Talk:Inertia group I (Ex)

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First observe, that instead of forming a connected sum M\#\Sigma_f we may think of it as cutting a disc M-D^n=M^\bullet and glueing the disc back identifying boundary spheres via diffeomorphism f. By the theorem of Cerf definition of M^\bullet does not depend on the embedding D^n \hookrightarrow M.

If \Sigma\in I(M) and G\colon M\#\Sigma\to M is a diffeomorphism, find a decomposition M\#\Sigma=M^\bullet\cup_f D^n and set F=G|_{M^\bullet}.

Conversely suppose that there exist a diffeomorphism F\colon M^\bullet\to M^\bullet such that F|_{\partial(M^\bullet)=S^{n-1}}=f. Then glue a disc via identity to the source and target M^\bullet and extend F as identity. Composition of inclusion and F, S^{n-1}\to M^\bullet\to M^\bullet is equal to f, hence (by definition) the target is equal to M\#\Sigma. This proves that \Sigma \in I(M).

(Is the Cerfs thm properly applied?)

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