# Calculate the fundamental group of $S^1/\mathbb Z_n$

Calculate the fundamental group of $S^1/\mathbb Z_n$ ,where $\mathbb Z_n$ acts naturally on $S^1$ by rotations of $2\pi /n$

The origin of this problem is the following unclear solution of another problem:

I restate the question in picture here:

PROBLEM Compute $\pi_1 (X)$, where $X$ is the quotient space of a torus $S^1\times S^1$ obtained by identifying points on the circle $S^1 \times\{y_0\}$ that differ by rotations of the circle by $\frac{2\pi}{n}$ for $y_0$ a base point in the second $S^1$ factor and $n\ge2$ a fixed integer

My question is how to give a rigor and systematical answer, not just some intuitive or geometrical answers. What puzzled me is that in the above picture, it claims that the fundamental group of $S^1/\mathbb Z_n$ is $\mathbb Z_n$ . However, I guess $S^1/\mathbb Z_n$ seems to be homeomorphic to $S^1$; hence, the fundamental group should be $\mathbb Z$.

I think here we cannot use Van-Kampen theorem and we must back to the definition. Please help!

• Isn't $S^1/\mathbb Z_n\cong S^1$ anyway? – Hagen von Eitzen Jun 13 '15 at 13:05
• I do think so.Please see my edit. – Hang Jun 13 '15 at 13:50
• I point out that Chapter 11 of the book "Topology and Groupoids" (available from amazon) determines the fundamental groupoid of an orbit space under a properly discontinuous action, e.g. that of a finite group on a Hausdorff space. An older version of this chapter is available at arXiv:math/0212271. I can't work out more now as I am travelling tomorrow! – Ronnie Brown Jun 13 '15 at 15:06
• Where is this text excerpt from? The claim that "Since the homotopy class of $a$ serves as the sole generator and identity of $\pi_1(S^1,x_0)$, ..." is just stupid. The identity of a group is never a generator for anything except the trivial group. The identity of $\mathbb{Z}$ is $e=0$, the generator of $\mathbb{Z}$ (up to sign) is $1$. Believe it or not, $1\neq 0$. "...this implies that $q_*([a]^n)=e$ in $\pi_1(S^1/\mathbb{Z}^n)$" is then also patently false. – Dan Rust Jun 15 '15 at 14:17
• What the author has instead shown is that $q_*([a])=[a]^n$ (obviously this only makes sense after one identifies $\pi_1(S^1)$ with $\pi_1(S^1/\mathbb{Z}_n)$ via an isomorphism). Not to mention that $\mathbb{Z}^n$ should be $\mathbb{Z}_n$, unless there is some important context missing. There are several other typos, such as claiming $q_*\colon S^1\to S^1/\mathbb{Z}^n$ when really this should be $q$, and claiming that a map $S^1\vee S^1\twoheadrightarrow S^1$ somehow induces a map on fundamental groups in the other direction? – Dan Rust Jun 15 '15 at 14:18