Notation. Given $n\in \Bbb{N}$, $r > 0$ and $x\in \Bbb{R}^n$, let denote $B_r^n(x) = \{ z \in \Bbb{R}^n : d(x,z) < r\}$ where $d$ is the Euclidean metric.
I have to prove the following:
Let $M=(X,\mathcal{A})$ a smooth manifold where $\mathcal{A}$ is a smooth atlas and such that $n:=\dim M$. Then, for every $p\in M$ there exists a chart $(U, \phi)$ of $p$ such that $\phi(U)=B_1^n(0)$ .
I know that there are posts that comment about this, but I have to do a complete proof and I want to know is what I made is correct.
First, let $(V,\varphi)\in \mathcal{A}$ be a chart of $p$ and $f:\Bbb{R}^n \hookrightarrow \Bbb{R}^n$ the map defined by $f(x)=x-\varphi(p)$. This is a homeomorphism since is polynomial, then its restrictions are also homeomorphisms on their images, so $f \circ \varphi $ is a homeomorphism on its image. Hence, $f \circ \varphi (V)$ is open subset of $\Bbb{R}^n$, so noticing that $f \circ \varphi (p) = 0$, there exists $r > 0$ such that $B_r^n(0) \subseteq f \circ \varphi (V)$ .
Let $g_r: B_r^n(0) \hookrightarrow B_1^n(x)$ the map defined by $g_r(x)=\frac{x}{r}$ . This map is homeomorphism since is polynomial and $\forall x \in B_r^n(0): \|g_r(x)\|=\frac{\|x\|}{|r|}<1$ . Therefore, if we take $U:= \varphi^{-1} \circ f^{-1} \circ g_r^{-1} ( B_1^n (0) )$ and $\phi = g_r \circ f \circ \varphi : U \hookrightarrow B_1^n (0)$ the chart $(U, \phi)$ satisfies the requirements: if $(W,\psi)\in \mathcal{A} \Rightarrow \psi \circ \varphi^{-1} \circ f^{-1} \circ g_r^{-1}$ is $\mathcal{C}^{\infty}$ because $\psi \circ \varphi^{-1}$ is a charts change on a smooth atlas and $f$ and $g$ are polynomial.
Is this fine? Thanks in advance :)