Zero Sectional Curvature implies exp is a local isometry Im studying DoCarmo's book Riemannian Geometry, the first problem of the chapter 5 (Jacobi Fields) states that
If $(M,g)$ is a riemannian manifold with sectional curvature identically zero, show that for every $p \in M$, $exp_p: B_\varepsilon(0) \subseteq T_pM \rightarrow B_\varepsilon(p)$ is an isometry.
I do not figure out how to relate Jacobi Fields with this question in particular, I appreciate any hint.
(PD this is not a homework, Im studying Riemannian geometry for my own just for research purposes)
 A: Here's another possible way.
Let $v\in B_\epsilon(0)$ and $w\in T_v(T_pM)\cong T_pM$. Consider the
  variation of geodesics given by
  \begin{equation*}
    \gamma_s(t)=\exp_p\bigl(t(v+sw)\bigr)
  \end{equation*}
It follows that
  \begin{equation*}
    J(t)=\frac{\partial\gamma_s}{\partial
      s}\biggr|_{s=0}=(d\exp_p)(tw)
  \end{equation*}
  is a Jacobi field along $\gamma_0$. Since sectional curvature is
  identically zero, $J$ satisfies the following PDE
  \begin{equation*}
    \frac{\partial^2J}{\partial t^2}=0
  \end{equation*}
Let $\{e_i\}$ be an orthnormal basis for $T_pM$ and extend it to a
  parallel frame along $\gamma_0$ such that $e_i(0)=e_i$. In terms of
  the frame, $J(t)=a^i(t)e_i(t)$ and $w=w^ie_i$. Solving the above equation, we have $J(t)=(w^it)e_i(t)$, so that
  \begin{equation*}
    w^ie_i(1)=(d\exp_p)(w)
  \end{equation*}
  Following the same argument, if $u\in T_v(T_pM)\cong T_pM$ and $u=u^ie_i$,
  then
  \begin{equation*}
    u^ie_i(1)=(d\exp_p)(u)
  \end{equation*}
  Therefore,
  \begin{align*}
    \langle(d\exp_p)(w),(d\exp_p)(u)\rangle
    &=w^iu^j\langle e_i(1),e_j(1)\rangle\\
    &=w^iu^j\langle e_i,e_j\rangle\\
    &=\langle w,u\rangle,
  \end{align*}
  proving that $\exp_p\colon B_\epsilon(0)\to B_\epsilon(p)$ is an
  isometry. Note that we used the fact that $\langle
  e_i(t),e_j(t)\rangle$ is indepdent of $t$.
A: What I am gonna do is to show that in normal coordinates the metric is the euclidean one
Given a vector $v$ of $T_{q}M,$ for some $q$ in a normal neighborhood of $p,$  we are going to compute $g(v,v).$  To this end, we decompose $v$ orthognally as $v=a\frac{\partial}{\partial r}+u,$ where the second part, $u,$  is tangente to the geodesic sphere (we know the decomposicition is orthogonal thanks to Gauss' lemma).  The norm of the radial part is just $a^{2}.$  I will now show how to compute the tangential part.
Let $\gamma(s)$ be the unique geodesic joining $p$ with $q$ (Is unique beacause we are in a normal neighborhood).  Notice that in normal coordinates the componentes of a Jacobi field $J$ along a geodesic $\gamma(s)$ such that $J(0)=0$ and $J^{\prime}(0)=V,$ are expresed as $J^{i}(s)=sV^{i}.$ Hence, if we call $d$ the distant from $p$ to $q,$ $u$ is the value at $q$ of the Jacobi field whose components in normal coordinates are: $J^{i}=\frac{s}{d}u^{i}.$
Fuerthermore, letting $n$ be a normal vector to $\gamma^{\prime}(0)$ and $n(s)$ its pararllell transport along $\gamma,$ we will have $J(s)=sn(s),$ because our manifold has curvature zero.
Taking covariant derivatives, $D J (0)=n(0).$ But $n(s)$ is pararlell; that is, in particular, its lenght is constant.
$$\vert u \vert ^{2}=\vert J(d) \vert ^{2}=d^{2}\vert n(d) \vert ^{2}=d^{2}\vert D J(0) \vert ^{2},$$
  But also, form the expresion in normal coordinates we see:
  $$ (D J(0))^{i}=\frac{1}{d}u^{i}.$$  
Now well, the metric coincides with the euclidean at $p$ (we are in normal coordinates given by $\text{exp}_{p}$).
  Therefore:
  $$\vert u \vert_{eucl}= d \vert DJ(0) \vert,$$ 
  Where the subscript means we are taking the eculidean norm of the components.
Substituting, we reach:
  $$\vert u \vert ^{2}=d^{2}\vert DJ(0) \vert ^{2}=d^{2}\frac{1}{d^{2}}\vert u \vert_{eucl} ^{2}=\vert u \vert_{eucl}^{2},$$
  An we are done
