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I have been solving some past exam questions and I came across the following question. Let $X=R^3-${(x-axis)U (y-axis)} be the complement of the x and y-axes in $R^3$. Compute the fundamental group of $X$. I am not really sure how to go about this. My guess is that if I could find a retract of this space whose fundamental group is known then the question is solved. I am also wondering if the Seifert-van Kampen theorem would work but I have no clue. Any help would greatly be appreciated.

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Do you think that this space has the same homotopy type of a cylinder? –  Sigur Sep 1 '12 at 3:12
    
@Sigur. Honestly I don't know if that is the case, but even if it is I would have to show that they are homotopy equivalent. That does not seem easy for me. –  smanoos Sep 1 '12 at 3:15
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Think instead about the unit ball, with two perpendicular holes drilled through it... –  user641 Sep 1 '12 at 3:18
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It might be useful to deformation retract $X$ onto its intersection with the unit sphere. –  Julian Rosen Sep 1 '12 at 3:45
    
@PinkElephants. Could you please provide some details if possible? –  smanoos Sep 1 '12 at 4:06
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3 Answers

up vote 5 down vote accepted

You can show that $\mathbb{R}^{3}$ minus a coordinate cross deformation retracts onto the 2-sphere minus four points. I would try and do this as follows: For any point in your space, put a line through the given point and the origin then do a straight line homotopy. Now pick one of the removed points on the 2-sphere as a pole and stereographically project onto $\mathbb{R}^{2}$ with three points removed. The fundamental group of this space is the same as a wedge of three circles: The free group on three generators.

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We can use van Kampen's theorem repeatedly. We use cylindrical coordinate and divide your space $X=\mathbb R^3\setminus \{(r,\theta,z): \lnot[z=0\land \theta=\frac{n\pi}{2}]\}$ into four pieces. Let $$A_1=\{(r,\theta,z)\in X: 0\leq \theta\leq \frac{\pi}{2}\},\ldots,A_4=\{(r,\theta,z)\in X: \frac{3\pi}{2}\leq \theta\leq 2\pi\}$$ and note that $X=A_1\cup\cdots\cup A_4$. Furthermore, each $A_i$ is a wedge-shaped slice of $\mathbb R^n$ minus a line through it, so is homotopy equivalent to a circle thus has fundamental group $\mathbb Z$. We have that $A_1\cap A_2$ is a closed half-plane minus a point on its boundary hence is contractible, so $\pi_1(A_1\cup A_2)\cong \mathbb Z*\mathbb Z=F_2$. Since $(A_1\cup A_2)\cap A_3=A_2\cap A_3$ is also a closed half-plane minus a point on its boundary (as $A_1\cap A_3$ is the $z$-axis minus the origin, which sits inside $A_2$), we have $\pi_1(A_1\cup A_2\cup A_3)\cong F_2*\mathbb Z=F_3$. Finally, $(A_1\cup A_2\cup A_3)\cap A_4$ is a bent plane minus a point, and $A_4$ can be contracted onto $(A_1\cup A_2\cup A_3)\cap A_4$ so the inclusion map $(A_1\cup A_2\cup A_3)\cap A_4\to A_4$ induces an isomorphism of fundamental groups, while the inclusion $$(A_1\cup A_2\cup A_3)\cap A_4\to A_1\cup A_2\cup A_3$$ induces a nontrivial map of fundamental groups which by freeness must be an inclusion. We call the generators of these images $a$ and $b$ respectively. Thus we have $$\pi_1(X)\cong F_3*_{\mathbb Z} \mathbb Z\cong F_4/N$$ where $N$ is the smallest normal subgroup containing $a^nb^{-n}$ for all $n$. But this is clearly the kernel of the map $F_4\to F_3$ which identifies $a$ and $b$, thus $\pi_1(X)=F_3$, the free group on three generators.

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Building on Steve D's comment, we can retract your space into the union of 2 circles, identifying 2 pairs of points, e.g. if you traced two distinct circles on a sphere that go through both the north and south pole. This suggests that 3 loops generate the fundamental group, one circling around each half-axis. More concretely, each generator is homotopic to a path that only intersects the plane (z=0) in two quadrants, and so the 4 generators correspond to each unordered pair of adjacent quadrants. For simplicity, we'll say that the generators are the paths that first intersect the lower numbered quadrant.

Then you can think of the relations between these paths. For example, if you take the loop going through quadrant 1 and quadrant 2, and then the path going around quadrant 2 and quadrant 3, that's homotopic to the path going from quadrant 1 to quadrant 3. Then take the path going from quadrant 3 to 4, and you get the path from 1 to 4. This is the only type of relation that emerges. I'm going to use the notation (i->j) for the path that goes first through quadrant i then j, and so our first relation is $(1->2)(2->3)(3->4) = (1->4)$. If you want to see this type of relation in a different case, e.g. $(2->3)(3->4)(4->1) = 2->1 = (1->2)^{-1}$, you can see that that this is follows from our first relation, $(1->2)^{-1}(1->2)(2->3)(3->4)(1->4)^{-1} = (1->2)^{-1}(1->4)(1->4)^{-1}$

So a presentation of the group might look like this:

$\langle a,b,c,d | abc=d \rangle.$

where $a = (1->2), b=(2->3)$, etc. But then, you don't need $d$ at all, so it's just the free group on 3 generators.

I just saw Alex's post, by the way, and it's the same thing without all my handwaving.

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