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So I am currently going through some lecture notes where the fundamental group of a torus is calculated by van Kampen's theorem:

The torus is decomposed into its characteristic fundamental polygon and a circle $o$ inside. Clearly, this circle has $\pi_1(o)=0$ and the intersection between the polygon and the circle is the circle.

So by van Kampen's theorem: The fundamental group of my torus is given by $\pi_1(T^2)= \frac{\pi_1(char.poly)}{N(Im \ (i))}$, where $i: \pi_1(o\cap \ char.poly)=0 \rightarrow \pi_1(char.poly)$ is the homomorphism corresponding to the characteristic embedding and $N(Im(i))$ is the normal subgroup induced by the image of this embedding(as a subgroup of $\pi_1(char.poly)$.

Now, there are two things I don't understand: It is claimed that $\pi_1(char.poly)= \pi_1(S^1 \vee S^1)$( I don't see the relationship between this fundamental polygon and 'an eight') and I don't know how to calculate this normal subgroup there. Is there anybody able to help me a little bit?

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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.

Deforming a punctured torus into a wedge of two circles

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  • $\begingroup$ thank you for your nice answer. it means that $\pi_1(U=S^1 \vee S^1) = \mathbb{Z}*\mathbb{Z}$. But how exactly do you get the eight. All I see is the polygon with no interior. What does it have to do with an eight? $\endgroup$
    – user66906
    Jun 9, 2014 at 11:01
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    $\begingroup$ As Ted said above, the "polygon with no interior" is the figure eight, since all four vertices are identified to a single point after we quotient the square to make it $P$ (i.e. $T^2$). Picture pulling the vertices (but not the entirety of the edges themselves) to one point in the center. This gives you $S^1 \vee S^1 \vee S^1 \vee S^1$. But two pairs of those loops/edges are identified already, so that space is actually the figure eight $S^1 \vee S^1$. $\endgroup$
    – Kyle
    Jun 9, 2014 at 12:23
  • $\begingroup$ ah, thanks now I understand how that was meant, but isn't it true that the fundamental polygon of the sphere also has an interior which would be missing in your contraction? $\endgroup$
    – user66906
    Jun 9, 2014 at 12:30
  • $\begingroup$ When you puncture a polygon, it deformation retracts to its edges. If there are no edges identified together, the space left over is a circle $S^1$. If the edges are identified as $A^{-1} B^{-1}AB$, the space left over is a wedge of two circles. In the case that the polygon represents a sphere, the edges become a pair of line segments, which further deform retract to a single point. This makes sense, since a punctured sphere also deform retracts to a point. $\endgroup$
    – Kyle
    Jun 9, 2014 at 14:54
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Briefly: First, you need to distinguish between a disk (which is contractible) and a circle, which has $\pi_1(S^1)\cong\Bbb Z$. When you remove an open ball from the torus (the square with identifications on its edges), what's left deformation retracts to the edges of the square. This is $S^1\vee S^1$, because of the identifications: All four vertices are the same point, and the top and bottom edges become one circle and the left and right edges become the other.

The intersection of the disk and the complement of a smaller disk deformation retracts to a circle, and when we include this circle in the larger piece, it wraps (in $\pi_1$) once around the boundary of the square, giving you the word $aba^{-1}b^{-1}$, which generates the commutator subgroup of $\pi_1(S^1\vee S^1)\cong \langle a,b\rangle$.

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  • $\begingroup$ Thank you for your answer. Sorry, why isn't $\pi_1 (S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z}$? I cannot really follow your reasoning: Especially, this all four vertices are the same point? $\endgroup$
    – user66906
    Jun 9, 2014 at 11:08
  • $\begingroup$ Sure, it is. The $\Bbb Z$'s are generated by $a$ and $b$, respectively. $\endgroup$
    – Ted Shifrin
    Jun 9, 2014 at 11:25
  • $\begingroup$ That's different from your last line isn't it? $\endgroup$
    – user66906
    Jun 9, 2014 at 12:03
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    $\begingroup$ That notation is reasonably standard for the free group generated by $a$ and $b$. If I wanted to indicate any relations, I'd write something like $\langle a,b\mid aba^{-1}b^2\rangle$. $\endgroup$
    – Ted Shifrin
    Jun 9, 2014 at 12:54

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