# Talk:Structure set (Ex)

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Revision as of 08:03, 28 August 2013 by Marek Kaluba (Talk | contribs)

**Solution**:

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 But equality in the simple structure set is just the existence of a diffeomorphism making the following diagram commute.