Let $M$ be a smooth Riemannian manifold which is complete. For $p\in M$, let $U_pM$ denote the set of all unit vectors in $T_pM$. We define a function $s:U_pM\to [0,\infty]$ such that $s(v)$ denotes the distance function on $T_pM$ between the zero vector to its first conjugate point along a geodesic $\gamma_v$.
I am having difficulty writing down this function explicitly and how I prove that this function is continuous.
Any help or reference will be appreciated.
Suppose $(M,g)$ is a complete Riemannian manifold. Let $UM=\{(x,\xi)\in TM:|\xi|_g=1\}$, and let $\tau_c:UM\to[0,\infty]$ denote the conjugate distance function, i.e., $$\tau_c(x,\xi)=\inf\{t>0:d(\exp_x)_{t\xi}\text{ is degenerate}\}.$$
Suppose $\{(x_j,\xi_j)\}\subset UM$ is such that $(x_j,\xi_j)\to(x,\xi)\in UM$ and $T$ be any limit point of $\{\tau_c(x_j,\xi_j)\}$ in $[0,\infty]$. For each $j$, let $\eta_j\in U_{x_j}M$ be such that $$d(\exp_{x_j})_{\tau_c(x_j,\xi_j)\xi_j}(\eta_j)=0.$$ Since $x_j\to x$, $\{(x_j,\eta_j)\}$ is contained in a compact subset of $UM$, and so after possibly passing a subsequence, we may assume that $\eta_j\to\eta$ for some $\eta\in U_xM$. Then by continuity of the exponential map, it follows that $$d(\exp_x)_{T\xi}(\eta)=0,$$ and hence for the geodesic emanating from $(x,\xi)$, we conclude $\gamma_{x,\xi}(T)$ is conjugate to $x$, thus showing that $T=\tau_c(x,\xi)$ as desired.