Talk:Structure set (Ex)

Solution:

We begin with the map $\mathcal{S}^s(M) \to \mathcal{M}^s(M)$$<wikitex>; '''Solution''': We begin with the map $\mathcal{S}^s(M) \to \mathcal{M}^s(M)$, from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things: * If manifolds $N$ and $N'$ simply homotopy equivalent to $M$ are diffeomorphic then their images by the map belong to the same orbit of $\mathcal{E}^s(M)$-action on $\mathcal{S}^s(M)$. * If two elements of $\mathcal{S}^s(M)$ belong to the same orbit of $\mathcal{E}^s(M)$-action, then they are diffeomorphic. Let $N$ be a smooth manifold and $f\colon N\to M$ a simple homotopy equivalence. Consider a map which takes $N$ to $[(N,f)]\in \mathcal{S}^s(M)$. Suppose now that $N'$ is a manifold diffeomorphic to $N$, and $f'\colon N'\to M$ a simple homotopy equivalence (possibly $N'=N$ and $f\nsim f'$). Then there exists $h\colon M\to M$ such that the following diagram commutes. $$ \xymatrix{ N \ar[r]^{f} \ar[d]^d_{\cong} & M \ar@{.>}[d]^{h}\ N'\ar[r]^{f'}& M } $$ Map $h$ is given by composition $f'\circ d\circ f^{-1}$ (the homotopy inverse) and hence is a simple homotopy equivalence. The commutativity of the diagram tells us that in $\mathcal{S}^s(M)$ we have the following equalities. $$h\cdot [(N,f)] = [(N,h\circ f)]=[(N,f'\circ d\circ f^{-1}\circ f\simeq f'\circ d)]=[(N',f')],$$ where $\cdot$ denotes the $\mathcal{E}^s(M)$-action. Therefore $[(N,f)]$ and $[(N',f')]$ belong to the same orbit. Suppose now, that $[(N,f)],[(N',f')]\in \mathcal{S}^s(M)$ belong to the same orbit of $\mathcal{E}^s(M)$-action. It means, that there exist a simple homotopy equivalence $h\colon M\to M$ such that $$h\cdot[(N,f)]=[(N,h\circ f)]=[(N',f')].$$ But equality in the simple structure set $\mathcal{S}^s(M)$ is just the existence of a diffeomorphism $d\colon N\to N'$ making the following diagram commute. $$ \xymatrix{ N \ar[r]^{f} \ar@{.>}[d]^d & M \ar[d]^{h}\ N'\ar[r]^{f'}& M } .$$ </wikitex>\mathcal{S}^s(M) \to \mathcal{M}^s(M)$, from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things:

• If manifolds $N$$N$ and $N'$$N'$ simply homotopy equivalent to $M$$M$ are diffeomorphic then their images by the map belong to the same orbit of $\mathcal{E}^s(M)$$\mathcal{E}^s(M)$-action on $\mathcal{S}^s(M)$$\mathcal{S}^s(M)$.
• If two elements of $\mathcal{S}^s(M)$$\mathcal{S}^s(M)$ belong to the same orbit of $\mathcal{E}^s(M)$$\mathcal{E}^s(M)$-action, then they are diffeomorphic.

Let $N$$N$ be a smooth manifold and $f\colon N\to M$$f\colon N\to M$ a simple homotopy equivalence. Consider a map which takes $N$$N$ to $[(N,f)]\in \mathcal{S}^s(M)$$[(N,f)]\in \mathcal{S}^s(M)$. Suppose now that $N'$$N'$ is a manifold diffeomorphic to $N$$N$, and $f'\colon N'\to M$$f'\colon N'\to M$ a simple homotopy equivalence (possibly $N'=N$$N'=N$ and $f\nsim f'$$f\nsim f'$). Then there exists $h\colon M\to M$$h\colon M\to M$ such that the following diagram commutes.

$\displaystyle \xymatrix{ N \ar[r]^{f} \ar[d]^d_{\cong} & M \ar@{.>}[d]^{h}\\ N'\ar[r]^{f'}& M }$

Map $h$$h$ is given by composition $f'\circ d\circ f^{-1}$$f'\circ d\circ f^{-1}$ (the homotopy inverse) and hence is a simple homotopy equivalence. The commutativity of the diagram tells us that in $\mathcal{S}^s(M)$$\mathcal{S}^s(M)$ we have the following equalities.

$\displaystyle h\cdot [(N,f)] = [(N,h\circ f)]=[(N,f'\circ d\circ f^{-1}\circ f\simeq f'\circ d)]=[(N',f')],$ $$\displaystyle h\cdot [(N,f)] = [(N,h\circ f)]=[(N,f'\circ d\circ f^{-1}\circ f\simeq f'\circ d)]=[(N',f')],$$

where $\cdot$$\cdot$ denotes the $\mathcal{E}^s(M)$$\mathcal{E}^s(M)$-action. Therefore $[(N,f)]$$[(N,f)]$ and $[(N',f')]$$[(N',f')]$ belong to the same orbit.

Suppose now, that $[(N,f)],[(N',f')]\in \mathcal{S}^s(M)$$[(N,f)],[(N',f')]\in \mathcal{S}^s(M)$ belong to the same orbit of $\mathcal{E}^s(M)$$\mathcal{E}^s(M)$-action. It means, that there exist a simple homotopy equivalence $h\colon M\to M$$h\colon M\to M$ such that

$\displaystyle h\cdot[(N,f)]=[(N,h\circ f)]=[(N',f')].$ $$\displaystyle h\cdot[(N,f)]=[(N,h\circ f)]=[(N',f')].$$

But equality in the simple structure set $\mathcal{S}^s(M)$$\mathcal{S}^s(M)$ is just the existence of a diffeomorphism $d\colon N\to N'$$d\colon N\to N'$ making the following diagram commute.

$\displaystyle \xymatrix{ N \ar[r]^{f} \ar@{.>}[d]^d & M \ar[d]^{h}\\ N'\ar[r]^{f'}& M } .$