Let $(M,g)$ be a connected Riemannian manifold or pseudo-Riemannian manifold of dimension $n$ ( without boundary ). Let $U := \operatorname{exp}_p(B_{\delta}(0))$ be a geodesic ball in $M$ around $p\in M$. Every orthonormal basis $(b_i)$ for $T_p M$ determines a basis isomorphism $B : \mathbb{R}^{n} \to T_pM$ by $B(x_1 , \dots, x^n) = x^ib_i.$ We then get a coordinate chart $$ \varphi := B^{-1} \circ (\operatorname{exp}_p|_{B_{\delta}(0)})^{-1} : U \to T_pM \to \mathbb{R}^n$$
Q. My question is, can we choose $\varphi$ such that its image is a ball in $\mathbb{R}^{n}$ ( of radius $\delta$ ) with $\varphi(p)=0$?
My fist attempt is, note that $(\operatorname{exp}_p|_{B_{\delta}(0)})^{-1}(U)=B_{\delta}(0) \subseteq T_p M$. So if we can take $B$ such that $B^{-1}$ is an isometry, then we maybe show.. Such $B$ exists? My attempt works?
EDIT : I think that such existence of $B$ is possible : Show the inner product spaces are isometric. Am I argue well?