[proofplan]
We compute $(d\exp_p)_0(v)$ directly from the definition of the differential as the velocity of a curve. The natural curve to choose is $t \mapsto tv \in T_p M$, whose image under $\exp_p$ is — by the definition of $\exp_p$ — the geodesic with initial velocity $v$. The chain $\exp_p(tv) = \gamma_p(1, tv) = \gamma_p(t, v)$ uses the rescaling identity for geodesics; differentiating at $t = 0$ gives $\dot\gamma_p(0, v) = v$. The argument is one line of calculus once the rescaling lemma is in hand.
[/proofplan]
[step:Identify $T_0(T_pM)$ with $T_pM$ via the canonical isomorphism]
The tangent space $T_pM$ is a finite-dimensional real vector space, and for any such vector space $V$, the tangent space $T_0 V$ at the origin is canonically isomorphic to $V$ via the map
\begin{align*}
\iota: V &\to T_0 V \\
v &\mapsto \dot\alpha_v(0), \qquad \alpha_v(t) := tv.
\end{align*}
Here $\alpha_v: \mathbb{R} \to V$ is the straight-line curve through the origin with constant velocity $v$, and $\dot\alpha_v(0) \in T_0 V$ is its tangent vector at $t = 0$. The map $\iota$ is a linear isomorphism: it is the standard identification of a vector space with its tangent space at any point, available because $V$ has a canonical flat affine structure.
Throughout the rest of the proof, we use $\iota$ to identify $T_0(T_pM)$ with $T_pM$. Under this identification, a vector $v \in T_pM$ corresponds to the velocity at $0$ of the straight-line curve $t \mapsto tv$ in $T_pM$.
[/step]
[step:Differentiate $\exp_p$ along the straight-line curve $t \mapsto tv$]
Fix $v \in T_pM$. The differential $(d\exp_p)_0: T_0(T_pM) \to T_p M$ is, by definition, the linear map sending $\dot\alpha_v(0)$ to the velocity at $0$ of the composed curve $\exp_p \circ \alpha_v$:
\begin{align*}
(d\exp_p)_0(v) &= \frac{d}{dt}\bigg|_{t=0} \exp_p(tv).
\end{align*}
Here we have used the identification $\iota(v) = \dot\alpha_v(0)$ from the previous step, so $(d\exp_p)_0(v)$ stands for $(d\exp_p)_0(\iota(v))$.
By the [definition of the exponential map](/page/Exponential%20Map), $\exp_p(w) = \gamma_p(1, w)$ for all $w$ in the domain of $\exp_p$, where $\gamma_p(\cdot, w)$ is the geodesic with $\gamma_p(0, w) = p$ and $\dot\gamma_p(0, w) = w$. Substituting $w = tv$:
\begin{align*}
\exp_p(tv) &= \gamma_p(1, tv).
\end{align*}
Now apply the [Geodesic Rescaling](/theorems/2710) lemma with $\lambda = t$ and initial velocity $v$: it states $\gamma_p(\lambda s, a) = \gamma_p(s, \lambda a)$, so taking $s = 1$, $\lambda = t$, $a = v$ gives
\begin{align*}
\gamma_p(1, tv) &= \gamma_p(t, v).
\end{align*}
The rescaling lemma applies because for $|t|$ sufficiently small, $tv$ lies in the maximal domain of $\gamma_p(\cdot, v)$ via reparametrisation; equivalently, both $\gamma_p(1, tv)$ and $\gamma_p(t, v)$ are defined for $t$ in some open interval around $0$. Combining,
\begin{align*}
\exp_p(tv) &= \gamma_p(t, v).
\end{align*}
[/step]
[step:Evaluate the velocity at $t = 0$ using the geodesic initial condition]
Differentiating both sides at $t = 0$:
\begin{align*}
(d\exp_p)_0(v) = \frac{d}{dt}\bigg|_{t=0} \exp_p(tv) = \frac{d}{dt}\bigg|_{t=0} \gamma_p(t, v) = \dot\gamma_p(0, v).
\end{align*}
By the defining property of $\gamma_p(\cdot, v)$, namely $\dot\gamma_p(0, v) = v$, we conclude
\begin{align*}
(d\exp_p)_0(v) &= v.
\end{align*}
Since $v \in T_pM$ was arbitrary, $(d\exp_p)_0 = \operatorname{id}_{T_pM}$ as linear maps $T_p M \to T_p M$ (under the identification $T_0(T_pM) \cong T_pM$). This completes the proof.
[guided]
We have arrived at the identity $\exp_p(tv) = \gamma_p(t, v)$ from the previous step. The remaining task is to differentiate at $t = 0$ and read off the answer. Why is this the moment of payoff? Because the right-hand side is a single geodesic with a known initial velocity, while the left-hand side is the curve whose velocity at $0$ defines $(d\exp_p)_0(v)$. Differentiating both sides equates these two velocities.
By the definition of the differential applied in the previous step, $(d\exp_p)_0(v) = \frac{d}{dt}|_{t=0} \exp_p(tv)$. Substituting the rescaled identity $\exp_p(tv) = \gamma_p(t, v)$ inside the derivative — valid because the two functions of $t$ agree on an open neighbourhood of $0$, hence have the same derivative at $0$ — we obtain
\begin{align*}
(d\exp_p)_0(v) = \frac{d}{dt}\bigg|_{t=0} \exp_p(tv) = \frac{d}{dt}\bigg|_{t=0} \gamma_p(t, v) = \dot\gamma_p(0, v).
\end{align*}
What is $\dot\gamma_p(0, v)$? This is exactly the place where the definition of $\gamma_p(\cdot, v)$ is consumed. Recall: $\gamma_p(\cdot, v)$ is *defined* as the unique geodesic with initial position $\gamma_p(0, v) = p$ and initial velocity $\dot\gamma_p(0, v) = v$. The initial-velocity condition is not a computation — it is the defining property, baked into the existence-and-uniqueness theorem for geodesics. So
\begin{align*}
(d\exp_p)_0(v) &= v.
\end{align*}
This holds for every $v \in T_pM$, so $(d\exp_p)_0$ agrees with the identity map on all of $T_pM$. As linear maps $T_p M \to T_p M$ (under the identification $T_0(T_pM) \cong T_pM$ from Step 1), $(d\exp_p)_0 = \operatorname{id}_{T_pM}$.
The conclusion has a strong geometric meaning: at the origin, the exponential map is "first-order Euclidean". This is the foundational fact behind normal coordinates, geodesic polar coordinates, and the entire local geometry of Riemannian manifolds — the Taylor expansion of the metric in normal coordinates begins $g_{ij}(x) = \delta_{ij} + O(|x|^2)$, with the linear term killed precisely by $(d\exp_p)_0 = \operatorname{id}$. This completes the proof.
[/guided]
[/step]
