# Regular Covering Map corresponding to a normal subgroup of a free group

This question is related to one I asked here: Degree 4 Regular based covering maps of the Torus with corresponding normal subgroup.

But this is a far shorter question, and focuses on a slightly different point, than my previous question.

If we let $X = (B_{2},v)$ denote the bouquet of two circles identified at $v$ with edges oriented as in this diagram:

Then $G = \pi_{1}(X,v) = \left< x, y \right> = F_{2}$ is the free group on two generators. Then we define the map:

\begin{align*} \phi : G \rightarrow C_{4} = \left< \xi \mid \xi^4 \right>,\ x\mapsto\xi,\ y\mapsto \xi \\ \end{align*}

Then since $\phi$ is a surjective group homomorphism on to a group of order 4, $K = \operatorname{ker}(\phi)$ is an index four normal subgroup of $G$. Hence there is a unique regular degree $4$ covering $(\tilde{X},\tilde{v})$ of $(X,v)$ corresponding to $K$.

Then we know that $(\tilde{X},\tilde{v})$ is a regular graph on 4 vertices with 4 edges coming out of each vertex such that $x^4, y^4, xy^{-1}, y^{-1}x, xy^{3}, y^{3}x, x^{3}y, yx^{3}, x^{2}y^{2}, y^{2}x^{2}$ are all loops at every vertex. In the previous question, this is what I came up with:

I believe that this satisfies the regularity requirement, and the loop requirements. However if we look carefully at a small neighbourhood around a vertex, $v_{1}$ say, we see the following picture:

Then, if we were to close the loops we would see that the edge labeled $y$ is in the opposite orientation to that of $X$. Meaning that my solution is not correct.

However, suppose we try another solution. We draw $4$ dots on the vertices of a square, with $4$ edges leaving each vertex all in the correct orientation, as in the following picture:

Where the red edges correspond to edges labelled $y$, and the cyan ones correspond to edges labelled $x$. However, whatever I try, I can't seem this connect these edges correctly, without edges intersecting.

Would anyone be able to help me see where I'm going wrong?

I have since talked to my tutor, who told me that my answer is correct. The reason I was getting confused is that I neglected the fact that there is a homeomorphism of $B_{2}$ leaving the orientation of the edge labelled $x$ fixed, whilst swapping the orientation of the edged labeled $y$. So the covering spaces will be blind to the absolute orientation of these edges.