Why is the fundamental group of the Hopf link abelian but a two component unlink isn't

Question

Title says it all basically. I'm trying to understand why the fundamental group of the hopf link (or really, the compliment of the Hopf link) is abelian. I mean, in a certain way I understand it, but on the other hand, just considering paths $a$ and $b$ where $a$ goes through one hoop and $b$ the other, I had previously convinced myself that $ab=ba$ based on the fact that the trace of their images is the same. Why is this thinking flawed? looking for general insight really

Answer 1

Using Wirtinger's presentation, you should see that there is no relation in the fundamental group of the unlink, because there are no crossings. So, you should always get the free group on the same number of generators as components of your unlink.

Now, for the Hopf link, again with Wirtinger's presentation. Since there are two overarcs, you get one generator for each one, $x$ and $y$. The relation you get around the crossings, which you can see in the image below, comes from the four strands at one of the crossings. Follow the red strand on the left pass under $x$ and then $y$, so you get $xy$. And that is the same as following the red dotted strand on the right, $yx$. Hence, $xy = yx$ and the group is $\langle x,y \vert xy=yx\rangle$. (Note, you can always exclude one crossing's relation.)

I should add an intuitive point here too. The complement of the Hopf link in $S^3$ is really a thickened torus, $T^2 \times I$, and therefore deformation retracts to the torus. And the torus is known to be abelian.

Answer 2

N. Owad already mentions how the complement of a Hopf link deformation retracts onto a torus, but I'd like to add a bit about the geometry of the complement of a split link $L$ of two unknots. Being a split link means there is an embedded sphere in the complement separating the two components. Since spheres are simply connected, the van Kampen theorem says that $\pi_1(S^3-L)$ is the free product of two copies of $\pi_1(B^3-U)$, where $U$ is an unknot in the ball $B^3$. Gluing another $B^3$ to the boundary does not change the fundamental group (since the boundary $S^2$ is simply connect as is $B^3$), so $\pi_1(B^3-U)=\pi_1(S^3-U)$. Since $S^3-U$ is a solid torus, $\pi_1(S^3-L)=\mathbb{Z}*\mathbb{Z}$.