# Talk:Structure set (Ex)

Solution:

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

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

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$$h\colon M\to M$ such that
$\displaystyle h\cdot[(N,f)]=[(N,h\circ f)]=[(N',f')].$
But equality in the simple structure set $\mathcal{S}^s(M)$$\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.
$\displaystyle \xymatrix{ N \ar[r]^{f} \ar@{.>}[d]^d & M \ar[d]^{h}\\ N'\ar[r]^{f'}& M } .$