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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?

AG learner
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    You're collapsing the boundary to a single point? If $g \neq 0$, that's not a 3-manifold. It has a point with a neighborhood homeomorphic to the come $C(\partial M)$. The cone point is not a manifold point unless $\partial M$ is a sphere. –  Jan 10 '16 at 17:49

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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 $H_n(X)=H_n(M,\partial M)=H^0(M)=\Bbb Z$.

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  • could you please explain this line " the local homology of the cone point is the homology of $\Sigma N$, so $N$ must have the same homology as a sphere" – Anubhav Mukherjee Jan 10 '16 at 18:33
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    @Anubhav.K: Let $p$ be the cone point. I mean that $H_(CN,CN-p)$. If this was a manifold of dimension $n$, you check by excision that the homology of this is $H_(S^n)$. But by excision you actually calculate $H_(CN,CN-p)=H_(CN,N)=H_(CN/N)=H_(\Sigma N) = H_{*+1}(N)$. All homology here is taken to be reduced. –  Jan 10 '16 at 18:36
  • In order to use Whitehead theorem, you need a map which induced isomorphism in homology groups, in this case what is the the map?? – Anubhav Mukherjee Jan 10 '16 at 18:48
  • @Anubhav.K: Pick a small ball in $N$ and collapse the complement to get a map $N \to S^n$ of degree 1. –  Jan 10 '16 at 18:49
  • yes, degree 1 map will work – Anubhav Mukherjee Jan 10 '16 at 18:49
  • @MikeMiller Interesting. Thanks for nice answer and valuable reference, but with regard to the computation of local homolgy in your comment, I find some result different: by considering the pair $(CN,CN-p)$, one has the long exact sequence $0=H_(CN)\to H_(CN,CN-p)\to H_{-1}(CN-p)\to H_{-1}(CN)=0$ All homology are taking relative homology. Therefore one has the isomorphism $H_(CN,CN-p)\cong H_{-1}(CN-p)\cong H_{-1}(N)$ But notice that the subscript is $-1$ compared to your result having subscript as $*+1$. I see you are using the triple $(N,CN-p,CN)$ but why we get different result? – AG learner Jan 11 '16 at 04:54
  • @Yilong: I was going off memory. I agree with your calculation. –  Jan 11 '16 at 07:55