Is this space $S^3$?

Question

I learned here that if we glue two copies of the solid tori together along their boundary, we get $S^3$. What happens if the gluing map is more complicated? In particular, recall that each $A \in \mathrm{GL}_2(\mathbb{Z})$ gives a homeomorphism $T^2 \rightarrow T^2$. Suppose we glue two solid tori along this map, would we still get $S^3$? If not, what would we get? Can its homotopy or homology groups be computed?

Answer 1

Actually, given any $A\in GL_2(\Bbb Z)$, if $[p,q]^T$ is the first column of $A$, then the space you get using $A$ is just the lens space $L(p,q)$, whose fundamental group is $\Bbb Z_p$; that is, the space you're describing is determined by the homotopy class of the image, for any homeomorphism you're using, of the meridian of the torus, so the only case when you get $S^3$ is when $[p,q]=[0,1]$. See Hatcher's notes on $3$-manifolds, where he shows this in the section "Classification of Lens Spaces".

Edit: It is easy to show the manifold $M$ you describe is orientable; if the homeomorphism reverses orientation take two copies of the solid torus with opposite orientations. Then by Poincaré's Duality $H_2(M)\cong H^1(M),$ however as $H_1(M)\cong \Bbb Z_p$, we get by the universal coefficient theorem that $H^1(M)\cong 0$, and thus $H_2(M)\cong 0$. Finally as $M$ is a closed orientable $3$-manifold we get $H_3(M)\cong\Bbb Z$, and $H_i(M)\cong 0$ for all $i>3$