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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)).

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    $Y$ou should say what your definition of "equivalence" of two smooth curves is.2012-04-24

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I assume that you mean "reparamaterization" of $C^1$-curves. Namely, if $\alpha:I\to\mathbb{R}^n$ and $\beta:J\to\mathbb{R}^n$ are two $C^1$ curves we call $\alpha$ a reparametrization of $\beta$ if there there exists a homemorphism $f:I\to J$ such that $\beta\circ f=\alpha$.

Clearly this is reflexive since we can take $f:I\to I$ to be $\text{id}$.

It's transitive for the composition of homeos is a homeo.

Now, you work to prove it's symmetric. If $f:I\to J$ is a homeo such that $\beta\circ f=\alpha$ we want a homeo $g:J\to I$ such that $\beta=\alpha\circ g$. Any ideas?

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    No, it's not. I used homeomorphism instead of $C^1$ map. That said, my same advice goes with the swapping of those two words and the fact that the inverse of a positive injective $C^1$ map is necessarily $C^1$ (as they indicated).2012-04-24