What is the space obtained by identifying boundary $\mathbb T^2$ of a solid torus By identifying boundary of solid $\mathbb T^2$, one obtains a 3-manifold, but what is the space "looks like"? For example, can we understand it through Heegaard splitting? More generally I want to ask how to describe the 3-manifold obtained by identifying the boundary of a solid genus $g$ surface (Denote the space as $X$)?
So far I have computed its homology groups: Let $M_g$ be genus $g$ surface and let $R$ be the space bounded by $M_g$, so we have maps  $$M_g \xrightarrow{i} R\xrightarrow{\pi} X=R/M_g.$$ with the inclusion map $i: M_g \to R$ identified with the boundary, and $\pi:R\to X$ identify the boundary to a point. 
By knowing that $\tilde{H}_2 (M_g)=\mathbb Z$, $\tilde{H}_1 (M_g)=\mathbb Z^{2g}$, $\tilde{H}_0 (M_g)=\tilde{H}_3(M_g)=0$; $\tilde{H}_1 (R)=\mathbb Z^g$, and $\tilde{H}_0 (R)=\tilde{H}_1(R)=\tilde{H}_3(R)=0$, therefore one has the long exact sequence:
$$ 0\to \tilde{H}_3(X)\to \\ \mathbb Z\, \to \, 0\, \to \, \tilde{H}_2(X)\to \\  \mathbb Z^{2g}\xrightarrow{\phi}\mathbb Z^g \xrightarrow{\psi}\tilde{H}_1 (X)\to 0$$
with $\phi$ sending $g$ generators to generators and the other $g$ generators to zero. 
By this one can show that $\tilde{H}_3(X)=\mathbb Z$, $\tilde{H}_2(X)=\mathbb Z^g$, $\tilde{H}_1(X)=\tilde{H}_0(X)=0$. Actually, this computation is a part of Hatcher's section 2.2 problem 29 .
But still it is hard to imagine the space $X$, can we find further information of the space, like its Heegaard pliting?
 A: I would tend to think of this as "Attaching the cone $C(\partial M)$ to $M$" as opposed to "collapsing the boundary" - I can visualize the former far better than the latter. 
The first thing to note is that this is not actually a manifold. Actually, the cone $C(N)$, $N$ a manifold, is a manifold if and only if $N$ is a sphere. (Proof: the local homology of the cone point is the homology of $\Sigma N$, so $N$ must have the same homology as a sphere; and the arguments here show that if $N$ is dimension at least 3, then it must be simply connected; all of this plus a version of Whitehead shows $N$ is homotopy equivalent to a sphere; the topological Poincaré conjecture, now known in all dimensions, shows $N$ is homeomorphic to a sphere.)
You might be confused about the term $H_3(X)=\Bbb Z$ if it's not a manifold. Think of this in your case as an artifact of Poincare duality with boundary (sometimes known as Lefschetz duality), here manifesting in the form $H_n(X)=H_n(M,\partial M)=H^0(M)=\Bbb Z$.
