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I'm trying to get my head round the following calculation of the fundamental group of the torus, using Seifert Van-Kampen (I know it's easier to do this by considering covering spaces, but I'm trying to learn the Seifert Van-Kampen method).

Consider the torus $T$ as the unit square in $\mathbb R^2$ with opposite edges identified. Let $A$ be an open disc (say of radius $1/4$ about the origin), and $\overline{A}$ be it's closure. Let $B$ be complement of $A$, so that $A \cap B = S^1$. Then we have $T = A \cup B$ where $A$ and $B$ are a neighbourhood deformation retract pair, so we can use Van Kampen. It is clear that $\pi_1(A) = \{e \}$ and $\pi_1(A \cap B) = \mathbb Z$. So we need to find $\pi_1(B)$. Observe that $B$ deformation retracts onto the boundary of the square which, after identification, is homeomorphic to $S^1 \vee S^1$; in this way we see that $B$ and $S^1 \vee S^1$ are homotopy equivalent (inclusion in one direction, the retraction in another), and so $\pi(S^1 \vee S^1) = F_2$ (which can be calculated by Van Kampen, or by using the theory of covering spaces). So, by Van Kampen, $\pi_1(T) \cong F_2 *_{\mathbb Z} \{e\}$. This is the quotient of $F_2$ by the normal closure of the image of $l_* : \pi_1(A \cap B) \to \pi_1(B)$ (where $l_*$ is the homomorphism induced from the inclusion map $l : A \cap B \hookrightarrow B$). The following is causing me confusion: The generator of $\pi_1(A \cap B)$ is homotopic in $B$ to a loop of the form $u.v.\overline{u}.\overline{v}$, where $u$ and $v$ represent the two generators of $F_2$. So $\pi_1(T) \cong \langle a,b ; aba^{-1}b^{-1} \rangle$, the free abelian group on two generators.

Any explanation or geometric intuition would be really appreciated. Thanks

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2 Answers 2

The key here is that you want to analyze the image of $l_*$. The generator of $\pi_1(A\cap B)$ is one loop around the 'circle' $A \cap B$. When this generator of $\pi_1(A \cap B)$ is mapped into $\pi_1(B)$, it maps onto the boundary of the circle, and so it maps onto $u.v.\bar{u}.\bar{v}$. In other words, pushing the 'circle' that is $A \cap B$ out onto the boundary of the square gives you a certain loop in $B$, and that loop is equivalent to $u.v.\bar{u}.\bar{v}$.

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Think of the unit square in $\mathbb{R}$ with the top and bottom edges identified with $a$ pointing right, and the left and right edges identified with $b$ pointing down. Now say your generator $c$ for $\pi_1(A\cap B)$ traverses the circle formed by $A\cap B$ once in the clockwise direction. When you include this into $B$, where does it go? By your retraction it goes to the boundary, and as it goes clockwise it runs along $aba^{-1}b^{-1}$.

Geometrically, picture $A$ as a disk on the surface of the torus, and $B$ as the rest of the torus (with a little overlap to make $A\cap B$). Then picture the circle $A\cap B$ staying still on the surface of the torus as you retract $B$ to the "frame" formed by $S^1VS^1$. If you were to include this circle into the frame by stretching it out, it would run along one circle, then the other, then the first in the reverse direction, then the second in reverse direction.

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