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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!

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    Have you tried playing around with the homotopy long exact sequence?2011-09-19
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    In the tag merging and synonyms thread on the meta there was a decision that [3-manifolds] should be merged into [low-dimensional-topology]. No reason to use both tags on the same question.2011-09-19
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    @AsafKaragila: Ah, thank you! I was wondering why I couldn't find it :)2011-09-19
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    You use the Seifert - VanKampen theorem. It was designed precisely for situations like this.2011-09-20
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    @RyanBudney: I would like to do that! But for example, look at the annulus example I gave in the bottom of my post: there I used van Kampen's theorem, thinking that attaching this 2-disk is simply adding a relation, and the rest of the 3-ball doesn't affect the group. However, the presentation I end up with is not correct. So my question is: what went wrong?2011-09-20
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    I don't understand, what is going wrong? You claim something is not true in your computation but I don't see why. If $S$ is an annulus you still get a lens space (or $S^1 \times S^2$), and that's what you're getting.2011-09-20
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    @RyanBudney: Suppose I have this annulus cross $S^1$, and I am going to attach two solid tori to its two boundary components, let's say with slopes $2/1$ and $3/1$. Then the presentation I get for $\pi_1$ is $\langle t,d\ |\ d^2=t, d^3=t\rangle$ which is the cyclic group of order 2. However this is the same space (via a fiber-preserving diffeo.) as if I had attached with slopes $0/1$ and $5/1$, which gives me $\langle t,d\ |\ t=1, d^5=1\rangle$, a cyclic group of order 5.2011-09-20
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    Oh, it looks to me like you're losing track of a convention for how to label your generators. I suspect your relation $d^3=t$ should be $d^{-3}=t$. I've made this mistake before, thinking of an annulus as either $S^1 \times [0,1]$ vs. thinking of it as a twice punctured sphere.2011-09-20
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    @RyanBudney: ah, yes, you're right! If I think of this space before the attaching as a having an "inside" and "outside" toral boundary, then one should have generator $d$ and one $d^{-1}$. Much appreciated.2011-09-20

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