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Let $M$ be an orientable $3$-manifold with connected boundary $\Sigma_g$, a surface of genus $g>0$.

I would like to find a reference to the following two statements.

1) $\pi_1(M)\ne 0$.

2) $\pi_1(M)\ne \pi_1(\Sigma_g)$.

2)' If 2) is too hard I would be happy just to know that the map $\pi_1(\Sigma_g)\to M$ is not an isomorphism.

I think I can prove the first statement by contradiction. If $\pi_1(M)$ were trivial it would stay so after gluing a handlebody to $\Sigma_g$. In the resulting simply-connected manifold one can chose loops inside $M$ that have non-zero linking number with loops inside the handlebody hence they are not null-homologous in $M$. Hence we get a contradiction with $\pi_1(M)=0$.

But I don't see how to prove 2), and this might be hard. I would be grateful for some tips.

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You can see both by considering 1st homology groups instead. Thus, you will see that the map of abelianizations of fundamental groups is not an isomorphisms. If you do not like homology, look at the universal covers. If you had an isomorphism, boundary of universal cover would be a single plane, preserved by the group of deck-transformations. Then the quotient of the universal cover could not be compact. – Misha Jul 23 '13 at 12:18
Dear Misha, thank you very much for this comment! I like homology :) . I would like to ask you, if you don't mind to extend the this comment to an answer – aglearner Jul 23 '13 at 12:22
up vote 10 down vote accepted
  1. The long exact sequence of the pair $(M, \partial M)$ combined with Poincare duality immediately imply that the natural map $$ i_*: H_1(\partial M)\to H_1(M) $$ cannot be an isomorphism, unless $H_1(\partial M)=0$, which means that the boundary is a sphere. Thus, the map of fundamental groups cannot be an isomorphism either. The same argument implies that $\pi_1(M)$ cannot be trivial as $b_1(M)\ne 0$.

  2. In $M$ has compressible boundary, then either $\pi_1(M)$ is infinite cyclic or splits as a free product (use the loop theorem). In either case, you are done. Suppose there exists an isomorphism $\pi_1(\partial M)\to \pi_1(M)$ and the boundary is incompressible. This means that you have a monomorphism $j: F=\pi_1(\partial M)\to \pi_1(M)\cong \pi_1(\partial M)$, which is not an isomorphism (see 1). This is impossible unless $\partial M$ is the torus and the image of $j$ has index 2 in $\pi_1(M)$ (finite index is clear; to get index 2 use again the homology argument). Thus, the associated 2-fold cover of $M$ is $T^2\times I$ and $M$ itself is a nontrivial $I$-bundle over the Klein bottle. But, Klein bottle group is not isomorphic to $Z^2$.

Edit: An easier argument, as Ian says, is to say that once we know that the boundary is a torus and index is finite, then the map $i_*$ on $H_1$ with rational coefficients is an isomorphism, contradicting 1.

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Misha, thank you very much for writing the detailed answer! – aglearner Jul 23 '13 at 13:09
In the case that $g>1$, one may use the fact (by the sphere theorem) that $M$ is aspherical, and $\chi(M)=\frac12 \chi(\partial M)$. – Ian Agol Jul 23 '13 at 14:33
Also, any finite-index subgroup of $\mathbb{Z}^2$ will induce an isomorphism of $H_1(;\mathbb{Q})$, giving an immediate contradiction. – Ian Agol Jul 23 '13 at 14:38
@IanAgol: Right, I should have thought about it, I will modify the answer. – Misha Jul 23 '13 at 15:57

The magic words are "half lives, half dies". See Hatcher's notes, Lemma 3.5 (and read the rest of the notes, while you are at it :))

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Igor, thank you very much for the reference :) ! – aglearner Jul 23 '13 at 20:55

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