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](https://mathoverflow.net/questions/172599/bases-of-free-groups).



Bases of surface groups
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.




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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: 


1. there exists a surface basis for $\pi\_1(S',p')$ containing $[\gamma\_1],\ldots,[\gamma\_k]$
2. 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 $\xi\_1,\ldots,\xi\_k$.




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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.


