Let $M$ be a Riemannian manifold, and let $V\subset TM$ denote the domain of the exponential map $\exp:V\to M$, which is an open subset of $TM$ containing the zero section of $TM\to M$.
Let $\sigma:(-\epsilon,\epsilon)\to M$ be a smooth curve in $M$ with a smooth vector field $W$ along $\sigma$. Suppose $(\sigma(0),W(0))\in V$, so that $\exp_{\sigma(0)}(W(0))$ is defined. Then $\exp_{\sigma(0)}(tW(0))$ is defined for all $t\in [0,1]$, since $V\cap T_pM$ is star-shaped respect to $0$ (in the vector space $T_pM$) for each $p\in M$.
What I want to show is, that $\exp_{\sigma(s)}(tW(s))$ is defined for all $s\in (-\delta, \delta)$, for sufficiently small $\delta \in (0,\epsilon)$. I was reading the book Lee's Introduction to Riemannian manifolds, and Lee says that this is because $[0,1]$ is compact, but I can't understand, and I think I need more details. Any hints? Thanks in advance.
We wish to define a map $\Gamma(s,t)=\text{exp}_{\sigma(s)}(tW(s))$ on a domain $(s,t) \in (-\delta,\delta) \times I$, for some $\delta>0$. To see how this is possible, we write $\Gamma(s,t)$ as composition $$ (s,t) \longmapsto (\sigma(s),tW(s)) \longmapsto \text{exp}_{\sigma(s)}(tW(s)). $$ Now, we know that compactness of $I$ implies compactness of the set $\{(p,tv): t \in I\}$ and this set is contained in the domain of exponential map (which is open in $TM$). By continuity of the map $(s,t) \mapsto (\sigma(s),tW(s))$, we have an open subset $B\subseteq \mathbb{R}^2 $ contain $\{0\} \times I$ and so $\Gamma(s,t)$ defined on $B$. Can you find some $\delta>0$ such that $B \supseteq (-\delta,\delta)\times I$ ?