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.
Fundamental group of a solid torus with a tunnel
algebraic-topology
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The approach is correct, but you can use a simpler method. The space you care about is homotopic to the wedge sum of two spheres and one circle. Thus the fundamental group is Z. To see why this space is homotopic to the space I said, you just check first that you can contract the upper half-sphere, thus the lower half-sphere becomes to a sphere, and the torus becomes a sphere with its south and north poles glued together. A sphere with its south and north poles glued together is homotopic to a sphere with a line connect its south and north poles. After these deformations you have a space which is what I said above.
Perhaps we're thinking of different notions of "fundamental polygons", but I believe the torus $T^2$ is the fundamental polygon $P$ (obtained as the quotient of a square). The two spaces are certainly homeomorphic, if not the same by definition. Thus $\pi_1(T^2)\cong\pi_1(P)$. We'll decompose and apply van Kampen's theorem to $P$. For convenience, we'll call the horizontal edges $A$ and the vertical edges $B$.
We'll decompose $P$ in almost the same way you suggested: fix a point $x_0$ in the middle of $P$, and let $U$ be $P \setminus \{x_0\}$ and $V$ be a small open disk around $x_0$. Then $\pi_1(T^2) \cong \pi_1(P)\cong\big( \pi_1(U) * \pi_1(V)\big)/N$, where $N$ is the subgroup generated by those "words" in $\pi_1(U) * \pi_1(V)$ that represent loops that are actually nullhomotopic (that is, can be shrunk down to points). In particular, think about the "boundary" word $A^{-1}B^{-1}AB$. It is nontrivial in $U$, but we know that it can actually be shrunk down to a point when it lives in $P$.
The image below depicts a deformation of $U$ to the figure eight $S^1 \vee S^1$. What does this imply for $\pi_1(U)$? I'm also happy to provide more hints.
Best Answer
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.