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I'm getting my feet wet with Seifert Fibered Spaces in Hatcher's 3-manifold papers. Elsewhere, it is said that this follows easily from the definition. I am not seeing it. I think we would need to know that $T^2$ and $K^2$ are the only surfaces which have a foliation of circles.

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Both surfaces can be described foliating over a (the yellow) circle as a base (the Zerlegung) by circles (in reds and blues). In the picture enter image description here the "leaves" (in reds and blues) meet the base in yellow. The edges of the squares are identified regarding the specified letters and the directions of them the get $T$ and $K$ respectively.

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  • $\begingroup$ This is a nice illustration of the fibration of these surfaces, thank you. Is there an argument as to why Seifert Fibered 3-manifolds cannot have other kinds of boundary components? $\endgroup$ – Prototank Mar 9 '18 at 2:46
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    $\begingroup$ The boundaries of a SFS come from the boundary components of the orbit surface which are circles, over them -fibering- gives these two kind of borders. $\endgroup$ – janmarqz Mar 9 '18 at 2:53

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