For $n\geq1$, let $p_1,\ldots,p_n$ be distinct points in $S^2$. Let $f\colon S^3\to S^2$ be the Hopf fibration, let $B_n=S^2\setminus\{p_1,\ldots,p_n\}$ be the $n$-times punctured sphere, and let $E_n=f^{-1}(B_n)$ be the $n$-fold Hopf link complement. I'm interested in an algebraic way of calculating $\pi_1(E_n)$. I know from here that $\pi_1(E_n)\cong\mathbb{Z}\times F_{n-1}$ and that this can be shown by coming up with a clever deformation retract. However, I'd like to know if the following algebraic argument is correct. Bold text signifies things I'm not entirely sure about.
By construction, we have an $S^1$-principal bundle $S^1\to E_n\to B_n$. This gives rise to a long exact sequence of homotopy groups, which ends as follows:
$$\ldots\to\pi_2(B_n)\to\pi_1(S^1)\to\pi_1(E_n)\to\pi_1(B_n)\to\pi_0(S^1)=1.$$
We have $\pi_1(S^1)=\mathbb{Z}$, $\pi_1(B_n)=F_{n-1}$, and I think $\pi_2(B_n)$ is trivial, so we in fact get a short exact sequence:
$$1\to\mathbb{Z}\to\pi_1(E_n)\to F_{n-1}\to1.$$
More than this, I think any Hopf fibre in $E_n$ lies in the centre of the fundamental group, and thus the inclusion $\mathbb{Z}\to\pi_1(E_n)$ is central. Therefore, $\pi_1(E_n)$ is a central extension of $F_{n-1}$ by $\mathbb{Z}$. These are, I think, in bijective correspondence with elements of $H^2(F_{n-1},\mathbb{Z})$, which is trivial. The extension must therefore be trivial, i.e.,
$$\pi_1(E_n)\cong\mathbb{Z}\times F_{n-1}.$$
Are there any problems with this argument? If not, could you give quick reasons or references for the statements preceded by "I think"?
Best Answer
$\pi_2(B_n)$ is trivial since $B_n$ is homotopy equivalent to a wedge sum of circles, the universal cover of which is contractible, so the group is isomorphic to $\pi_2(*)=1$. (That is, $B_n$ is an Eilenberg-Maclane space.)
Yes, $\pi_1(S^1)\to \pi_1(E_n)$ is central. A way to see this is that if $\gamma\in\pi_1(S^1)$ is the generator and $g\in\pi_1(E_n)$ is any element, then we can go from $g\gamma$ to $\gamma g$ by dragging the basepoint of $\gamma$ along $g$ while transporting $\gamma$ from fiber to fiber -- the element $g$ itself is parameterizing basepoints of fibers through this homotopy. (More generally, $E_n$ is an example of a Seifert fibration with only regular fibers and with an orientable base manifold and with $\pi_1(S^1)\to\pi_1(E_n)$ injective.)
So we have a central extension of $F_{n-1}$ by $\mathbb{Z}$, which is classified by a class in the group cohomology $H^2(F_{n-1};\mathbb{Z})$, where the abelian group $\mathbb{Z}$ is given the structure of a trivial $F_{n-1}$-module. A fun way to compute group cohomology with trivial coefficients is to compute $H^2(K(F_{n-1},1))$ for an Eilenberg-Maclane space. We already established that $B_n\simeq K(F_{n-1},1)$, so we want $H^2(B_n)$. Since $B_n$ is homotopy equivalent to a wedge sum of circles, which has no two cells, then $H^2(B_n)=0$. (Note: $E_n$ is an oriented circle bundle over $B_n$, which is classified by the first Chern class, an element of $H^2(B_n)$. Nice to see things coincide.)
Topologically, a reason the group splits is that $E_n\to B_n$ has a section $B_n\to E_n$. In knot theory, this is a kind of Seifert surface for the $n$-fold Hopf link. There's a way to explicitly construct one as the Milnor fiber of a complex singularity associated to the Hopf link.