In finding the fundamental group of the torus by using the Van Kampen theorem, I was reading this: "Let us choose the covering consisting of the punctured torus U, an open disk V that covers the puncture, and their intersection (a punctured disk).... $\pi _1 (V) =0$ and $\pi _1 (U)$is free group of two generators $\alpha ,\beta $..... Thus, $\pi _1(U\cap V)$ is infinite cyclic on one generator $\gamma$. Visualizing the punctured torus once again as its fundamental polygon, any lift of $\phi _1(\gamma )$ is a loop around the puncture (where $\phi _1:\pi _1(U\cap V)\rightarrow \pi _1 (U)$ ). Deforming this loop to the edge of the square, it can be seen that$\phi _1(\gamma )=\alpha \beta \alpha^{-1} \beta ^{-1}$ "

I need to understand the last sentence, I mean why $\phi _1(\gamma )=\alpha \beta \alpha^{-1} \beta ^{-1}$?

here is the link which i was reading from http://people.reed.edu/~jerry/332/projects/colley.pdf page 4


1 Answer 1


Here is the fundamental polygon of the torus.

enter image description here

The punctured torus simply has a point $P$ removed. It's best to think of $P$ having been removed from the gray area. Your proof claims any lift of a generator of $\pi_1(U\cap V)$ is a loop (with winding number 1) around $P$. Recalling that elements of fundamental groups are invariant under homotopies, we can retract this loop to run around the boundary. Starting at the bottom right of our diagram and traveling counterclockwise, we see this look $ABA^{-1}B^{-1}$.

  • $\begingroup$ @ Kevin Carlson thank a lot :) $\endgroup$
    – Ronald
    Commented Apr 10, 2013 at 21:39

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