Exercise 1.2.10

Proof. Recall that we have the following diagram for $X$:

where $A_1,A_2$ describe the range of the two open sets we use for our decomposition for van Kampen’s theorem. Omitting the cylinder for clarity, we get that the three sets $A_1,A_1\cap A_2,A_2$ are represented by the “missing” segments as follows:

Note that the identifications given by the line segments for $A_1\cap A_2$ in the center follow from following the path of the loops in our original picture, and that we get inverses because any loop around, say, $b_1$ is homotopy equivalent to one around $b_1^{-1}$ looping the other way, just by moving the loop around the horseshoe-shaped segment labeled $b_1,b_1^{-1}$. Since we can deformation retract each of $A_1$, $A_1\cap A_2$, and $A_2$ to circles going around each of the labeled segments, we see that the loops around the segments $a_i,b_i,c_i$ act as our generators for the fundamental groups ${\pi }_1(A_1),{\pi }_1(A_1\cap A_2),{\pi }_1(A_2)$ (note that we omit base points since all our sets are path connected, and so the fundamental groups are not dependent on the choice of base point). Now we see that

and so by van Kampen’s theorem, we have that

Now we consider the loop $\gamma$ in our original picture, which is in the region $A_1\cap A_2$. Drawn looking at a vertical cross-section of our original cylinder, we see that the $\gamma$ loops around the segments of $A_1\cap A_2$ in the following fashion:

Choosing our base point $x_0$ to be the center of this circle, we then have that $\gamma$ is homotopy equivalent to the loop $a_1{b}_1{c}_1{b}_1^{-1}$ in the diagram below:

Since $a_1{b}_1{c}_1{b}_1^{-1}=a_1{a}_2{c}_1{a}_2^{-1}\ne 1$ in ${\pi }_1(X)$, $\gamma$ is not represented by $1\in {\pi }_1(X)$, and so is not nulhomotopic in $X$. □