Let X be an n-dimensional differentiable manifold. A Riemannian metric in X is a family $g = \{ g_p | p \in X \}$ with the following properties: for each $p \in X$ , the map $g_p : T_pX \times T_pX \longrightarrow ℝ$ is an inner product in $T_pX$, and for every differentiable chart (U, h, V) for X with coordinates $(x_1, . . . , x_n)$ in V , the functions $$g_{ij}: U \longrightarrow ℝ \; \; g_{ij}(p):= g_p (\frac{\partial}{\partial x_i}(p), \frac{\partial}{\partial x_j}(p)) \; \, 1 \leq i,j \leq n$$ are differentiable. A curve $\gamma : [a, b] \longrightarrow X$ defined on a closed interval [a, b] is called differentiable if there exist an $\epsilon > 0$ and a differentiable curve $\gamma':(a - \epsilon,b+\epsilon) \longrightarrow X$ such that $\gamma'|_{[a,b]}=\gamma$
(a) Show that a differentiable curve $\gamma[a, b] \longrightarrow X$ defines a geometric tangent vector $[\gamma](t) \in T_{\gamma (t)}X$ for all $t \in [a,b]$ (Be careful in case t = a or b.)
(b) Let g be a Riemannian metric in X. For a differentiable curve $\gamma : [a, b] \longrightarrow X$ , its length (with respect to g) $L(\gamma)$ is defined as $$L(γ) := \int_a^b \sqrt{g_{\gamma (t)} ([\gamma](t), [\gamma](t))} dt$$ where $[\gamma](t) \in T_{\gamma(t)}X$ is as in a). Show that the length of a differentiable curve $\gamma : [a, b] \longrightarrow X$ is independent of the parametrization, i.e. that if $f : [c, d] \longrightarrow [a, b]$ is a diffeomorphism and $\gamma':=γ\circ f$, then $L(\gamma)=L(\gamma') $.}
For a), my answer would be: For every $t \in [a,b]$, we can define $\gamma^{t}: \gamma' |_{(t-\epsilon, t+\epsilon)}: (-\epsilon, + \epsilon) \longrightarrow X$. This is a differentiable curve in X. But this is surely not enough, I do not understand what is to show here.
For b) I don't know how to formulate this. It seems quite obvious, but I don't know how the diffeomorphism interacts with the metric. I would start with $$L(\gamma') = \int_c^d \sqrt{g_{\gamma'(t)} ([\gamma'](t), [\gamma'](t)) dt}$$