times \{0,1\}$ is a homeomorphism. Two such objects $(W,\partial W,f,\partial f)$ and $(W',\partial W',f',\partial f')$ are equivalent in the structure set if there exists a homeomorphism $h:W\rightarrow W'$ such that the maps commute up to simple homotopy equivalence $f\sim_s f'\circ h$ and commute strictly on the boundary $\partial f=\partial f'\circ h|_{\partial W}$. Take a representative $(W,\partial W,f,\partial f)$. Since $f:W\rightarrow M\times I$ is a simple homotopy equivalence and $M\times I$ is an s-cobordism, the same is true for $W$ and by the s-cobordism theorem $W$ is homeomorphic to a cylinder $M_0\times I$ and every element of the structure set is represented by an element of the form $(M_0\times I,M_0\times\{0,1\},f,\partial f)$. $\partial f|_{M_0\times\{1\}}$ defines a homeomorphism $g:M_0\rightarrow M$ and precomposing the map $f:M_0\times I\rightarrow M\times I$ with $g^{-1}\times\id_I:M\times I\rightarrow M_0\times I$ shows that every element of the structure set is even represented by an element of the form $(M\times I,M\times\{0,1\},f,(f_0,id))$ with $f_0$ a self homeomorphism $f_0:M\rightarrow M$. Using this we will denote these representatives by $(f,f_0)$. Define the map $F:S^s(M\times I,M\times\{0,1\})\rightarrow \tilde \pi_0 \textup{SHomeo}_+(M)$ by sending $[(f,f_0)]$ onto $[f_0]$. To see that $F$ is well defined assume that $(f,f_0)$ and $(f',f_0')$ are equivalent. Then there exists a self homeomorphism $h$ of $M\times I$ such that $h|_{M\times\{1\}}=\id$ and $h|_{M\times\{0\}}=f\circ g^{-1}$. So $k:=h\circ(g\times \id)$ is a self homeomorphism of $M\times I$ which is $f$ on one side and $g$ on the other side. So $f$ and $g$ are pseudo-isotopic and $F$ is well-defined. Take an element $f\in \tilde \pi_0 \textup{SHomeo}_+(M)$ and choose a homotopy $H$ to the identity. This induces a map $K:M\times I\rightarrow M\times I,(m,t)\mapsto(H(m,t),t)$ with $K|_{M\times\{0\}}=f$ and $K|_{M\times{1}}=\id$. So $[(K,f)]\in S^s(M\times I, M\times\{0,1\})$ is a preimage of $f$ under $F$, i.e. $F$ is surjective. Since $S^s(M\times I, M\times\{0,1\})$ is abelian it remains to check that $F$ is a homomorphism of groups to show that $\tilde \pi_0 \textup{SHomeo}_+(M)$ is abelian. Let $(f,f_0),(f',f_0')$ be two representatives as above. $(f',f'_0)$ is in the structure set equivalent to $(M\times I,M\times\{0,1\},((f_0\times \id))\circ f',(f_0\circ f'_0,f_0))$ and the addition $(g,g_0)$ of $(f,f_0)$ and $(f',f_0')$ is defined by gluing together $(M\times I, M\times\{0,1\},f,(f_0,id))$ and $(M\times I,M\times\{0,1\},((f_0\times \id))\circ f',(f_0\circ f'_0,f_0))$ at the 1- respectively 0-boundary. So $g_0=f_0\circ f_0'$, which is the product of $f_0,f_0'$ in $\tilde \pi_0 \textup{SHomeo}_+(M)$. So $F$ is a group homomorphism. A proof of this exercise can also be found in \cite{Wang1974}. <wikitex>S^s(M\times I,M\times\{0,1\})$ in the topological category is represented by a manifold $W$ with boundary together with a simple homotopy equivalence $(f,\partial f):(W,\partial W)\rightarrow (M\times I,M\times\{0,1\})$ such that $\partial f:\partial W\rightarrow M\times \{0,1\}$ is a homeomorphism.

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

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

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

Define the map

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

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

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

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

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

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