Every point $p\in\ M$ has an open neighborhood $U \subset M$ which is diffeomorphic to $R^n$

Question

Let M be a differentiable manifold of dimension n. Show that every point $p\in\ M$ has an open neighborhood $U \subset M$ which is diffeomorphic to $R^n$.

I know by definition of topological manifold that U is homeomorphic to $R^n$. I have to show that the homeomorphism $f$ and $f^{-1}$ are differentiable (where $f: U \rightarrow\mathbb\ R^n$), but I dont have any ideas how to show that.

Answer 1

Each chart $\phi : U \to V \subset \mathbb R^n$ belonging to the differentiable structure (= maximal differentiable atlas) $\mathfrak A$ on $M$ is a diffeomorphism.

To see that, note that

1. Each open $V \subset \mathbb R^n$ has a canonical differentiable structure which is generated by the single chart $id_V : V \to V$. A map $f : V \to V'$ between open subsets of Euclidean spaces is differentiable as a map of manifolds with these canonical structures iff it is differentiable in the usual sense of multivariable calculus.

2. Each open subset $U \subset M$ has a canonical differentiable structure inherited from $M$. Its differentiable structure $\mathfrak A_U$ is given by all charts $\phi' : U' \to V'$ belonging to $\mathfrak A$ with the property $U' \subset U$.

3. If $\phi : U \to V$ belongs to $\mathfrak A$, then the differentiable structure on $\mathfrak A_U$ is generated by the single chart $\phi$.

To prove that $\phi$ is a diffeomorphism we have to check that $\phi$ and $\phi^{-1}$ are diferentiable maps between the manifolds $U, V$. But we have two differentiable single chart atlases on $U$ and $V$. Hence it suffices to show that $id_V \circ \phi \circ \phi^{-1}$ and $\phi \circ \phi^{-1} \circ id_V$ are differentiable maps between open subsets of $\mathbb R^n$. This is trivial since both maps are nothing else than $id_V$.

Now take any chart $\phi : U \to V$ in $\mathfrak A$ with $p \in U$. There exists an open ball $B$ such that $\phi(p) \in B \subset V$. The restriction $\phi_B : \phi^{-1}(B) \to B$ is a chart in $\mathfrak A$ around $p$. It is well-known that there exists a diffeomorphism $h : B \to \mathbb R^n$. Then $h \circ \phi_B$ belongs to $\mathfrak A$ and we are done.