Finding the fundamental group of the complement of a certain graph

I'm having trouble solving problem 12 from Section 1.2 in Hatcher's "Algebraic Topology".

Here's the relevant image for the problem:

I'm trying to find $\pi_1(R^3-Z)$, where $Z$ is the graph shown in the first figure. The answer (according to the problem statement) is supposed to be $\langle a,b,c| aba^{-1}b^{-1}cb^\epsilon c^{-1}=1\rangle$, where $\epsilon=\pm 1$.

I attempted to use Van Kampen's theorem, using a cover of two open sets, depicted in the lower image. The first open set is the area above the bottom horizontal line, minus the graph, and the second open set is the region below the top horizontal line, minus the graph. The intersection is the area in between the two horizontal lines minus the graph.

Call the top set $A$, and the bottom set $B$. With applications of Van Kampen, I got that $\pi_1(A)\cong \pi_1(B)\cong Z*Z*Z$, and that $\pi_1(A \cap B)\cong Z*Z*Z*Z*Z$. After using Van Kampen's theorem again, I got that $\pi_1(R^3-Z)\cong Z*Z$, which is wrong.

I remember doing this problem and having similar difficulties, until I realized that it was actually way easier to find $\pi_1(Y)$ than it was to find $\pi_1(\mathbb{R}^3-Z)$. That these are the same is more or less a proof by picture (remember that fundamental group is an invariant of homotopy type). I suggest drawing a "fundamental domain" for $Y$ which is modeled on the usual one for $X$, but of course you'll need to modify it to take account of the removed disk. I can say more if you're still not getting it...

More:

There is a (sort of) evident deformation retraction of $\mathbb{R}^3-Z$ onto $Y$, which implies that the two spaces are homotopy equivalent, which implies that they have the same fundamental group.

To find $\pi_1(Y)$, start by drawing the usual square fundamental domain for a Klein bottle:

To make things easy for ourselves, let's assume (without loss of generality) that the b-circles are attached at the circle which in the projection to $\mathbb{R}^3$ bounds the disk that we'll cut out. When we do so, we get this:

It'll be easiest to think of this as a single disk whose boundary is identified to itself. That way we can consider this as a graph (in fact, a bouquet of circles) with a single disk attached, and it's easy to find the fundamental group of such a thing. To do this, we add in an edge:

So now we're looking at this space as a bouquet of circles $a,b,c$, and we've sewn on a disk according to the relation $aba^{-1}b^{-1}cb^{-1}c^{-1}$. If you do this slightly differently you can switch to $\epsilon=1$.

This is admittedly not an answer, but it seems to me like you have already done a lot of the work. So here are my suggestions:

1) Name the generators of each copy of the integers, think about $\pi_1(A \cup B)$ in terms of generators and relations by using van Kampens' theorem as you are doing.

2) Remember what van Kampen says, you are taking the free product of the two over the intersection, so you need to name the generators of $\pi_1(A \cap B)$ in terms of how they include into $\pi_1(A)$ and $\pi_1(B)$ respectively. Think about how you take tensor products.