Draw a cover of $S^1 \vee S^1$ whose $\pi_1$ is isomorphic to $\langle a^2,b^3,aba^{-1}b^{-1} \rangle \leq F_2$ Draw a cover of $S^1 \vee S^1$ whose $\pi_1$ is isomorphic to $\langle a^2,b^3,aba^{-1}b^{-1} \rangle \leq F_2$.
This is a follow-up to my post from yesterday regarding the kernel K of a map $\Phi: F_2\to \mathbb{Z}_2 \bigoplus \mathbb{Z}_3$ which sends $a\mapsto ([0]_2,[1]_3)$ and $b\mapsto ([1]_2, [0]_3)$.
Having obtained $K=\ker{\Phi}=\langle a^2,b^3,aba^{-1}b^{-1} \rangle \leq F_2$ I am now tasked with drawing a covering of $S^1 \vee S^1$ such that the fundamental group of the covering space maps isomorphically to $K$ under the homomorphism induced by $\Phi$.
I know that any covering of $S^1 \vee S^1$ will be a 4-regular graph (each vertex will have 4 half-edges), and I have seen many examples of interesting covers of $S^1 \vee S^1$ such as in this post.
Using these as inspiration my first attempt at drawing a covering is this:

However, I then noticed that the outer vertices do not have 4 half-edges. So my next thought is just to extend the graph to infinity, making a fractal of the pattern (i.e. each outer 'red' vertex would be joined to a vertex of another 'blue' triangle). Though I don't know how to TeK that just yet.
I reckon the fundamental group of such an infinite graph would be (choosing any base point) $\langle a^2, ba^2b^{-1}=a^2, b^{-1}a^2b=a^2, b^3 \rangle$ assuming we have a commutativity relation.
Also, I'm not sure how $aba^{-1}b^{-1}$ could be a loop in the covering space, maybe I am confused about that generator. Is it effectively the same as a commutivity relation?
Is this correct, or am I missing something?
edit: Was able to finish up the graph correctly in Tikz thanks to the answer below. Here it is:

 A: I'm pretty sure the diagram below (which was suggested by Zeekless) is the covering space you're looking for. Firstly, it's clearly a covering space since each vertex has the required pairs of arrows coming in and out. Call the covering space $X$, and let $p:X \to S^1 \vee S^1$ be the covering map.
Secondly, all the generators you want are in the image of the fundamental group. For example, if you start at any vertex and trace out the paths "$a$", "$b$", "$a^{-1}$", and then "$b^{-1}$", you get back where you started. Therefore, the word $aba^{-1}b^{-1}$ is in $p_*\pi_1(X)$. Similarly, following "$a$" twice and "$b$" three times (respectively) gets you back where you started, so we have $\langle a^2, b^3, aba^{-1}b^{-1}\rangle \subseteq p_*\pi_1(X)$.
We claim that actually this inclusion is an equality. To show this, we can consider $p_*\pi_1(X)/K$, where $K = \langle a^2, b^3, aba^{-1}b^{-1}\rangle$, and show that the quotient is trivial. Note that you already showed that $K$ is a kernel, hence a normal subgroup of $F_2$, and therefore also of $\pi_1(X)$, so we are allowed to quotient by it. Let $[w] \in p_*\pi_1(X)/K$, where $w$ is a word in $a$ and $b$. Since $aba^{-1}b^{-1}\in K$, the quotient group is abelian, so we may assume that $w = a^ib^j$ for some $i,j$. Since $a^2,b^3\in K$, we may assume that $0 \leq i \leq 1$ and $0 \leq j \leq 2$. There are then six possibilities for $w$. Of these six, only the identity is actually in $p_*\pi_1(X)$, which means that $[w]$ is the identity in $p_*\pi_1(X)/K$. It follows that $p_*\pi_1(X)/K$ is trivial, so $K = p_*\pi_1(X)$.

