I study differential geometry and do not understand the choice of environment of $(-2,2)$ for the $\gamma$ curve in the following proof. Are there geodesics in arbitrarily small intervals? Why is this interval explicitly desired? I would be very happy if I could get an intuitive explanation, because this proof haunts my head and I can't work it out by my own.

Be $M$ a smooth manifold and $\nabla$ an affine connection on $M$. Then for every point $p \in M$ there exists an open environment $V \subseteq TM$ of $\mathbb{0}_p \in T_p M \subseteq TM$, so that for every tangential vector $\mathfrak{v} \in V$ there is exactly one geodetic $\gamma_{\mathfrak{v}}:(-2,2) \rightarrow M$ with $\gamma_{\mathfrak{v}}(0)=\pi(\mathfrak{v})$ (where $\pi: TU \rightarrow U$ ist the natural projection) and $\gamma'_{\mathfrak{v}}(0) = \mathfrak{v}$.

Proof. Be $(U,\varphi)$ a map around $p$ with local coordinates $x_1, \ldots, x_n$. Then $T\varphi:TU \rightarrow \varphi(U) \times \mathbb{R}^n$ identifies each tangential vector $\mathfrak{v}\in TU$ with a point $(x_1,\ldots,x_n,v_1,\ldots,v_n) \in \varphi(U) \times \mathbb{R}^n$. A curve $\gamma: (-\epsilon,\epsilon) \rightarrow U$ is a geodetic with $\gamma(0) = \pi(\mathfrak{v})$ and $\gamma'(0)=\mathfrak{v}$, if the coordinates $\gamma_i = x_i \circ \gamma$ of $\gamma$ solve the following initial value problem ($1 \leq k \leq n$):

$$ \gamma'_k(t) = \eta_k(t) \quad \gamma_k(0) = x_k\\ \eta'_k(t) = - \sum_{i,j=1}^{n} \Gamma_{ij}^k(\gamma(t))\eta_i(t)\eta_j(t) \quad \eta_k(0) = v_k $$

So there is a $\eta >0$ and an open environment $\Omega_1 \subseteq \varphi(U)$ of $\varphi(p)$ and $\Omega_2 \subseteq \mathbb{R}^n$ of $0$, so that for all $(x,v) \in \Omega_1 \times \Omega_2$ exactly one solution $(\gamma_1,\ldots,\gamma_n):(-\epsilon,\epsilon) \rightarrow \varphi(U)$ exists. Then

$$ \tilde\gamma_k:(-2,2)\rightarrow \mathbb{R}\quad \tilde\gamma_k(t) = \gamma_k(\frac{\epsilon}{2}t) $$

defines a solution $(\tilde\gamma_1,\ldots,\tilde\gamma_n)$ with

$$ \tilde\gamma_k(0)=x_k \quad \tilde\gamma'_k(0) = \frac{\epsilon}{2}v_k $$

By replacing $\Omega_2$ with $\frac{\epsilon}{2}\Omega_2$ we can assume that $\epsilon = 2$.

Is it true that it doesn't matter what environment I choose for my curve? Can I always get a geodesic this way?