Definition. Let germ $\phi\colon (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$, is germ of diffeomorphism, if there exist $\psi\colon (\mathbb{R}^n,0)\to (\mathbb{R}^{n},0)$, such that $\psi\circ \phi = \phi \circ \psi = 1_{\mathbb{R}^n,0}$.
I need prove that, the set $\operatorname{Dif}( \mathbb{R}^n,0)$ is a group under the composition.
I'm confused with the definition of it being a germ of diffeomorphism, I am trying to test the lock, but I have not been able to.
Let $f,g\in Dif(\mathbb{R}^n,0)$ then there exist $\varphi\colon(\mathbb{R}^n,0)\to(\mathbb{R}^n,0)$ such that $f\circ \varphi=\varphi\circ f=1_{\mathbb{R}^n,0}$. Again there exist $\psi\colon (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ such that $g\circ \psi=\varphi\circ g=1_{\mathbb{R}^n,0}$ . Now I've to show that $f\circ g\in Dif(\mathbb{R}^n,0)$, I have not been able to show that the closure is hold.
any suggestion or help is welcome