Trying understand the part of Hopf-Rinow's theorem that states that if $M$ is complete as a metric space, then it's geodesic complete I am reading Riemannian Geometry by Do Carmo and I would like to understand why $g$ extends $\gamma$ beyond $s_0$ in the proof below:

I can't see this easily. I understand that $g$ extends $\gamma$, but I can't see why extends beyond $s_0$ and not extend to $s_2$ for some $s_2 < s_0$ once that $\gamma$ can be defined on $[0,s_1]$ with $s_1 < s_2 < s_0$. Is $\gamma$ defined on $[0,s_0)$?
Thanks in advance!
 A: I'll present a (very slightly - but essentialy the same) different argument with some more details (which honestly I think should have been in the book in the first place) which I think makes things clearer. Suppose $\gamma$ is a unit speed geodesic defined on $[t_0, t_1)$, with $t_0 < t_1$. We want to prove $\gamma$ extends beyond $t_1$. Pick a sequence $(t_n)_{n \in \mathbb{N}}$ such that $t_n \nearrow t_1$. By the arguments in the book and by the hypothesis that $M$ is complete as a metric space, we know the sequence $\{\gamma(t_n)\}_{n \in \mathbb{N}}$ is a Cauchy sequence and therefore we can assume without loss of generality (just to avoid writing subsequences indexes) that there exists $p_0 \in M$ such that $\gamma(t_n) \to p_0$.
Choose a totally normal ball of radius $\epsilon$ around $p_0$, let's call it $B_{\varepsilon}(p_0)$. Pick $n, m \in \mathbb{N}$ and points $\gamma(t_n), \gamma(t_m)$ on this ball such that  $\max\{{\operatorname{dist}(\gamma(t_n), p_0)}, \operatorname{dist}(\gamma(t_m), p_0)\} < \frac{\varepsilon}{2}$ (the $\frac{\varepsilon}{2}$ here is just to ensure that from $t_n$ to $t_m$, $\gamma$ does not leave the ball, it could be anything larger than $2$ in the denominator - notice we're using the triangle inequality here). Since we're in a totally normal ball, there exists a unique geodesic $g: I \to M$ such that $g(0) = \gamma(t_n)$ and $g(\delta) = \gamma(t_m)$ (for some $0 < \delta < T$), where $I = [0, T)$ for some $T \geq \varepsilon$ (again because we're in a totally normal neighbourhood, this $\epsilon$ is uniform, in the sense that it works for any point in the ball - i.e, every geodesic starting at a point $q \in B_{\varepsilon}(p_0)$ is defined at least until time $\varepsilon$ - which, to be a little pedantic, means that the length of any interval of definition of any such geodesic has length $\geq \varepsilon$ ). Since we're in a totally normal ball, $g$ coincides with $\gamma$ wherever $g$ is defined (you could question why it couldn't coincide only on the ball, but geodesics are uniquely determined by a starting point and an initial condition). In particular this means that $g'(0) = \gamma'(t_n)$. Without loss of generality (picking $n$ large enough) we can assume that $t_n > t_1 - \varepsilon$, so we can define the following geodesic $\tilde{\gamma}: [t_0, t_1 + \tilde{\varepsilon}) \to M$ (where $t_1 + \tilde{\varepsilon} = t_n + \varepsilon > t_1$):
$$
\tilde{\gamma}(t)= \begin{cases}\gamma(t), & t \in[t_0, t_n), \\ g\left(t-t_{n}\right), & t \in\left(t_{n}-\varepsilon, t_{n}+\varepsilon\right)\end{cases}
$$
This geodesic is well defined since both expressions on the right-hand side are geodesics that have the same position and velocity at $t_n$ - therefore, by the uniqueness of geodesics, they agree where they overlap. It's clear that $\tilde{\gamma}$ is then an extension of $\gamma$, which means $\gamma$ can actually be defined on all of $\mathbb{R}$ (but didn't this argument only prove $\gamma$ can be defined on $(t_0, +\infty)$? Well, yes, but I think you can see why this is enough to conclude it can be defined on all of $\mathbb{R}$ - just take the argument I started with with a backwards parametrization).
Observation: The proof in the book is yet another instance of a "contradiction proof" that isn't actually a proof by contradiction. What we did here was actually to show directly that every geodesic is defined on all of $\mathbb{R}$. Also, why do we need to pick a totally normal ball instead of just a normal one? Well, let's try to argue like before and see where it breaks down. Choose a normal ball of radius $\varepsilon$ around $p_0$. For each $\gamma(t_n)$, there exists a geodesic $g_n$ connecting $\gamma(t_n)$ to $p_0$. But since the normal neighbourhood we picked is not necessarily a totally normal neighbourhood, we can't ensure that it will be a normal neighbourhood around each of its points, which was a crucial fact we used in the proof above. You could say "but $g_n$ is in a sense converging to $\gamma$"... yes, but so what? There's still something missing, which a totally normal neighbourhood takes care of.
