Bases of surface groups Let $\Gamma_g$ be a surface group of genus $g \geq 2$. A $2g$-tuple  $(x_1,y_1, \dots,x_g,y_g) \in \Gamma_g^{2g}$ will be called a Surface Basis if we have the presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$ Take some $1 \leq k \leq g$ and $H \leq \Gamma_g$ of finite index such that $x_1, \dots, x_k \in H$. One can show that $H$ is a surface group of genus $(g-1)[\Gamma_g : H] + 1$, and that it has a surface basis. My question is:

Is there a surface basis for $H$ containing $x_1, \dots, x_k$?

The analogous question for free groups has a positive answer as shown in Bases of free groups.
 A: There is indeed a surface basis for $H$ containing $x_1,\ldots,x_k$. I'll give a topological proof, basically the same as the proof suggested in the comment of @HJRW. First, I'll give a topological re-interpretation of your problem, by formulating a topological property which is equivalent to the property "there is a surface basis of the fundamental group containing such-and-such a list of elements". Then I'll say why this topological property is preserved under the kind of covering maps in your question.

For each genus $g$ fix a ``standard'' surface $S_g$ with base point $p$ and an isomorphism $\pi_1(S_g,p) \approx \Gamma_g$, and with an embedded rank $2g$ rose consisting of loops $\xi_1,\eta_1,\ldots,\xi_g,\eta_g$ representing $x_1,y_1,\ldots,x_g,y_g$. When $S_g$ is cut open along the rose, the result is a $2g$-gon whose boundary is attached to the rose using the word given by the standard relator in your question. Let $N_k$ be a regular neighborhood of $\xi_1 \cup \cdots \cup \xi_k$. Note that $N_k$ is a connected planar surface with connected complement; it follows that $N_k$ has $k+1$ boundary components, and that $S_g - N_k$ is a connected surface of genus $g-k$ with $k+1$ boundary components.
Consider now an abstract surface $S'$ of some genus $p'$, and a rank $k$ rose embedded in $F$ consisting of loops $\gamma_1,\ldots,\gamma_k$ all with common base point $q$. Then the following are equivalent: 


*

*there exists a surface basis for $\pi_1(S',p')$ containing $[\gamma_1],\ldots,[\gamma_k]$ 

*the regular neighborhood of the rose $\gamma_1 \cup \ldots \cup \gamma_k$ is a connected planar surface with connected complement. 


I've already proved 1$\implies$2. For 2$\implies$1, it follows the regular neighborhood of the rose is a planar surface with $k+1$ boundary components and that its complement has genus $g'-k$ with $k+1$ boundary components, and this allows us to construct a homeomorphism between $S'$ and the standard surface $S_{g'}$ that takes $\gamma_1,\ldots,\gamma_k$ to $\xi_1,\ldots,\xi_k$.

Okay, so, consider a finite index subgroup of $\pi_1(S,p)$ containing $x_1,\ldots,x_k$. Let $f : S' \to S$ be the corresponding finite degree covering map with base point $p'$ lifting $p$. The loops $\xi_1,\ldots,\xi_k$ lift one-to-one to loops $\xi'_1,\ldots,\xi'_k$ based at $p'$. The regular neighborhood $N_k$ of $\xi_1\cup\cdots\cup\xi_k$ lifts one-to-one to a regular neighborhood $N'_k$ of $\xi'_1\cup\cdots\cup\xi'_k$. Since the restricted covering map $N'_k \to N_k$ is a homeomorphism, this proves that $N'_k$ is planar. It remains to prove that $S'-N'_k$ is connected. 
Arguing by contradiction, assuming that $S' - N'_k$ is disconnected, consider a component $F$ of $S'-N'_k$ which meets $N'_k$. It follows that $\partial F = F \cap N'_k$ consists of a proper nonempty subset of the $k+1$ components of $\partial N'_k$. Let those components be enumerated $c'_1,\ldots,c'_j,c'_{j+1},\ldots,c'_{k+1}$ so that $\partial F = c'_1 \cup \cdots \cup c'_j$. Push this enumeration down to $N_k$ whose boundary circles are therefore enumerated $c_1,\ldots,c_j,c_{j+1},\ldots,c_{k+1}$.
Let the surface $F$ be decomposed as $F = F_1 \cup F_2$ where the covering map $S' \to S$ restricts to a covering map $F_1 \to S - N_k$ of some degree $\delta_1$, and it restricts to a covering map $F_2 \to N_k$ of some degree $\delta_2$. Let $\partial F_1$ be decomposed into two subsets $\partial F_1 = \partial_{\le j} F_1 \cup \partial_{>j} F_1$, with restricted covering maps $\partial_{\le j} F_1 \to c_1 \cup \cdots \cup c_j$ and $\partial_{>j} F_1 \to c_{j+1} \cup \cdots \cup c_{k+1}$ each of degree $\delta_1$ (the base spaces of these covering maps are not connected, nonetheless these covering maps each have constant degree $\delta_1$ over each component). It follows that the restricted map $\partial F_1 - (c'_1 \cup \cdots \cup c'_j) \to c_1 \cup \cdots \cup c_{k+1}$ is a covering map of unequal degrees over different components: it has degree $\delta_1-1$ over each component of $c_1,\ldots,c_j$ and it has degree degree $\delta_1$ over each component of $c_{j+1},\ldots,c_{k+1}$. However, we have an equation 
$$\partial F_1 - (c'_1 \cup \cdots \cup c'_j) = \partial F_2
$$ 
and the degree $\delta_2$ covering map $F_2 \to N_k$ restricts to a map $\partial F_2 \to c_1 \cup \cdots \cup c_{k+1}$ of equal degree $\delta_2$ over each component. This is a contradiction.
