Kernels on $\pi_1$ from inclusions as manifold boundary

Question

Suppose that $M$ is a compact connected 3-manifold with connected boundary $F$. Then is the kernel of the map $\pi_1(F) \to \pi_1(M)$ always nontrivial? Can anything else be said about the kernel (like its index)?

(Additionally, I am curious about what happens to the homology but maybe that should be a separate question...)

Answer 1

Here is a couple of examples when the boundary is toral.

1. Take a compact connected surface $S$ with a single boundary circle such that $S$ is not homeomorphic to the disk. Then $\pi_1(\partial S)\to \pi_1(S)$ is 1-1. (Hint: Use the fact that $\pi_1(S)$ is free.) Now, take the product $M=S\times S^1$; $\partial M=T^2$. Then it follows that $\pi_1(\partial M)\to \pi_1(M)$ is 1-1.

2. A less trivial example: A polygonal knot $K\subset S^3$ nontrivial if and only if the homomorphism $$ \pi_1(\partial E(K))\to \pi_1(E(K)) $$ is 1-1. Here $N(K)$ is a regular neighborhood of $K$ in $S^3$ and $E(K)$ the the closure of $S^3 - N(K)$. For a proof see W.Jaco's "Lectures on three-manifold topology". (The proof is based on the loop theorem.)

As for homology, the story is very different. If $M$ is compact then $rank(Ker(H_1(\partial M)\to H_1(M)))$ is half of the rank of $H_1(\partial M)$. See Hatcher's notes on 3-dimensional topology, Lemma 3.5.

Answer 2

The kernel of the map $\pi_1(F)\to\pi_1(M)$ can be trivial. For instance, let $N$ be any closed connected $3$-manifold and remove an open ball from $N$ to get a manifold $M$ with boundary $F=S^2$. Then $\pi_1(F)$ is trivial, so the map $\pi_1(F)\to\pi_1(M)$ automatically has trivial kernel.