# Talk:Inertia group I (Ex)

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First observe, that instead of forming a connected sum $M\#\Sigma_f$$; First observe, that instead of forming a connected sum M\#\Sigma_f we may think of it as cutting a disc M-D^n=M^\bullet and glueing the disc back identifying boundary spheres via diffeomorphism f. By the theorem of Cerf definition of M^\bullet does not depend on the embedding D^n \hookrightarrow M. If \Sigma\in I(M) and G\colon M\#\Sigma\to M is a diffeomorphism, find a decomposition M\#\Sigma=M^\bullet\cup_f D^n and set F=G|_{M^\bullet}. Conversely suppose that there exist a diffeomorphism F\colon M^\bullet\to M^\bullet such that F|_{\partial(M^\bullet)=S^{n-1}}=f. Then glue a disc via identity to the source and target M^\bullet and extend F as identity. Composition of inclusion and F, S^{n-1}\to M^\bullet\to M^\bullet is equal to f, hence (by definition) the target is equal to M\#\Sigma. This proves that \Sigma \in I(M). ''(Is the Cerfs thm properly applied?)'' M\#\Sigma_f$ we may think of it as cutting a disc $M-D^n=M^\bullet$$M-D^n=M^\bullet$ and glueing the disc back identifying boundary spheres via diffeomorphism $f$$f$. By the theorem of Cerf definition of $M^\bullet$$M^\bullet$ does not depend on the embedding $D^n \hookrightarrow M$$D^n \hookrightarrow M$.

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

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

(Is the Cerfs thm properly applied?)