[proofplan] We apply the **inverse function theorem** to $\exp_p: \mathcal{D}_p \to M$ at the point $0 \in T_pM$. The hypotheses are: $\exp_p$ is smooth on an open neighbourhood of $0$ (this is the standard smoothness statement for the geodesic flow), and its differential at $0$ is invertible. The latter is exactly the content of the [Differential of the Exponential Map at the Origin](/theorems/2711), which gives $(d\exp_p)_0 = \operatorname{id}_{T_pM}$ — invertible because the identity map is invertible. The Inverse Function Theorem then produces the open ball $B(0, \delta)$ on which $\exp_p$ restricts to a diffeomorphism onto an open neighbourhood of $p$. [/proofplan]

[step:Verify smoothness of $\exp_p$ on an open neighbourhood of $0 \in T_pM$] Let $\mathcal{D}_p \subseteq T_pM$ denote the set of $v \in T_pM$ for which the geodesic $\gamma_p(\cdot, v)$ is defined at $t = 1$, i.e., the natural domain of $\exp_p$. By the [definition of the exponential map](/page/Exponential%20Map), \begin{align*} \exp_p: \mathcal{D}_p &\to M \\ v &\mapsto \gamma_p(1, v). \end{align*} The set $\mathcal{D}_p$ is open and contains $0$: by standard ODE theory applied to the geodesic spray (a smooth second-order vector field on $TM$), the map $(t, v) \mapsto \gamma_p(t, v)$ is smooth on its open natural domain, which contains $\{(t, 0) : t \in \mathbb{R}\}$ since the constant geodesic at $p$ is defined for all $t$. Smoothness of $\exp_p$ on $\mathcal{D}_p$ follows because $\exp_p$ is the composition of $v \mapsto (1, v)$ with the smooth flow of the geodesic spray. In particular, $\exp_p$ is a smooth map from the open neighbourhood $\mathcal{D}_p$ of $0 \in T_pM$ into $M$. Both manifolds have dimension $n$, so the map is between manifolds of the same dimension — a hypothesis for the Inverse Function Theorem to deliver a local diffeomorphism. [/step]

[step:Identify the differential at $0$ as the identity, hence invertible] By the [Differential of the Exponential Map at the Origin](/theorems/2711), under the canonical identification $T_0(T_pM) \cong T_pM$, \begin{align*} (d\exp_p)_0: T_0(T_pM) &\to T_pM \\ v &\mapsto v. \end{align*} That is, $(d\exp_p)_0 = \operatorname{id}_{T_pM}$. The identity map on a finite-dimensional vector space is invertible (its inverse is itself), so $(d\exp_p)_0$ is a linear isomorphism $T_0(T_pM) \to T_pM$. [/step]

[step:Apply the Inverse Function Theorem] The **inverse function theorem** for smooth maps between manifolds of equal dimension says: if $F: U \to N$ is smooth on an open set $U$ containing a point $q$, and $dF_q: T_q U \to T_{F(q)} N$ is a linear isomorphism, then there exist open neighbourhoods $U' \ni q$ in $U$ and $V' \ni F(q)$ in $N$ such that $F|_{U'}: U' \to V'$ is a diffeomorphism. We verify the hypotheses for $F = \exp_p$, $U = \mathcal{D}_p$, $N = M$, $q = 0$: - $\exp_p$ is smooth on the open set $\mathcal{D}_p$ (Step 1). - $0 \in \mathcal{D}_p$ (Step 1). - $(d\exp_p)_0: T_0(T_pM) \to T_pM$ is a linear isomorphism (Step 2). The Inverse Function Theorem yields open neighbourhoods $U' \subseteq \mathcal{D}_p$ of $0$ in $T_pM$ and $V' \subseteq M$ of $\exp_p(0) = p$ such that $\exp_p|_{U'}: U' \to V'$ is a diffeomorphism. [/step]

[step:Shrink the neighbourhood to a metric ball $B(0, \delta)$]The neighbourhood $U' \subseteq T_pM$ is open in $T_pM$ and contains $0$. With respect to the inner product $g_p$ on $T_pM$, choose \begin{align*} \delta &:= \tfrac{1}{2} \sup\{r > 0 : B(0, r) \subseteq U'\} \in (0, \infty], \end{align*} where $B(0, r) := \{v \in T_pM : |v|_{g_p} < r\}$ is the open metric ball; the supremum is positive because $U'$ is open at $0$, and we can take any positive value smaller than the supremum to obtain $0 < \delta < \infty$ with $B(0, \delta) \subseteq U'$. Since $\exp_p|_{U'}$ is a diffeomorphism onto $V'$ and $B(0, \delta) \subseteq U'$ is open in $T_pM$, the restriction \begin{align*} \exp_p|_{B(0, \delta)}: B(0, \delta) &\to \exp_p(B(0, \delta)) \end{align*} is a diffeomorphism onto its image. The image $\exp_p(B(0, \delta))$ is open in $V'$ (because diffeomorphisms map open sets to open sets), hence open in $M$ (because $V'$ is open in $M$), and contains $p = \exp_p(0)$. This produces the open ball $B(0, \delta) \subseteq T_pM$ and the open neighbourhood $\exp_p(B(0, \delta))$ of $p$ in $M$ with the desired properties. The proof is complete.[/step]

[guided]The Inverse Function Theorem from Step 3 produces an open neighbourhood $U' \subseteq T_pM$ of $0$ on which $\exp_p$ is a diffeomorphism — but $U'$ is just *some* open set, with no particular shape. To extract a clean statement about local geometry near $p$, we want a neighbourhood with a definite *size*: a metric ball. Why a ball? Because a ball is the most basic shape adapted to the inner product $g_p$ on $T_pM$, and because the eventual application — the construction of normal coordinates around $p$ — needs a domain on which $\exp_p^{-1}$ extends to a chart. So the task of this step is purely topological: shrink $U'$ to an open Euclidean ball centred at $0$, then check that $\exp_p$ remains a diffeomorphism on the smaller domain. How do we find a ball $B(0, r) \subseteq U'$? Recall that $T_pM$ is a finite-dimensional inner-product space under $g_p$, so its metric topology (induced by $|v|_{g_p} = \sqrt{g_p(v, v)}$) coincides with the manifold topology. The set $U'$ is open in this topology, and $0 \in U'$, so by the very definition of openness in a metric space there exists some $r > 0$ with $B(0, r) \subseteq U'$, where \begin{align*} B(0, r) := \{v \in T_pM : |v|_{g_p} < r\}. \end{align*} The set of admissible radii $\{r > 0 : B(0, r) \subseteq U'\}$ is non-empty and is bounded above only if $U'$ is bounded; in general the supremum may be $+\infty$. To pick a definite finite value strictly inside the admissible range — which guarantees $B(0, \delta) \subseteq U'$ regardless of whether the supremum is attained — we take half the supremum: \begin{align*} \delta &:= \tfrac{1}{2} \sup\{r > 0 : B(0, r) \subseteq U'\} \in (0, \infty]. \end{align*} The supremum is positive because $U'$ is open at $0$ (so at least one positive admissible radius exists), and any positive value strictly less than the supremum is itself admissible. Hence $0 < \delta < \infty$ and $B(0, \delta) \subseteq U'$. (If the supremum were $+\infty$, the formula $\delta := \tfrac{1}{2}\sup$ is informal — we instead pick any finite $\delta > 0$, which still satisfies $B(0, \delta) \subseteq U'$. The role of the formula is just to produce *some* explicit positive $\delta$.) Now we transport the diffeomorphism conclusion from $U'$ to $B(0, \delta)$. The restriction of a diffeomorphism to an open subset of its domain is a diffeomorphism onto its image: this is a general fact about smooth manifolds, since smoothness, bijectivity onto the image, and smoothness of the inverse all restrict. So \begin{align*} \exp_p|_{B(0, \delta)}: B(0, \delta) &\to \exp_p(B(0, \delta)) \end{align*} is a diffeomorphism onto its image. Why was this restriction step necessary? Because the conclusion we want is about a *ball* $B(0, \delta)$, not the unspecified set $U'$. Without explicitly restricting and re-checking, we would only know the diffeomorphism statement on $U'$. It remains to verify that the image $\exp_p(B(0, \delta))$ is open in $M$ and contains $p$. Diffeomorphisms map open sets to open sets (the image of an open set under a homeomorphism is open, and diffeomorphisms are in particular homeomorphisms), so $\exp_p(B(0, \delta))$ is open in $V'$. Since $V'$ is open in $M$, an open subset of $V'$ is open in $M$ — the topology on $V'$ is the subspace topology, and any open subset of $V'$ in this topology is the intersection of $V'$ with an open subset of $M$, which is itself open in $M$ as a finite intersection of open sets. Finally, $p = \gamma_p(1, 0) = \exp_p(0) \in \exp_p(B(0, \delta))$ because $0 \in B(0, \delta)$. We have produced the open ball $B(0, \delta) \subseteq T_pM$ centred at $0$ and the open neighbourhood $\exp_p(B(0, \delta)) \subseteq M$ of $p$, with $\exp_p|_{B(0, \delta)}$ a diffeomorphism between them. This completes the proof. A few interpretive remarks. First, the radius $\delta$ depends on $p$: different base points $p \in M$ give different admissible radii $\delta_p$, and there is no a priori uniform lower bound. On a compact manifold one can extract a uniform $\delta > 0$ by a compactness argument (this is the **injectivity radius**); on non-compact or curvature-unbounded manifolds, $\delta_p$ may shrink to zero as $p$ approaches bad regions. Second, $B(0, \delta)$ is a metric ball in the *vector space* $T_pM$ measured by the inner product $g_p$; it is **not** the same as the Riemannian distance ball of radius $\delta$ around $p$ in $M$, except in special cases (e.g., constant curvature). The statement here is purely topological — we only assert $\exp_p$ is a diffeomorphism on some Euclidean ball in $T_pM$. Third, the size $\delta$ controls the **normal-coordinate chart** centred at $p$: composing $\exp_p^{-1}$ with a linear identification $T_pM \cong \mathbb{R}^n$ yields the chart $(B(0, \delta), \exp_p^{-1})$ used in [Normal Coordinates](/theorems/2713) and throughout local Riemannian geometry.[/guided]