I am working on exercise 6, p23 in Rolfsen's Knots and Links:
Show that if $G$ is any simple closed curve in $\mathbb{C}-\{0\}$, then within any $\epsilon$-neighborhood of $G$ lies a simple closed curve $G'$, disjoint from $G$, homotopic to $G$, and transversal to all of the curves $q^{-1}(M)$.
Here $q:\mathbb{C}-\{0\}\to T^2$ is the covering map $q(re^{i\theta})=(e^{i\ln r},e^{i\theta})$, and $M$ is the meridian, a knot of class $<0,1>$. I think the disjoint and homotopy part is trivial, but I don't know how to get transversality. Can someone give me a hint? Much appreciated.
To show that $G'$ is transversal to all of the curves $q^{-1}(M)$, you can utilize the fact that $q$ is a covering map. In particular, the covering map $q$ induces a group isomorphism between the fundamental group of $T^2$ and the deck transformation group of the covering.
Consider the preimage of the meridian $M$ under the covering map $q$. Let's call this preimage curve $q^{-1}(M)$. Since $M$ is a simple closed curve on $T^2$, its preimage $q^{-1}(M)$ consists of disjoint simple closed curves in $\mathbb{C} - {0}$, which are homotopic to $G$. Focus on a single component of $q^{-1}(M)$, which is a simple closed curve in $\mathbb{C} - {0}$. Within any small neighborhood of this component, you can find a simple closed curve $G''$, disjoint from $G$ and homotopic to $G$, using the same argument you mentioned as being trivial. Now consider the deck transformation group associated with the covering map $q$. Since $T^2$ is simply connected, this deck transformation group is isomorphic to the fundamental group of $T^2$, which is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. The deck transformations act on the preimage curves $q^{-1}(M)$, and any curve $G''$ homotopic to $G$ within a small neighborhood is also acted upon by these deck transformations. By considering the action of the deck transformations, you can choose a specific curve $G'$ from the family of curves homotopic to $G$, such that $G'$ is transversal to all of the preimage curves $q^{-1}(M)$. Try to think about how the action of the deck transformations can help you select a curve that intersects $q^{-1}(M)$ transversally.