In Hatcher's notes on 3-Manifolds (available here), he constructs Seifert-fibred spaces in the following way:
Let $S$ be some surface, possibly with boundary (let's say with boundary for now). Let M' be the oriented circle bundle over $S$. Now to some (or all) of the torus boundary components of M', we can "sew" a solid torus $N$ as follows. First, call the boundary component in M' we're focusing on $T$. Let $D$ be a meridional disk in $N$ with boundary $\delta D$, and attach $\delta D$ to a simple closed curve in $T$ of slope $\frac{\alpha}{\beta}$ (here slope means when I lift this simple closed curve to the universal cover $\mathbb{R}^2$, the line has slope $\frac{\alpha}{\beta}$). I fill in the rest (a 3-ball) in an essentially unique way. After I've done all the attaching I want, I get a Seifert-fibred space $M$.
My question is: how can I get a presentation for $\pi_1(M)$ from the knowledge of $S$ and the slopes? For example, if $S$ is a disk, then M' is just a solid torus. Attaching another solid torus with slope $\frac{p}{q}$ gives the lens space $L_\frac{p}{q}$. If $t$ represents a longitudinal circle in M', then \pi_1(M')=\langle t\rangle and for $M$ we simply add the relation $t^q=1$. But things can't be that easy!
Because if $S$ is an annulus, then M' is just $S\times S^1$; now suppose we want to fill in the two boundary components of M' with two solid tori $N_1$ and $N_2$, using slopes $\frac{a}{b}$ and $\frac{e}{f}$. If $t$ represents a longitudinal circle in M' and $d$ a meridional circle, then \pi_1(M') = \langle t,d\ |\ [t,d]=1\rangle. I want to believe that I then get $\pi_1(M)=\langle t,d\ |\ [t,d]=1, d^a=t^b, d^e=t^f\rangle$, but this is not true.
So what is the appropriate way to get the fundamental group of $M$ from the fundamental group of M'? To be overly specific, where does the Euler number come in?
Thanks!