Let $G$ a Lie group (multiplicatively written) and take $\mathfrak{g}$ the Lie algebra of $G$. I was looking for a general definition of exponential map but I'm not sure that I've understood the definition so I'm here.
Take $X \in \mathfrak{g}=T_eG$, where $e$ is the unit of the group. Using the definition of tangent space using curves I can say that $\begin{equation}exp(X)=\gamma(0)\end{equation}$ where $\gamma :\mathbb{R} \rightarrow G$ is a smooth curve such that $\gamma _{*,0}(\frac{\partial}{\partial t}\vert_0)=X$ where $\gamma_{*,0}:T_0\mathbb{R} \rightarrow T_eG$ is the differential of $\gamma$ in $0$. Is this right?
Here's the basic reasoning. We regard $\mathfrak{g}$ as $T_eG$. For each $X \in \mathfrak{g}$, we have the left and right-invariant extensions $X^L$ and $X^R$ of $X$ to vector fields on $G$. They are characterized by $$X^L_g = {\rm d}(L_g)_e(X) \quad \mbox{and} \quad X^R_g = {\rm d}(R_g)_e(X).$$Those fields have flows $\Phi_{X^L}$ and $\Phi_{X^R}$. These flows have the following invariance properties $$\Phi_{X^L}(t,gh) = g\Phi_{X^L}(t,h) \quad \mbox{and}\quad \Phi_{X^R}(t,gh) = \Phi_{X^R}(t,g)h,$$whenever both sides are defined.
For example, for $X^L$, we consider the curve $\alpha(t) = g\Phi_{X^L}(t,h)$. For $t=0$ we have $\alpha(0) = g\Phi_{X^L}(0, h) = gh$ and also $$\begin{align}\dot{\alpha}(t) &= {\rm d}(L_g)_{\Phi_{X^L}(t,h)}\left(\frac{{\rm d}}{{\rm d}t}\bigg|_t \Phi_{X^L}(t,h)\right) \\ &= {\rm d}(L_g)_{\Phi_{X^L}(t,h)}\left(\frac{{\rm d}}{{\rm d}s}\bigg|_{s=0} \Phi_{X^L}(s+t,h)\right) \\ &={\rm d}(L_g)_{\Phi_{X^L}(t,h)}\left(\frac{{\rm d}}{{\rm d}s}\bigg|_{s=0} \Phi_{X^L}(s, \Phi_{X^L}(t,h))\right) \\ &= {\rm d}(L_g)_{\Phi_{X^L}(t,h)}(X^L_{\Phi_{X^L}(t,h)}) \\ &= X^L_{g\Phi_{X^L}(t,h)} \\ &= X^L_{\alpha(t)}, \end{align}$$so the result follows from uniqueness of solutions to initial value problems.
This in particular implies that $X^L$ and $X^R$ are complete (because of the following escape lemma: a vector field $X$ on a manifold $M$ is complete if and only if the flow domain $D_X\subseteq \Bbb R \times M$ -- which is an open neighborhood of $\{0\}\times M$ -- actually contains some strip $I\times M$, where $I$ is an open interval around $0$; we take $I$ to be the interval where the integral curve of $X^L$ starting at $e$ is defined, and same for $X^R$).
Now, the flows of $X^L$ and $X^R$ are, in general, different. But the above shows that when we choose the preferred point $e \in G$, we do have $\Phi_{X^L}(t,e) = \Phi_{X^R}(t,e)$ for all $t \in \Bbb R$.
Then one defines $\exp(X) = \Phi_{X^L}(1,e)$ (or $\Phi_{X^R}(1,e)$, as we now know that choosing $X^L$ or $X^R$ doesn't matter).
To define the exponential, one does not use arbitrary curves. One uses the integral curve through $e$ of the left-invariant (or right-invariant) extension of $X$.