Could someone help me through this problem? Prove that “equivalence” of smooth curves has the familiar reflexive, symmetric, and transitive properties of an equivalence relation HINT:Use the facts that a 1-1 $C^1$ mapping $λ(t) : [c, d] →[a, b]$ with $λ>0$ has a 1-1 $C^{1}$ inverse $λ^{−1} : [a, b]→[c, d]$ with $(λ^{−1})>0$ and, if $λ:[c, d]→[a, b]$ and λ2 : [e, f ] → [c, d], then $λ1 ◦ λ2 : [e, f ] → [a, b]$ with all the desired properties of λ1 and λ2.
The two curves C1 : z(t), a ≤ t ≤ b and C2 : ω(t), c ≤ t ≤ d are smoothly equivalent if there exists a 1-1 C1 mapping λ(t) : [c, d]→[a, b] such that λ(c) = a, λ(d) = b, λ(t) > 0 for all t, and ω(t) = z(λ(t)).