Let $\alpha,\beta:I\subset\mathbb{R}\rightarrow\mathbb{R}^3$, $\alpha$ is p.b.a.l. and that $\beta(s)=\alpha(-s)$ and $I$ is a symetric relative to the origin interval. I must prove that the curvatures $k$ and torsion $\tau$ satisfies $k_{\beta}(s)=k_{\beta}(-s)$ and $\tau_{\beta}(s)=-\tau_{\alpha}(-s)$.
My try:
$T_{\beta}(s)=-T_{\alpha}(-s)$ and $T_{\beta}^{'}(s)=T_{\alpha}^{'}(-s)$
so we have that $k$ satisfies $k_{\beta}(s)=k_{\beta}(-s)$ and $N_{\beta}(s)=N_{\alpha}(-s)$,
the binormal vector, $B$, is given by $B_\beta=(-T_{\alpha}(-s))\times N_\alpha(-s)$ and $B_\beta^{'}=(-T_{\alpha}(-s))\times (-N_\alpha^{'}(-s))$, so as conclusion
$\tau_{\beta}(s)=<(-T_{\alpha}(-s))\times (-N_\alpha^{'}(-s)),N_\alpha(-s)>=<T_{\alpha}(-s)\times N_\alpha^{'}(-s),N_\alpha(-s)>=\tau_{\alpha}(-s)$ i.e
$\tau_{\beta}(s)=\tau_{\alpha}(-s)$
where is my gap is this steps?