I was reading Milnor's topology from the differential point of view.page 24
In the Homogeneity lemma,which state that for any two interiror point $y,z$ of smooth connected manifold $N$,exist a diffeomorphism $h:N\to N$ that is smooth isotopic to identity and carries $y$ to $z$.
He prove it as follows:
- for $N = S^{n-1}$ the result clear
- construct a smooth diffeomorphism $F_t$ on $\mathbb{R}^n$ to itself,such that carries origin to any point inside open disk,nad $F_t$ isotopic to identity map.
Finally we need to complete the proof.I don't know how to use the 2 facts above to complete the proof?
Let $x\in N$, conisder $N_x=\{y:f(x)=y$ and $f$ is isotopy to the identity $\}$.
We are going to show that $N_x$ is open and closed. Let $y\in N_x$ there exists a diffeomorphism $f_y$ connected to the identity such that $f_y(x)=y$. There exists a chart $U_y$ of $N$ containing $y$ which is homeomorphic to $\mathbb{R}^n$. This implies that for every $z\in U_y$, there exists $g_z$ isotopy to the identity, $g_z(y)=z$. $g_z\circ f_y(x)=z$ and is isotopic to the identity. This implies that $N_x$ is open. $N_x$ is closed, if $y=limy_n, y\in N_x$, there exists a neighborhood $U_y$ diffeomorphic to $\mathbb{R}^n$ which contains $y_n$. We deduce that there exists a diffeomorphism $h_n$ isotopic to the identity such that $h_n(y_n)=y$, if $f_n$ is a diffeomorphism isotopic to the identity such that $f_n(x)=y_n$, $h_n\circ f_n$ is isomorphic to the identity. We deduce that $N_x$ is closed.