Each picture in Problem 2 of the source sheet represents a solid polygon whose boundary segments are identified, not necessarily in pairs, according to the labelled arrows. Describe the fundamental group in each case.
The three polygon diagrams appear on page 1 of the source sheet.
Solution
1. In the left pentagon, the two \(a\)-edges and the two \(b\)-edges are paired as indicated, while the fifth edge remains boundary. The quotient has one vertex, three \(1\)-cells (the classes \(a,b\) and the unpaired edge \(c\)), and one \(2\)-cell.
Following the arrows identifies all five polygon vertices. The paired edges give the two edge classes \(a,b\), and the remaining edge gives \(c\).
2. The attaching word contains the unpaired edge \(c\) exactly once, so the single relation eliminates \(c\) and leaves\[\boxed{\pi_1\cong F(a,b)\cong F_2.}\]
Van Kampen gives a presentation with generators \(a,b,c\) and one relator in which \(c^{\pm1}\) occurs exactly once. Solve that relation for \(c\); no relation on \(a,b\) remains.
3. In the middle octagon, the \(a\)-pair and \(b\)-pair identify the eight vertices into four classes. The quotient has six \(1\)-cells (the two paired classes and four unpaired boundary edges), four \(0\)-cells, and one \(2\)-cell, so its Euler characteristic is \(-1\).
Number the vertices cyclically. The arrows identify the endpoints of the two \(a\)-edges in pairs and likewise for the two \(b\)-edges, producing four vertex classes. The four diagonal sides are left unpaired.
4. This quotient is a connected compact surface with nonempty boundary, hence its fundamental group is free. Since for a connected surface with boundary \(\chi=1-r\) when \(\pi_1\cong F_r\), one gets\[\boxed{\pi_1\cong F_2.}\]
The quotient is obtained by pairwise gluing boundary intervals of a disc and leaves boundary intervals unglued, hence is a compact surface with boundary. From 3, \(-1=1-r\), so \(r=2\).
5. In the right triangle all three boundary edges are identified to one oriented edge \(a\). The attaching word is freely equal to \(a\) (up to inversion), so\[\boxed{\pi_1=1.}\]
Traversing the boundary with the arrow orientations gives a word with two occurrences agreeing and one opposing, hence \(a^{\pm1}\) after free reduction. Van Kampen gives \(\langle a\mid a\rangle=1\). This is the standard dunce-cap edge identification.