algebraic-topologygeneral-topology
On pg. 47 of Hathcer's Algebraic Topology, the author discusses the fundamental group of $\mathbf R^n-(A\cup B)$, where $A$ and $B$ are circles in $\mathbf R^3$ which are linked.
The author writes that $\mathbf R^3-A\cup B$ deformation retracts to the wedge sum of a torus and a $2$-sphere.
I was unable to see how this is so. Can somebody please help me visualize a deformation retract.
Not a complete answer
Hatcher shows two unlinked circles have a fundamental group $\mathbb{F}_2$ which is humungous - it's the free group of all possible wors in two letters. The fundamental group of two linked circles is simply $\mathbb{Z} \oplus \mathbb{Z}$ - we can exploit the linking of the rings in order to express any entanglement as a special type.
In order to find the homotopy group of the Borromean rings you may need to draw as many as 4 rings.
3 pairwise linked circles will have three generators, looping ones around each of the possible three circles. Do these generators commute? Sure we can imagine one the circles disappearing and getting two circles. We have shown:
$$ \pi_1(3\text{ linked circles}) \subseteq \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z}$$
It does indeed, and the proof is exactly the same as in the orientable case.
Let $P$ be the $2g$-gon with its gluing pattern as you described, and let $\pi : P \to N_g$ be the quotient map. Let $\mathcal O$ be the point at the center of $P$. Then $X = N_g - \pi(\mathcal O)$.
Next, observe that the gluing pattern of $P$ has one vertex cycle. To put this another way, $\pi$ takes all vertices of $P$ to a single point. It follows $\pi(\partial P) \subset \pi(P)=N_g$, which is the image in $N_g$ of the boundary $\partial P$, is a wedge of $g$ circles in $N_g$, because $\pi(\partial P)$ is a 1-dimensional CW complex with one vertex and $g$ edges. Removing the point $\pi(\mathcal O)$, it follows that $\pi(\partial P)$ is a wedge of $g$ circles in $X$.
Now consider the retraction map $r : P - \mathcal O \to \partial P$ which takes each "radial segment" of $P - \mathcal O$ to the unique endpoint of that segment on $\partial P$. There is an obvious deformation retraction $$H : (P-\{\mathcal O\}) \times [0,1] \to P - \{\mathcal O\} $$ from the identity map of $P - \mathcal O$ to the retraction map $R$ defined as follows: if $x$ lies on a radial segment having endpoint $q \in \partial P$ then $H(x,t) = (1-t)x + t q$.
The final observation is that this deformation retraction on $P - \mathcal O$ respects all of the identifications made by the restricted quotient map $\pi : P - \mathcal O \to N - \pi(\mathcal O) = X$, and therefore it descends to a deformation retraction from $X$ to $\pi(\partial P)$.
Best Answer
This is a partial answer.
To see the deformation retraction, see the flow lines in the diagram below. It shows the flow lines in the vertical and horizontal slices. For the intermediary slices, we gradually move from horizontal slice to the vertical slice.
(In the diagram, dashed circles are the slices of the torus; bold circles and points are the slices of the removed circles).