Contact equivalence is chart independent and an equivalence relation

Suppose $(x\circ {\gamma }_1{)}^{\mathrm{\prime }}(0)=(x\circ {\gamma }_2{)}^{\mathrm{\prime }}(0)$ in some chart $(U,x)$ around $p$, and let $(V,y)$ be any other chart around $p$. Write $a:=x(p)$ and ${\alpha }_r:=x\circ {\gamma }_r$. For each coordinate function $h^j:=y^j\circ x^{-1}$, [L1] gives smooth functions $g_1^j,\dots ,g_n^j$ near $a$ such that $h^j(u)-h^j(a)={\sum }_{i=1}^n(u^i-a^i)g_i^j(u)$ and $g_i^j(a)={\mathrm{\partial }}_i{h}^j(a)$. Substituting $u={\alpha }_r(t)$, dividing by $t$, and letting $t\to 0$ shows $(y^j\circ {\gamma }_r{)}^{\mathrm{\prime }}(0)={\sum }_{i=1}^n{g}_i^j(a)({\alpha }_r^i{)}^{\mathrm{\prime }}(0)\mathrm{.}$ Because the vectors ${\alpha }_1^{\mathrm{\prime }}(0)$ and ${\alpha }_2^{\mathrm{\prime }}(0)$ are equal, the right-hand sides agree for $r=1,2$. Hence $(y\circ {\gamma }_1{)}^{\mathrm{\prime }}(0)=(y\circ {\gamma }_2{)}^{\mathrm{\prime }}(0)$, so the relation is chart independent.

Reflexivity and symmetry are immediate from the defining equality in [F1], and transitivity holds because equality of coordinate velocity vectors in any chart is transitive.

Hence contact equivalence is a chart-independent equivalence relation.