Compute the fundamental group

Question

Let X be the quotient of $S^1 \times [0,1]$ by the identification $(x,y) \sim (e^{2\pi i/3}x,y), y \in \{0,1\}$. Isn't it just a rotation of the cylinder's top and bottom by the same degree? Shouldn't the fundamental group just be the fundamental group of the cylinder?

Edit: It shouldn't be the fundamental group of the cylinder since it is not the rotation. It identifies every three points on the top and bottom circle of the cylinder.

For every point in the circle, you are gluing two other points to it, and leaving the interval alone, if I'm understanding this correctly. — A. Thomas Yerger, Mar 23 '16 at 05:02
Yes, do you know how to calculate its fundamental group? @AlfredYerger — Keith, Mar 23 '16 at 14:04
Fundamental group of a product is product of the groups, which is just Z. Now you're taking a quotient. Based on the equivalence relation, I suspect it's the group of order 3, but I don't have a proof immediately. — A. Thomas Yerger, Mar 23 '16 at 14:17

score 3 · Accepted Answer · answered Mar 23 '16 at 15:32

3

Use Seifert-van Kampen on the pair of open sets $U = S^1\times[0,1)/{\sim}$ and $V = S^1\times(0,1]/{\sim}$ which each deformation retract onto a circle, but whose intersection has a generator of $\pi_1$ that gets mapped to $3$ times a generator under the inclusion into either of these subspaces. I leave the details to you.

answered Mar 23 '16 at 15:32

Dan Rust

30,108

1

so we have $\pi_1(X)=\langle a,b|a^3=b^3\rangle\ne\mathbb Z$. – Andrey Ryabichev Mar 23 '16 at 15:39
Can you explain why does each $U$ and $V$ retract to the circle? They are identified, shouldn't them be different from the circle? – Keith Mar 25 '16 at 02:22

Compute the fundamental group

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