Talk:Structure set (Ex)

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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.

\displaystyle  \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.

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




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