Talk:Structures on M x I (Ex)
(Created page with "<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 togeth...") |
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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 $(f,f_0)$. | 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) | + | Define the map |
+ | $$F:S^s(M\times I,M\times\{0,1\})\rightarrow \tilde \pi_0 \textup{SHomeo}_+(M), \quad [(f,f_0)] \mapsto[f_0].$$ | ||
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. | 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. | ||
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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 | 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. | 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. | ||
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+ | A proof of this exercise can also be found in \cite{Wang1974}. | ||
<wikitex> | <wikitex> |
Latest revision as of 14:06, 31 May 2012
Recall that an element of the simple structure set in the topological category is represented by a manifold with boundary together with a simple homotopy equivalence such that is a homeomorphism.
Two such objects and are equivalent in the structure set if there exists a homeomorphism such that the maps commute up to simple homotopy equivalence and commute strictly on the boundary .
Take a representative .
Since is a simple homotopy equivalence and is an s-cobordism, the same is true for and by the s-cobordism theorem is homeomorphic to a cylinder and every element of the structure set is represented by an element of the form . defines a homeomorphism and precomposing the map with shows that every element of the structure set is even represented by an element of the form with a self homeomorphism . Using this we will denote these representatives by .
Define the map
To see that is well defined assume that and are equivalent. Then there exists a self homeomorphism of such that and . So is a self homeomorphism of which is on one side and on the other side. So and are pseudo-isotopic and is well-defined.
Take an element and choose a homotopy to the identity. This induces a map with and . So is a preimage of under , i.e. is surjective.
Since is abelian it remains to check that is a homomorphism of groups to show that is abelian.
Let be two representatives as above. is in the structure set equivalent to and the addition of and is defined by gluing together and at the 1- respectively 0-boundary. So , which is the product of in . So is a group homomorphism.
A proof of this exercise can also be found in [Wang1974].