Fundamental group of a solid torus with a tunnel?

Question

Imagine that you have a donut and a worm inside. This worm takes two turns around the solid torus, going back to the starting point after two laps. How could I find out what the fundamental group of this space is? Can I use Van-Kampen's theorem to solve it or maybe exact sequences? Thanks.

Answer 1

I call $A$ the space that you are talking about. $A$ is of the form $A=\Lambda-T$ where $\Lambda$ is the original donut, and $T$ is the "tunnel" (it is a specific subspace of $\Lambda$ homeomorphic to a torus). I claim that $A$ deforms retract onto a subset $X$ such that

$$X\simeq \Bbb T^2 \bigsqcup \Bbb M^2/\sim$$

where $\Bbb M^2$ is the mobius band and $\sim$ is the relation attaching the boundary of $\Bbb M^2$ along one "canonical generator" of the fundamental group of the torus $\Bbb T^2$. The space $X$ is just "a mobius band which has his usual boundary replaced by a torus". To see that, I did the following drawing:

In my drawing, the big torus $\Lambda$ is in green, the tunnel $T$ created by the worm is in red and $X$ is the union of $T$ and the hatched grey mobius band. The mobius band is created by drawing the lines as in the picture.

In order to understand that $A$ deforms retract onto $X$, think of it the way around. If you take a neighborhood of $X$ in $A$, it looks just like $A$.

The only thing left to do is to compute the fundamental group of $X$. The space $X$ can be seen as follows, with the identifications by colors:

I don't know what is the best way to compute $\pi_1(X)$. Here is two ways that I can think of:

• You can use the "classic trick" that consists in doing a hole in the space and then using Van Kampen to get the fundamental group of the original space. To be more precise, take a disk $D^2\subset X$ and $p\in D^2$. Then if $\gamma$ is a generator of $D^2-\{p\}$ and if $i:D^2-\{p\}\to X-\{p\}$ is the inclusion, the theorem of Van-Kampen implies $$\pi_1(X)\simeq\pi_1(X-\{p\})/_{\langle i_*(\gamma)\rangle}$$

• More interestingly, in the picture we see $X$ as the quotient space $$X= Y\times [0,1]/_{Y\times 0 \stackrel{\varphi}{\sim} Y\times 1}$$ where $Y$ is two circles joined by a straight line and $\varphi$ is a homeomorphism that "turns $Y$ upside down". This is perfect to use the quotient version of Van Kampen. This version is not very famous, and the only reference I have for it is in French (but it's actually a great reference if you can understand French, there is a very comprehensive video going with it!).

I'm sure there is some alternative ways to compute $\pi_1(X)$. Anyway using the first method I found $$\pi_1(X)\simeq \langle a,b~\vert~ bab^{-2}a^{-1}b=1\rangle.$$

In my opinion, this was not obvious at all, I can add details/drawings if needed. But I feel that you should give it a try using Van Kampen now that the situation is more clear!

I hope this helps!

Edit: I added some detail changed the names of the spaces (the previous names were implicitly saying that $\Lambda$ and $T$ had dimension 2 instead of 3)

Edit 2: As smartly suggested by Kyle Miller in the comments, there is an easy way to compute $\pi_1(X)$. If $a$ and $b$ are two generators of the torus such that $X$ is obtained by identifying the boundary of the mobius band with $a$, and if $c$ is a (well chosen) generator of the fundamental group of the mobius band, the theorem of Van Kampen gives

$$\pi_1(X)\simeq \langle a,b,c~\vert~ ba=ab,~c^2=a\rangle,$$

which can be rewritten

$$\pi_1(X)\simeq \langle b,c~\vert~ bc^2=c^2b \rangle,$$

which is the same as in the other presentation.

Answer 2

This answer is Adam's second approach in more detail.

The space in question is a nice example of what is called a mapping torus. We have a surface $\Sigma$ which is a disk minus two disks (a "pair of pants"), and then we can construct the space as a quotient $$X=\Sigma\times [0,1]/((f(x),1)\sim(x,0))$$ where $f:\Sigma\to\Sigma$ is a homeomorphism from swapping the holes by dragging them around while keeping the outer boundary fixed:

Through the van Kampen theorem, one can show that $\pi_1(X)$ is $\pi_1(\Sigma)\ *_{f_*}$, which is what is called an HNN extension. Concretely, the group has the presentation $$\pi_1(X) = \langle \pi_1(\Sigma),t\mid tgt^{-1}=f_*(g)\text{ for all }g\in\pi_1(\Sigma)\rangle.$$ The $t$ generator (the stable letter) corresponds to a loop that goes around in the $[0,1]$ direction of $X$. (See Example 1B.13 in Hatcher for more about mapping tori and HNN extensions, where the two homomorphisms are $\operatorname{id}$ and $f$.)

The following illustrates the computation of $f_*$, using the fact that $\pi_1(\Sigma)=\langle a,b\rangle$ is a free group on two generators:

Therefore, \begin{align*} \pi_1(X)&=\langle a,b,t\mid tat^{-1}=b, tbt^{-1}=bab^{-1}\rangle \\ &=\langle a,t\mid t^2at^{-2} = tat^{-1}ata^{-1}t^{-1}\rangle \\ &=\langle a,t\mid tat^{-1}at^{-1}a^{-1}ta^{-1} \rangle. \end{align*}

(If $L$ is the link L4a1, then $X$ is homeomorphic to $S^3-\nu(L)$, where $\nu(L)$ is a thickened-up $L$, a tubular neighborhood. This is not the unique link whose complement is homeomorphic to $X$: basically, you can twist the holes in $\Sigma$ around any odd number of times and get the same space!)