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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 $\alpha\cup \beta$.
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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.

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  • $\begingroup$ To use van Kampen's theorem, you need $A \cup B$ to be the space $X$ whose fundamental group you want to compute, but in your case you have $A \cup B = D^2 \times I$, the whole cylinder. So you'll need to pick different $A$ and $B$. $\endgroup$ Commented Feb 10, 2016 at 5:01
  • $\begingroup$ @TakumiMurayama Right. I was stupidly thinking that $A\cup B=X$. Haha. $\endgroup$ Commented Feb 10, 2016 at 5:17
  • $\begingroup$ Is it me or is this space just the connected sum of two solid tori? $\endgroup$ Commented Feb 10, 2016 at 11:29

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You can split the space $Y=D^2\times I \setminus \alpha\cup\beta$ in the following way : Hatcher 1.2.10

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 $\pi_1(Y)$

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  • $\begingroup$ Thanks. I need to think for some time on this idea. May be a long time. $\endgroup$ Commented Feb 10, 2016 at 5:25
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If you picture the solid cylinder as a solid ball, you will be able to deform $X$ into a solid ball without two parallel chords, which deformation retracts to the figure eight.

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  • $\begingroup$ I agree. And this gives us the fundamental group of $X$. But can you also prove using this that $\gamma$ is not null-homotopic in $X$? $\endgroup$ Commented Feb 11, 2016 at 3:10
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Let $X$ denote the complement of $\alpha$ and $\beta$ in $D^{2} \times I .$ since the two arcs $\alpha, \beta$ can deformation retract to two parallel lines. We see that $X$ is homeomorphic to a disk minus two distinct points, i.e. $X \cong D^{2}-\{a, b\} .$ The loop $\gamma$ is just the boundary of the disk. And $X$ deformation retracts to wedge of two circles(8). Hence the fundamental group of $X$ is free on two generators, i.e. $\pi_{1}(X) \cong \mathbb{Z} * \mathbb{Z}$. Therefore there is no nullhomotopy of $\gamma$ in $X$

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