TITLE: What 3-manifolds can be obtained by gluing $ S^1 \times P $ and two copies of $S^1 \times D^2$
QUESTION [5 upvotes]: Let P denote the pair of pants e.g. a sphere minus three small discs $D_1$,$D_2$,$D_3$ about marked points $x_1,x_2,x_3$. I then consider $P \times S^1$. We have boundary components $T_1$,$T_2$,$T_3$. Fix two homeomorphisms  $f_1: \partial(S^1 \times D^2) \to T_1 $ and 
$f_2: \partial(S^1 \times D^2) \to T_2$. 
What oriented three manifolds with boundary can be obtained by gluing $S^1 \times D^2$ along these maps (up to homeomorphism)?  Namely what are the possible homeomorphism types of manifolds:  
$S^1 \times D^2 \sqcup_{f_1} P \times S^1 \sqcup_{f_2} S^1 \times D^2 $

REPLY [9 votes]: You get:

 the solid torus,
 all the Seifert manifolds fibering over the disc with two exceptional fibers; 
 the connected sum $(D\times S^1) \# L(p,q)$ of a solid torus and a lens space $L(p,q)$, for all coprime $(p,q)$.
The proof goes as follows. Consider the filling meridians of the two solid tori you attach. If one meridian is attached along the fiber of $P\times S^1$, then the resulting manifold is the connected sum $(D\times S^1) \# (D \times S^1)$ of two solid tori: this is a nice exercise in 3-dimensional topology. The attaching of the second solid torus then produces $(D^2 \times S^1) \# L(p,q)$.
If both meridians are not glued along the fibers, then the fibration of $P\times S^1$ extends to a Seifert fibration over the disc with at most two exceptional fibers. If at least one fiber is not exceptional, then you actually get a solid torus.