# 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.

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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 $\epsilon=1$.

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For those who don't have Hatcher with them (or etched permanently into their brains) -- $X$ is the Klein bottle and $Y\subset X$ is obtained by deleting the disk bounded by the circle of self-intersection that you get when you take the usual 4-dimensional embedding and project down into 3 dimensions. – Aaron Mazel-Gee Jan 2 '11 at 19:20
I've found $\pi_1(Y)$, but could you explain why $\pi_1(Y)=\pi_1(\mathbb{R}^3−Z)?$ I'm having trouble picturing it. Edit: Actually, my solution to $\pi_1(Y)$ was wrong, so some assistance there would also be appreciated. – alephzero314 Jan 3 '11 at 8:33
Thanks very much for your detailed solution! I see what's going on now. – alephzero314 Jan 4 '11 at 0:03
@Aaron Mazel-Gee, i have a question!If we consider the van kampen thrm. if $U$ is the square$-\{y\}$, $y$ is the middle point,$V$the image of the interior of the square $\phi:\pi_1(U\cap V)\rightarrow \pi_1(U)$. how to compute the fundamental group? – Ronald Apr 11 '13 at 22:00
@Danial: This deserves to be its own question. – Aaron Mazel-Gee Apr 11 '13 at 22:06

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.

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The general kind of graph in $\mathbb R^3$ that might be involved is as follows:

Under the above circumstances, the fundamental group $\pi_1(\mathbb R^3\setminus i(K\times E^2),p)$ has a presentation with a generator for each overpass and relations of two types:

(a) at each vertex $v$ there is a relation $x^{\varepsilon_{1}}_{1} x^{\varepsilon_{2}}_{2}\dots x^{\varepsilon_{r}}_{r} = 1$, where the $x_{i}$ are the edges at the vertex and the sign $\varepsilon_{i}$ is $+1$ if the arrow for $x_{i}$ points towards the vertex, and $-1$ otherwise;

(b) at each crossing with overpass $x$ crossing $y$ and $z$ there is a relation of the form $y = xzx^{-1}$.

The latter are the usual Wirtinger relations for a knot or link. The following picture illustrates the relations, showing stages in the movement of a path under a vertex or a crossing:

(Figures from "Topology and groupoids" Chapter 9.)

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