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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?

goblin GONE
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    Note that by identifying two copies of the solid torus along their boundary, one gets $S^1\times S^2$ rather than $S^3$. So the result is certainly not always the sphere. – Mikhail Katz May 28 '18 at 09:28
  • @MikhailKatz, I see. But doesn't Nsunuk's answer glue across the identity? – goblin GONE May 28 '18 at 10:55
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    These are lens spaces. It is a useful exercise to compute the homology with Mayer-Vietoris. – Thomas Rot May 30 '18 at 06:29
  • In fact every compact, orientable, connected three manifold can be written as the union of two handlebodies, glued along their boundary. This is called a Heegaard splitting. – Cheerful Parsnip May 31 '18 at 00:13

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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 $i>3$

Camilo Arosemena Serrato
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