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"$\phi$ and $\psi$ are two smooth curves in $U$ with the same beginning and end points"

Does this mean:

(A) $\phi:[a,b]\to U$ and $\psi:[a,b]\to U$

(B) $\phi:[a,b]\to U$ and $\psi:[c,d]\to U$ s.t. $\phi(a)=\psi(c)$ and $\phi(b)=\psi(d)$

(C) Something else?

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    $\begingroup$ I would go with (B), absent further context (which might help). If they had used the word "loops," I would expect you would further need $\phi(a) = \phi(b)$. $\endgroup$
    – jdc
    May 13, 2014 at 23:33
  • $\begingroup$ Add to $B$ that $\phi,\psi$ are smooth, of course... $\endgroup$
    – Thomas Andrews
    May 13, 2014 at 23:50

2 Answers 2

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It must be (B), since (A) does not require that $\phi(a)=\psi(a)$ and $\phi(b)=\psi(b)$. The curves are subsets of $U$, and for them to have the same endpoints, the function values on the endpoints of the domain must be equal.

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If you consider both curves up to (order-preserving) reparametrization (i.e., the parametrization is a diffeomorphism with strictly positive derivative), then (A) and (B) are equivalent, since any two closed intervals in the Real line are homeomorphic (by an order-preserving homeomorphism). Now, take a homeomorphism $h: [c,d] \rightarrow [a,b]$. Then $\phi':=\phi o h$ is a reparametrization of $\phi$, with $\phi': [a,b] \rightarrow \mathbb R^2$.

So you can always reduce case (B) to case (A), and then, up to reparametrization you can just consider the case (A).

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