Suppose that $K$ is a simple closed curve in $M^3$. I'll assume that $M$ is orientable, compact, and without boundary. Let $V$ be a closed regular neighborhood of $K$; so $V \cong S^1 \times D^2$ is a solid torus. Let $X$ be the closure of $M - V$; so $X$ is the *exterior* of $K$. Let $T = X \cap V$; so $T$ is a two-torus. So $\partial X = \partial V = T$ and $M = X \cup_T V$. Note that $T$ is a two-torus. Let $D \subset V$ be a meridian disk; that is, a disk of the form $\lbrace \mbox{pt} \rbrace \times D^2$.

As Igor indicates, the map $\pi_1(T) \to \pi_1(X)$ induced by inclusion has a kernel if and only if there is a embedded disk in $E \subset X$ with boundary on $T$. If $\partial D$ and $\partial E$ meet once then $K$ bounds a disk in $M$.

To recap: the knot $K$ bounds an embedded disk in $M$ if and only if

- the map from $\pi_1(T) \to \pi_1(X)$ has kernel
*and*
- the curve that dies ($\partial E$) meets the meridian $(\partial D$) exactly once.