whether gluing the all faces of finite tetrahedrons in pairs would yield a manifold? In fact ,I have asked this question in this web.But after read the book recommended to me ,I found this question still can't be solved。
According to the Theorem 10.1.1 ,Theorem 10.1.2 and the Definition of proper in Ratcliffe's Foundations of Hyperbolic Manifolds,it is hard to say that gluing the all faces of finite tetrahedrons in pairs not always be 3-manifold.Besides, I think unproper is not mean that the result of gluing is not  locally homeomorphism $R^3$.
And I have  proofed that the result of gluing the all faces of finite tetrahedrons in pairs is not boundary  and bifurcate. But I don't know whether it is equal to 
locally homeomorphism $R^3$.
Sorry for my poor English.
ps:I have proof it is not 
 A: No, gluing faces of tetrahedra in pairs need not yield a 3-manifold. Stick two tetrahedra together to get a square pyramid, i.e. with a square face and four triangular faces in cyclic order $F_1,F_2,F_3,F_4$. Glue $F_1$ to $F_3$ and $F_2$ to $F_4$ so that a slice of the pyramid just below the top vertex is glued into a torus. Then the image of the vertices near this torus has all sufficiently small neighborhoods homeomorphic to a cone on a torus, which is not a 3-disk, and so the resulting identification space is not a manifold around the image of the vertex.
EDIT: This is incomprehensible in text, so I've sketched up a picture: 
We have here a square pyramid resulting from gluing two tetrahedra along faces that now share vertices $v,w,s$ as indicated by the dotted line. To construct a non-three-manifold from this pyramid, glue the two faces labelled $A$ and the two faces labelled $B$ so as to glue the red quadrilateral into a torus. (This isn't hard to do: identify $u$ to $s$ and $t$ to $w$ for the $A$-gluing and $u$ to $w$, $t$ to $s$ for the $B$-gluing.) Then a neighborhood of $v$ extending to the red quadrilateral is glued into the cone on a torus, which is not homeomorphic to a 3-ball.
