Let $\varphi:S_g\to S_g$ be a smooth homotopy equivalence on genus-$g$ surface, where $g\geq 2$. Let $C$ be an embedded circle in $S_g$ such that $\varphi$ is transversal to $C$. Therefore, $\varphi^{-1}(C)$ is an embedded $1$-dimensional submanifold of $S_g$. Using Stack-Record theorem one can show $\varphi\big|\varphi^{-1}(C)\to C$ is a finite covering map. Actually, $\varphi^{-1}(C)$ is disjoint union of finitely many embedded circles in $S_g$.
Is it possible to homotop $\varphi$ so that $\varphi$ is again transverse to $C$ with $\varphi^{-1}(C)$ is a single embedded circle?
Note that $\varphi$ is a degree $\pm1$ map. So, after homotopy, we can not expect $\varphi^{-1}(C)=\varnothing$ as homotopy preserves degree and degree $\pm 1$ maps are surjective.
Best Answer
Yes, this can be done, and in fact this is an important step in the proof of the theorem that every self-homotopy equivalence of $S_g$ is homotopic to a homeomorphism. I think you'll find at least a sketch of this proof (and that step) in the book of Farb and Margalit, and you should also find full references there. You might also look at the paper by Epstein, "Curves on 2-manifolds and isotopies", Acta Math. 115 (1966), 83–107.
Here's an outline.
Step 1: Prove that if $\varphi^{-1}(C)$ is not connected then one of two things occurs:
The reason this is true is that if two disjoint circles $\gamma_1,\gamma_2 \subset S_g$ are homotopic to each other then either they each bound discs or the pair of them bounds an annulus. I would say that the best modern proof (best both in intuition and rigor) uses hyperbolic geometry to prove that if a simple closed curve does not bound a disc then it is isotopic to a unique simple closed geodesic.
Step 2: Improve Step 1 so that:
To obtain a disc or annulus satisfying 3 or 4, simply take an innermost one (with respect to inclusion) of the collection of discs and annuli satisfying 1 or 2.
Step 3: If one obtains a disc $D$ as in 3, one may then homotope $\varphi$ on a small open neighborhood of $D$ so as to remove $\partial D$ from the set of components of $\varphi^{-1}(C)$. To do this, one uses transversality to find a disc $D'$ whose interior contains $D$, with $\gamma = \partial D'$, such that $\varphi(\gamma)$ lies on a circle $C'$ which is disjoint from but close to $C$ and is isotopic to $C$. The restricted map $\gamma \to C'$ is null-homotopic, so we may replace the restriction $\varphi \mid \hat D$ by a map whose image is contained in $C'$.
Step 4: Assuming that there are no such discs as in 3, take an annulus as in 4. Note in this case that $C$ itself does not bound a disc.
One may then homotope $\varphi$ to remove the two component of $\partial A$ from the list of component of $\varphi^{-1}(C)$. To do this, one first uses transversality to obtain an annulus $A' \subset S$ whose interior contains $A$, with $\partial A' = \gamma_1 \cup \gamma_2$, such that $\varphi(\gamma) \cup \varphi(\gamma_2)$ lies on a circle $C'$ which is disjoint from but close to $C$ and is isotopic to $C$. The circle $C'$, like $C$, is $\pi_1$-injective. Using this one then checks that the restrictions of $\varphi$ to the two boundary circles $\gamma_1,\gamma_2$ represent the same element of $\pi_1 C' = \mathbb Z$, and therefore the restricted map $\varphi : \partial A' \to C'$ may be extended to a map $A' \to C'$.