The fundamental group of a $3$-manifold with a boundary of genus $>0$ 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.
 A: 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 :))
A: *

*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 $b_1(M)\ne 0$. 

*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. 
