# Problem 10 of Section 1.2 from Hatcher.

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.

• 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 $B$. Commented Feb 10, 2016 at 5:01
• @TakumiMurayama Right. I was stupidly thinking that $A\cup B=X$. Haha. Commented Feb 10, 2016 at 5:17
• Is it me or is this space just the connected sum of two solid tori? Commented Feb 10, 2016 at 11:29

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 $\pi_1(Y)$
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.
• 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 $X$? Commented Feb 11, 2016 at 3:10
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 $$X$$