See page 53 of Hatcher's Algebraic topology for reference to image. Consider two arcs $\alpha$ and $\beta$ embedded in $D^2 \times I$ as shown in the figure. The loop $\gamma$ is obviously nullhomotopic in $D^2 \times I$, but show that there is no nullhomotopy of $\gamma$ in the complement of $\alpha \cup \beta$.
My reasoning is to consider the fundamental group of the space $(D^2 \times I) – (\alpha \cup \beta)$. To calculate the fundamental group of this space, we let $X = D^2 \times I, A = X – S^1$ and $B = X – S^1$ in such a way that $A \cap B = X – (\alpha \cup \beta)$. So by van Kampen's theorem, we have an isomorphism $$\frac{\pi_1(A) \ast \pi_1(B)}{N} \cong \pi_1(X).$$ This is given by $$\frac{\mathbb{Z} \ast \mathbb{Z}}{N} \cong 0$$ This obviously does not work and I'm not sure of how to proceed.
Best Answer
I'll add the figure from Hatcher for clarity.
The key here is to note that the knot can be undone by moving one edge of each arc from one side of the cylinder across to the other side. I actually figured this out playing with two strings.
Now, you have a full cylinder minus two straight lines. This has a deformation retract to a disk with two holes, which is in turn homeomorphic to a wedge of $2$ circles - so it has the fundamental group $F_2=\langle a,b\rangle$.
Now, throughout this process you need to keep track of where $\gamma$ is. At this point, it encompasses both holes. So after the retract to a wedge of $2$ circles, $\gamma$ is represented by $a\circ b$. In $F_2$, one has $a\circ b\neq e$. So $\gamma$ is not null-homotopic.
[I would like to thank @ChesterX and @Juggler for pointing out an error in my answer, hence the edit.]