Talk:Structure set (Ex)
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):Basically, the proof follows the same line. However in this case equality in (by definition) is existence of an -cobordism
Analogously, equality in implies existence of an -cobordism between and extending and .