Each figure in Problem 4 of the source sheet depicts a space made by gluing bands, each homeomorphic to \(I\times I\), to discs. For each space, compute its fundamental group and the fundamental group after a disc is glued to each boundary circle.
The two band-and-disc diagrams appear on page 1 of the source sheet.
Solution
1. The left-hand source figure is a disc with three untwisted bands attached in planar cyclic order. It is an orientable genus-\(0\) surface with four boundary components.
Starting from one disc (\(\chi=1\)), each band is a \(1\)-handle and decreases Euler characteristic by \(1\), so \(\chi=-2\). The cyclic order shown is planar, hence the surface has genus \(0\). Therefore \(2-b=-2\), so \(b=4\).
2. Hence before capping its boundary,\[\boxed{\pi_1\cong F_3,}\]and after attaching a disc to every boundary circle the result is \(S^2\), so\[\boxed{\pi_1=1.}\]
A genus-\(0\) surface with four boundary components deformation retracts to a wedge of three circles. Capping all four boundary components gives the sphere.
3. The right-hand source figure consists of two discs joined by three bands with the opposite cyclic ordering shown. It is an orientable genus-\(1\) surface with one boundary component.
Here \(\chi=2-3=-1\). The opposite cyclic order of the three bands is the standard thickening of a theta graph to a once-punctured torus rather than a pair of pants. Thus \(2-2g-b=-1\) with \(b=1\), giving \(g=1\).
4. Consequently before capping,\[\boxed{\pi_1\cong F_2,}\]and after attaching a disc to its boundary circle one obtains \(T^2\), so\[\boxed{\pi_1\cong\mathbb Z^2.}\]
A once-punctured torus deformation retracts to a wedge of two circles. Capping its unique boundary component gives the closed torus.