Let $M$ be a riemannian manifold (With the Levi-Civita connection), $c$ a curve in $M$ with $c(t_0)=c_0,c(t)=c_t$ $P_{c,t_0,t}$ denote the function $T_{c_0}M\to T_{c_t}M$ given by parallel transporting each tangent vector $v\in T_{c_0}M$ to $T_{c_t}M$ along the curve $c$.
I want to prove that $P=P_{c,t_0,t}$ is an isometry which preserves orientation, given that $M$ is orientable.
I've already proven that it's an isometry and its inverse is given by $P_{c,t,t_0}$. I now want to prove, that, with the orientation induced by $\phi_t:\mathbb{R}^n\to T_{c_t}M$ with $\phi_t(e_i)=\partial_{i,c_t}$ the orientation is preserved. In other words, the mapping $P':T_{c_t}M\to T_{c_t}M$ given by $\partial_{i,c_t}\mapsto P_{c,t_0,t}\partial_{i,c_0}$ has positive determinant with respect to those bases.
Since it's an isometry, then its determinant is $\pm 1$, so I just have to prove it's $1$.
If the function $d:\text{dom} c\to \mathbb{R}$ given by $t\mapsto \det P'$ is continuous, then since $d(t_0)=1$, if $d(t_1)=-1$ there would be some $t\in (t_0,t_1)$ such that $d(t)=0$ and this is impossible.
So the only thing left is to prove that $d$ is continuous. How could I do that?
An idea would be, that this function can be seen as a composition of continuous functions:
$t_0\stackrel{V}{\to} V_{c(t_0)}\stackrel{P}{\to} V_{c(t)}\stackrel{?}{\to}d(t_0)$ but I'm not sure what would go in ?.
Let $\{e_i^0\}$ be an orthonormal basis for $T_{c(t_0)}M$. Let $(U,\phi)$ be a chart about $c(t)$. Define $\det:TU \times TU \times \dots \times TU \longrightarrow \mathbb R $ by $$(u_1,\dots,u_n)\mapsto \det(u_1, \dots u_n) = \det( a_{ij}) $$ where $ u_j = \sum^n_{i=1} a_{ij} e_i $. This function is continuous . Let $V_i(t)$ be the parallel transport of $e_i^0$ to $T_{c(t)}M$ which is continuous. So we have the following diagram of the continuous function a $$(t_0,t_1) \xrightarrow{V=(V_1,V_2,\dots,V_n)} TU \times TU \times \dots \times TU \xrightarrow{\text{det}} \mathbb R. $$ Since vectors $ V_j(t)= \sum^n_{i=1} a_{ij}(t)e_i^t $ are linearly independent $ \zeta(t) = (\det (V(t)) \neq 0$. And since $\zeta$ is continuous if we choose $\{e_i^0\}$ such that $\zeta(0) >0$ it will remain so for every $t$. The results follows from this.