Fundamental group of R^3\two linked circles

Computation of the fundamental group of $X=\mathbb{R}^3\setminus (C_1\cup C_2)$, where $C_1$ and $C_2$ are two disjoint linked circles.

We apply the van Kampen theorem. 

Let $U$ be the open halfspace to the left of the green plane, and $V$ be the open halfspace to the right of the blue halfspace. Fix the basepoint $x_0$.

Clearly $\pi_1(U)=\pi_1(V)=\mathbb{Z}*\mathbb{Z}$, as both $U$ and $V$ deformation retract to $S^1\vee S^1$. Similarly, $\pi_1(U\cap V)=\mathbb{Z}*\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$, as $U\cap V$ deformation retracts to $S^1\vee S^1\vee S^1\vee S^1$. Let $p_1$ and $p_2$ be homotopy classes of loops in $\pi_1(U\cap V)$ encircling the missing arcs in $xy$ plane, with $p_1$ corresponding to the arc with positive $y$. Similarly $q_1$ and $q_2$ are classes of loops encircling the arcs in $xz$ plane with $q_1$ corresponding to the arc with positive $z$. Thus $p_1,p_2,q_1,q_2$ are generators of $\pi_1(U\cap V)$.

Let $i:U\cap V\hookrightarrow U$ and $j:U\cap V\hookrightarrow V$ be the inclusion maps. We might clearly take $a=i_*p_1$ and $b=i_*q_1$ as generators of $\pi_1(U)$ and $c=j_*p_1, d=j_*q_1$ as generators of $\pi_1(V)$. The amalgamation product tells us that $i_*r$ should be identified with $j_*r$ in $\pi_1(U)*\pi_1(V)$ for all $r\in \pi_1(U\cap V)$. Thus $a=c$ and $b=d$, so we reduce $\pi_1(U)*\pi_1(V)$ to $\mathbb{Z}*\mathbb{Z}$. But there are more identifications we need to take into account, in order to obtain $\pi_1(X)$. Observe for example that we must identify $i_*p_2$ with $j_*p_2$. But $i_*p_2=a$ and $j_*p_2=dcd^{-1}$, so $a=bab^{-1}.$ So we further reduce $\pi_1(U)*\pi_1(V)$ to $\mathbb{Z}\times \mathbb{Z}.$ One more identification needs to be taken into account, but it does not give anything new. Thus $\pi_1(X)=\mathbb{Z}\times\mathbb{Z}$.

$\mathbf{Another}$ approach is to use cover by the opened sets as on the next figure.

 

It is easy to see that $\pi_1(U)\cong\pi_1(V)\cong\mathbb{Z}\times\mathbb{Z}$ as they both deformation rectract onto a torus. Moreover, as in the previous situation, we see that we may identify the generators of $\pi_1(U)$ and $\pi_1(V)$ in the amalgamated product, via generators of $\pi_1(U\cap V)$. Thus $\pi_1(X)=\mathbb{Z}\times\mathbb{Z}.$

 Similar, but easier arguments, could be employed to show that $\pi_1(\mathbb{R}^3\setminus\text{circle})=\mathbb{Z}$ and $\pi_1(\mathbb{R}^3\setminus\text{two unlinked circles})=\mathbb{Z}*\mathbb{Z}$.

 

Join is simply connected

If $X$ is path-connected and $Y$ is arbitrary, then $X*Y$ is simply connected. 

 

$\textbf{Proof}.$

Define $U:=X*Y\setminus Y=X\times Y\times [0,1)/\sim,$ where $\sim: \ (x,y_1,0)\sim(x,y_2,0) $ for all $y_1,y_2\in Y.$  Thus $U$ is clearly open and deformation retracts onto $X\cong  X\times Y\times \{0\}/\sim.$ As $X$ is path-connected, so is $U$.

Fix $x_0\in X$. Let $V_1$ be the cone with vertex $x_0$ and base  $\{x_0\}\times Y\times\{1/2\}$. Also define $V_2:=X\times Y\times [1/2,1]/\sim$, where $\sim: \ (x_1,y,1)\sim(x_2,y,1) $ for all $x_1,x_2\in X.$ Set $V:=V_1\cup V_2$. Observe that $V_2$ deformation retracts onto  $\{x_0\}\times Y\times\{1/2\}$ (collapse everything to $Y$ and then push it to $\{x_0\}\times Y\times\{1/2\}$). But this is the base of the cone $V_1$, and $V_1$ deformation retracts to $x_0$. Hence $V$ is contractible.

Now argue as in the easier part of Seifert-van Kampen. The set $V$ is not open, but $U\cup \text{int}V=X*Y$. Thus every loop in $X*Y$ based at $x_0$ is a finite product of paths, each belonging to either $U$ or $V$. Moreover, since $X$ is path-connected, $U\cap V=V_1\cup (X\times Y\times [1/2,1))$ is path-connected. Thus every loop in $X*Y$ based at $x_0$ is homotopy equivalent to a finite product of loops at $x_0$, each belonging to either $U$ or $V$. Observe that $X$ is nullhomotopic in $X*Y$ (use a cone with base $X$ and vertex some point in $Y$), and hence $U$ is such as well. Thus the inclusion $U\hookrightarrow{} X*Y$ induces a trivial homomorphism $\pi_1(U)\to \pi_1(X*Y)$. Moreover, $\pi_1(V)$ is already trivial. Thus each loop in $X*Y$ is homotopy equivalent to a constant loop.

Homotopy type of R^3\(circles)

 

$\mathbf{1}$. The spaces $X=\mathbb{R}^3\setminus S^1$ and $S^{1}\vee S^{2}$ are homotopy equivalent.

First we rewrite $X\cong S^3\setminus (\{*\}\cup S^1)$, as $S^3$ is the one-point compactification of $\mathbb{R}^3$. Now observe $S^3\setminus S^1$ is homeomorphic to the open solid torus, $S^1\times e^2$, where $e^2$ is the open unit $2$-disc. An explicit homemorphism is as follows : $$(x,y,z,t)\mapsto (x\sqrt{1-z^2-t^2},y\sqrt{1-z^2-t^2},z,t).$$

 Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\setminus \{*\}$. Here $S^1\times e^2$ is open $3$-manifold. Removing a point from open $3$-manifold $M$ is wedging $M$ with $S^2$ $-$ we may think of enclosing the point with $S^2$ and retracting the enclosed region to the sphere (or blowing up the point to a sphere); then we may push the sphere away from the manifold, to the boundary, forming the wedge product. Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\vee S^2\sim S^1\vee S^2$. 

 

$\mathbf{2}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint unlinked circles, is homotopy equivalent to $S^{1}\vee S^1\vee S^{2}\vee S^2$.

As before, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is contractible in $S^1\times e^2$, i.e. it is not encircling the hole of the torus. Thus we may take a $2$-sphere which bounds a region containing the circle to be removed. This regions retracts to $S^2\vee S^1$, as in $\mathbf{1}$. Pushing it to the boundary of $S^1\times e^2$ we get a wedge with $S^2\vee S^1.$ Removing the point $*$ induces, as in $\mathbf{1}$, one more wedge, with $S^2$. Hence $X\sim (S^1\times e^2)\vee S^2\vee S^2\vee S^1$, which finishes the derivations as $S^1\times e^2\sim S^1$

 

$\mathbf{3}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint linked circles, is homotopy equivalent to $S^{2}\vee (S^1\times S^1)$.

As in $\mathbf{2}$, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is not contractible in $S^1\times e^2$, i.e. it encircles the hole of the torus. Removing the point $*$ induces, as in $\mathbf{1}$, a wedge with $S^2$.  Hence $X\sim ((S^1\times e^2)\setminus S^1)\vee S^2$. But $(S^1\times e^2)\setminus S^1$ is a solid torus with core circle removed, which clearly deformation retracts to a torus $-$ $S^1\times S^1$.