# Is exponential map locally a diffeomorphism w.r.t. base point?

Let $M$ be a riemannian manifold and $\exp_p: T_pM \rightarrow M$ the exponential map at $p \in M$.

At each point $p\in M$, $\exp_p$ can be restricted to a neighborhood $V$ of $0\in T_pM$ so that $\exp_p|_V$ is a diffeomorphism. Let $U\subset M$ denote image of $V$ under this map.

Choosing an orthonormal basis $\{E_i\}$ for $T_p M$ and using the isomorphism $E: \mathbb R^n \rightarrow T_pM$, where $E(x^1, \dots, x^n)= x^iE_i$, gives normal coordinates $(U, \phi)$, where $$\phi := E^{-1} \circ \exp^{-1}_p: U \rightarrow \mathbb R^n$$

My question is: can we fix some $z\in \mathbb R^n$ (small enough) and take some small neighborhood $\widetilde{U}$ of $p$ so that the function $\widetilde \phi$, defined as $$\widetilde{\phi}_z (p) := \exp_p ( E(z))$$ is a diffeomorphism when restricted to $\widetilde{U}$?

(In the definition of $\widetilde \phi$, the isomorphism $E$ will also vary with $p$.)

$\newcommand{\Reals}{\mathbf{R}}$"Yes." Here's a sketch.
Let $U$ be a coordinate neighborhood about a point $p$ and let $(E_{i})_{i=1}^{n}$ an orthonormal frame in $U$. The tangent bundle $TU$ is trivialized by the mapping $E:U \times \Reals^{n} \to TU$ defined by $$E(p, x) = \sum_{i} x^{i} E_{i}(p).$$ The smooth mapping $\Phi:U \times \Reals^{n} \to U \times M$ defined by $$\Phi(p, z) = \bigl(p, \exp_{p} E(p, z)\bigr) := \bigl(p, \widetilde{\phi}_{z}(p)\bigr)$$ is a diffeomorphism in some neighborhood of $p \times \{0\}$ by the inverse function theorem. Further, the restriction $p \mapsto \Phi(p, 0) = (p, p)$ is the diagonal embedding, so along the zero section, the vertical component of $\Phi$ is a submersion. Consequently, there exist neighborhoods $\widetilde{U}$ of $p$ and $V$ of $0$ such that for $z$ in $V$, the mapping $\widetilde{\phi}_{z}:\widetilde{U} \to M$ is a (smooth, bijective) submersion, i.e., a diffeomorphism to its image in $M$.