Talk:Inertia group I (Ex)

First observe, that instead of forming a connected sum $M\#\Sigma_f$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}M\#\Sigma_f$ we may think of it as cutting a disc $M-D^n=M^\bullet$$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?)