# Talk:Structure set (Ex)

(Difference between revisions)

Solution(simple structure set):

We begin with the map $\mathcal{S}^s(M) \to \mathcal{M}^s(M)$$; '''Solution'''(simple structure set): 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 } . '''Solution''' (homotopy structure set): Basically, the proof follows the same line. However in this case equality [(N_0,f_0)]=[(N_1,f_1)] in \mathcal{S}(M) (by definition) is existence of an h-cobordism (H,\partial_0H,\partial_0H)\colon (W,\partial_0W,\partial1W)\to (M\times I,M\times\{0\},M\times\{1\}) satisfying f_i=\partial_iH\circ g_i\colon N_i\to \partial_iH\to M\times \{i\} for some orientation preserving diffeomorphisms g_i, i=0,1. But the h-cobordism gives us a homotopy equivalence d\colon N_0\to N_1 and the same formula for h\colon M\to M applies. Analogously, equality h\cdot [(N,f)] = \cdot [(N',f')] in \mathcal{S}^h(M) implies existence of an h-cobordism between N and N' extending h\circ f and f'. \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/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_7IHgNZ$$M$ are diffeomorphic then their images by the map belong to the same orbit of $\mathcal{E}^s(M)$$\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. $\displaystyle h\cdot [(N,f)] = [(N,h\circ f)]=[(N,f'\circ d\circ f^{-1}\circ f\simeq f'\circ d)]=[(N',f')],$

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

Solution (homotopy structure set):

Basically, the proof follows the same line. However in this case equality $[(N_0,f_0)]=[(N_1,f_1)]$$[(N_0,f_0)]=[(N_1,f_1)]$ in $\mathcal{S}(M)$$\mathcal{S}(M)$ (by definition) is existence of an $h$$h$-cobordism $\displaystyle (H,\partial_0H,\partial_0H)\colon (W,\partial_0W,\partial1W)\to (M\times I,M\times\{0\},M\times\{1\})$
satisfying $f_i=\partial_iH\circ g_i\colon N_i\to \partial_iH\to M\times \{i\}$$f_i=\partial_iH\circ g_i\colon N_i\to \partial_iH\to M\times \{i\}$ for some orientation preserving diffeomorphisms $g_i$$g_i$, $i=0,1$$i=0,1$. But the h-cobordism gives us a homotopy equivalence $d\colon N_0\to N_1$$d\colon N_0\to N_1$ and the same formula for $h\colon M\to M$$h\colon M\to M$ applies.

Analogously, equality $h\cdot [(N,f)] = \cdot [(N',f')]$$h\cdot [(N,f)] = \cdot [(N',f')]$ in $\mathcal{S}^h(M)$$\mathcal{S}^h(M)$ implies existence of an $h$$h$-cobordism between $N$$N$ and $N'$$N'$ extending $h\circ f$$h\circ f$ and $f'$$f'$.