The definition I have been given for a smooth abstract surface, $S$, to be orientable is that given a continuous family of maps $f_t: D \to S$ that embed the closed unit disk into $S$ with $f_0(D) = f_1(D)$ then the boundaries $f_0(S^1)$ and $f_1(S^1)$ are oriented the same way. I'm now trying to show that the following is a sufficient condition for orientability: Let $\phi = F_i^{-1} \circ F_j$ be some transition map then $det(D\phi) > 0$. As part of the proof I'm trying show $f_0(S^1)$ is oriented the opposite way to $f_1(S^1)$ iff $det(\phi) < 0$, where $\phi$ is the transition $f_0^{-1} \circ f_1$. I am having some difficulties:
Firstly, I've never seen a rigorous definition of what it means for $f_0(S^1), f_1(S^1)$ to be oriented the same way. I'm guessing this means that $f: e^{it} \to e^{is(t)}$ where s is some increasing function? But I'm not sure on this.
For one implication, let's say the transition map is given by $\phi: (sin(t), cos(t)) \to (sin(s(t)), cos(s(t)))$ for some increasing function $s$. Now how do I get hold of the derivatives of $\phi$ in order to look at the matrix $D\phi$?
I have no idea how to go about the other implication but I think a lot of the difficulty arises from the fact I'm not sure how to get my hands on the matrix $D\phi$?
Apologies for the long question, and thanks for any help!