I have the following theorem and proof in my lecture, which I don't quite understand:
Let $M$ be a smooth manifold, $p \in M$. Then it holds that
$(d \exp _p)_0=: T_0 T_pM \cong T_pM \rightarrow T_pM$ is the identity.
On the proof it says:
Since $\exp(tv)= \gamma_v(t)$ for $v \in T_pM$, it holds
$$(d \exp _p)_0 (v)= \biggl.\dfrac{\partial}{\partial t} \biggr|_0 \exp (tv)=v$$
($\gamma_v$ is the unique maximal geodesic with $\gamma^{'}(0)=v$).
Now what I don't understand:
The exponential function is only defined on a neighboorhood of $0$ in $TM$ right, so $(d \exp_p)_0: T_0 D$ where $D \subset T_pM$, right?
I don't understand the proof at all, where does the $\frac{\partial}{\partial t}$ come from?
Can somebody explain this?
The exponential map is indeed defined in a neighborhood, but it is a neighborhood around the point $p\in M$, not $0$. Also note that $T_0 D = T_0 (T_p M)$, it is just the same tangent space.
First one just uses the definition of the differential. Note that $c:(-\epsilon, \epsilon) \to T_p M: t \mapsto tv$ is a curve in $T_p M$ and its derivative $c'(0) = v$ is the tangent vector $v$ in $T_0 (T_p M)$. If we plug this curve in the exponential function and then take the derivative, we obtain the differential. It might be more correct to write $\frac{d}{dt}$ instead of $\frac{\partial}{\partial t}$, since there is only dependency on $t$.
Secondly, note that $\exp(tv) = \gamma_v(t)$, i.e. the unique maximal geodesic with $\gamma'_v(0)=v$. So $$ \biggl. \frac{d}{dt}\biggr|_0 \exp(tv) = \biggl. \frac{d}{dt}\biggr|_0 \gamma_v(t) = \gamma'_v(0) = v. $$