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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.

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  • $\begingroup$ Whilst I know what you mean by $A=X-S^1$, note that you are dealing with arcs, not loops, and that you should specify the embedding (otherwise one may think $A=B$). As for your question, I suggest you try to choose $A$ and $B$ accordingly to compute the fundamental group of $X=(D^2\times I)-(\alpha \cup \beta)$ rather than just $X=D^2 \times I$, the latter of which is contractible anyway. $\endgroup$ – fixedp Nov 5 '16 at 10:10
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I'll add the figure from Hatcher for clarity.

enter image description here

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 is homeomorphic to a full ball ($D^3$) minus two lines, which in turn has a deformation retract to its boundary: a sphere with 4 holes.

Since the sphere is a one-point compactization of the plane, this is homeomorphic to $\Bbb{R}^2$ with 3 holes (another approach: a stereographic projection of the sphere through one of the holes).

$\Bbb{R}^2$ with 3 holes has a deformation retract to a wedge of $3$ circles, so it has the fundamental group $F_3=\langle a,b,c\rangle$.

Now, throughout this process you need to keep track of where $\gamma$ is. At this point, it is a line seperating 2 holes on the plane from the other hole. So after the retract to a wedge of $3$ circles, $\gamma$ is represented by $a\circ b$. In $F_3$, one has $a\circ b\neq e$. So $\gamma$ is not null-homotopic.

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  • $\begingroup$ $\mathbb{D}^3$ minus two non-intersecting lines deformation retracts onto a disc $\mathbb{D}^2$ minus two points, which is homotopy equivalent to wedge of $2$ circles. $\endgroup$ – ChesterX Jan 30 at 5:24

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