Computing fundamental group of the mapping torus $\pi_1(T_{z\mapsto z^r}(S^1))$ I am trying to compute the fundamental group of the mapping torus of $f(z) = z^r$ for $r\in \mathbb{R}$ on the domain $S^1$. 
So, the space is $S^1\times I$ with $(z, 0)$ identified with $(z^r, 1)$.
When $r$ is prime, I think this is just $\mathbb{Z} $. When it is square free, I think it is the direct product of the number of prime factors of $r$ of $\mathbb{Z} $s. Beyond this, I am struggling. 
 A: If you've seen fundamental polygons before, one way for me to answer your question would be to say "allow repeated edges on your fundamental polygon!" and claim that the answer is $\langle a, b | aba^{-r}b^{-1}\rangle$ (which it is).
One way to see that this relation must be true is to take the loop $a = S^1 \times \{0\}$, drag it around the "torus" (that is, conjugate it by $b$) towards $S^1 \times \{1\}$, and observe that you get $a^r$.  Perhaps slightly harder is to convince yourself that no other generators occur.
Another way to think about this is via the universal cover, which is very strangely built up.  Let's fix $r = 2$.  We can think about each cell in the universal cover as a square, just as for the torus.  Let's have the square have $a$ along the left, $b$ along the bottom, $a^{-2}$ along the right, and $b^{-1}$ along the top.
Glue the squares into columns; each square in the column has the same size.  Now shrink the column in half vertically, and glue it to the right edge of the previous column.  All the edges line up, so keep on doing this to the left (and to the right).  We get a simply connected space, which is thus the universal cover, but we've got this bizarre five-sided fundamental polygon that gives us the relation $aba^{-2}b^{-1}$.
Hopefully this helps!

Postscript: I think you mean $r \in \mathbb Z$, or even $r \in \mathbb N$.  If $r$ is irrational I'm not even sure that the space you get is semilocally simply connected, which is kinda convenient for defining the fundamental group.
