In his famous paper

Stallings, John, On fibering certain 3-manifolds. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1961) pp. 95–100 Prentice-Hall, Englewood Cliffs, N.J.

Stallings proves a theorem that (roughly stated, I'm ignoring some hypotheses) says that if $M$ is a 3-manifold, then every short exact sequence $$1 \longrightarrow G \longrightarrow \pi_1(M) \longrightarrow \mathbb{Z} \longrightarrow 1$$ with $G$ finitely generated comes from a fiber bundle $M \rightarrow S^1$. In particular, $G$ is a surface group.

This is an oft-quoted theorem, but I have trouble reading the paper. Are there any expository accounts of it anywhere?

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    $\begingroup$ You may find this recent MO question about Stallings’ paper useful. mathoverflow.net/q/377412/1463 $\endgroup$
    – HJRW
    Oct 1, 2021 at 7:19
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    $\begingroup$ Good question. I've spent time today looking, and I have failed to find an exposition. I'm sure that I've seen one (perhaps a master's thesis?) at some point... In any case, you could perhaps say where you are stuck, and we could (try to) help? $\endgroup$
    – Sam Nead
    Oct 1, 2021 at 21:04
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    $\begingroup$ @HJRW: Thanks! That clears up some of the confusing parts, and indeed I think I now know how to prove that $G$ is a surface group carried by a surface in $M$ (the first half of the paper). If no one finds an exposition, I'll try to isolate a specific question. $\endgroup$
    – Laura
    Oct 1, 2021 at 21:05

1 Answer 1


In the book knots by Gerhard Burde, Heiner Zieschang, Michael Heusener they give a proof for the special case of a knot exterior (Theorem 5.1). It becomes a bit easier, because if M is not closed, then G is a free-group.

Be careful and don't stumble on illegal copies of the book e.g. https://www.maths.ed.ac.uk/~v1ranick/papers/burdzies.pdf

It is not directly an exposition but perhaps it clarifies a missing part for you.


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