The key reference here is Jean Cerf, "Groupes d’automorphismes et groupes de difféomorphismes des variétés compactes de dimension 3", Bull. Soc. Math. France (1959). The full text is available here.
Let $M$ be a closed 3-manifold, $G$ its group of self-homeomorphisms, $H$ its group of self-diffeomorphisms. (All orientation preserving for convenience.) He proves that $\pi_n(G,H) = 0$ for all $n \geq 0$, thus that the inclusion $H \hookrightarrow G$ is a weak homotopy equivalence - assuming Smale's conjecture (now theorem, due to Hatcher) that the inclusion $SO(4) \hookrightarrow \text{Diff}^+(S^3)$ is a homotopy equivalence. (In fact, he shows that the inclusion is a homotopy equivalents, but the arguments are a bit more delicate.)
For a manifold $M$ with boundary, $H$ and $G$ as above should be the automorphisms restricting to the identity on the boundary. The idea is to show that, if you have a decomposition $M = M_1 \cup M_2$, with $M_1 \cap M_2$ a properly embedded surface, having $H_i$ be $(n-1)$-connected in $G_i$ (for both $i$) is equivalent to having $H$ be $(n-1)$-connected in $G$.
Now we want to induct. Say that a manifold homeomorphic to $D^3$ is order 0; and a manifold that decomposes into order $(n-1)$-pieces is order $n$ if we can't decompose it into pieces smaller than order $(n-1)$. But by Smale's theorem, it's true that $\pi_i(G,H) = 0$ for all $i \geq 0$ when $M = D^3$. Because every 3-manifold has finite order (look at Heegaard decompositions), this proves the theorem.
As for why the third paragraph is true, this is Cerf's "Lemma 0", at the very end of his paper.
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:
- Some component of $\varphi^{-1}(C)$ bounds an embedded disc in $S_g$; or
- Some pair of components of $\varphi^{-1}(C)$ bound an embedded annulus in $S_g$.
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:
- Some component of $\varphi^{-1}(C)$ bounds an embedded disc $D \subset S_g$ such that the interior of $D$ is disjoint from $\varphi^{-1}(C)$; or
- Some pair of components of $\varphi^{-1}(C)$ bound an embedded annulus $A \subset S_g$ such that the interior of $A$ is disjoint from $\varphi^{-1}(C)$.
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 $A' \to C'$.
Best Answer
Of course $f^{-1}(C)$ will never be empty: then $f$ is not surjective and so does not have degree 1. But you can make $f$ restrict to a homeomorphism on $f^{-1}(C)$.
Every degree-1 map $g: \Sigma_g \to \Sigma_h$ is homotopic to a map $f$ which collapses a subsurface $S \subset \Sigma_g$ to a point $f(S) \in \Sigma_h$ but is otherwise an oriented homeomorphism $f: \Sigma_g \setminus S \to \Sigma_h \setminus f(S)$. This follows from a result of Edmonds.
Now simply choose a representative for $C$ which does not include the single point $f(S)$.