# Talk:Structure set (Ex)

Marek Kaluba (Talk | contribs) (Solution to simple h. eq. version) |
Marek Kaluba (Talk | contribs) (Added solution for h-structure set) |
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− | '''Solution''': | + | '''Solution'''(simple structure set): |

− | We begin with | + | 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}^ | + | 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. |

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+ | '''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. | ||

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+ | 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 $f'$. | ||

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## Latest revision as of 13:14, 29 August 2013

**Solution**(simple structure set):

We begin with the map , from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things:

- If manifolds and simply homotopy equivalent to are diffeomorphic then their images by the map belong to the same orbit of -action on .
- If two elements of belong to the same orbit of -action, then they are diffeomorphic.

Let be a smooth manifold and a simple homotopy equivalence. Consider a map which takes to . Suppose now that is a manifold diffeomorphic to , and a simple homotopy equivalence (possibly and ). Then there exists such that the following diagram commutes.

Map is given by composition (the homotopy inverse) and hence is a simple homotopy equivalence. The commutativity of the diagram tells us that in we have the following equalities.

where denotes the -action. Therefore and belong to the same orbit.

Suppose now, that belong to the same orbit of -action. It means, that there exist a simple homotopy equivalence such that**Solution** (homotopy structure set):

Analogously, equality in implies existence of an -cobordism between and extending and .