Problem. 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 $\alpha\cup \beta$.
I tried to use van Kampen's theorem to find the fundamental group of $X=D^2\times I-\alpha\cup \beta$. Let $A=D^2\times I-\alpha$ and $B=D^2\times I-\beta$. To find the fundamental group of $A$, we note that after a homeomorphism, $A$ looks like a cylinder with it's axis removed. The fundamental group of $A$ is thus $\mathbf Z$. Similarly for $B$. Now we need to find the normal subgroup in $\pi_1(A)\sqcup \pi_1(B)$ generated by words of the form $i_{AB}(\omega)i_{BA}(\omega)^{-1}$ and quotient by it. Here $i_{AB}:A\cap B\to A$ and $i_{BA}:A\cap B\to B$ are the inclusion maps and $\omega$ is a loop in $A\cap B$. Intuitively, the only loops in $A\cap B$ whose image in $A$ in nontrivial in $A$ are the ones which link with $\alpha$. I am not sure how to say this precisely and I will be grateful if someone can help me with this. Similarly for $B$. I am not able to make progress from here.
Best Answer
You can split the space $Y=D^2\times I \setminus \alpha\cup\beta$ in the following way :
Here $X=A\cap B$. Carefully label all the "missing lines" in $A$ and $B$, with orientation. Then try to see the inclusion maps of $X\hookrightarrow A$ and $X\hookrightarrow B$. You'll see that $\gamma$ gives a non-trivial element in $\pi_1(Y)$