Let \(D^2\) be the closed unit disk in \(\mathbb R^2\) and let \(\sim\) be the smallest equivalence relation on \(D^2\) satisfying \((\cos\theta,\sin\theta)\sim(\cos(\theta+2\pi/3),\sin(\theta+2\pi/3))\) and \((\cos\theta,\sin\theta)\sim(\cos(-\theta),\sin(-\theta))\) for all \(\theta\in[0,2\pi/3]\). Let \(X:=D^2/\sim\) be the corresponding quotient space.
Compute a presentation for \(\pi_1(X)\).
Is \(X\) homotopy equivalent to \(S^1\)? (Prove your answer.)
Solution
Put \[ v_0=(1,0), \qquad v_1=\left(\cos\frac{2\pi}{3},\sin\frac{2\pi}{3}\right), \qquad v_2=\left(\cos\frac{4\pi}{3},\sin\frac{4\pi}{3}\right), \] and divide the boundary circle into the three counterclockwise arcs \[ e_0:[0,2\pi/3], \qquad e_1:[2\pi/3,4\pi/3], \qquad e_2:[4\pi/3,2\pi]. \]
1. In the quotient, the three boundary vertices \(v_0,v_1,v_2\) have one common image, and the interiors of \(e_0,e_1,e_2\) have one common image \(1\)-cell, which we orient by the counterclockwise orientation of \(e_0\) and call \(a\).
For \(0\le\theta\le2\pi/3\), the first generating relation identifies the point of \(e_0\) with parameter \(\theta\) with the point of \(e_1\) having parameter \(\theta+2\pi/3\). Thus it identifies \(e_0\) with \(e_1\) preserving their counterclockwise orientations.
The second generating relation identifies the same point of \(e_0\) with the point of \(e_2\) having angle \(-\theta\) modulo \(2\pi\), namely angle \(2\pi-\theta\). As \(\theta\) increases from \(0\) to \(2\pi/3\), the angle \(2\pi-\theta\) decreases from \(2\pi\) to \(4\pi/3\). Thus this identification reverses the counterclockwise orientation of \(e_2\).
At \(\theta=0\), the first relation gives \(v_0\sim v_1\). At \(\theta=2\pi/3\), it gives \(v_1\sim v_2\). Hence all three vertices have the same image.
For an interior parameter \(0<\theta<2\pi/3\), the only generating pairs involving the corresponding point of \(e_0\) identify it with the points at angles \(\theta+2\pi/3\) and \(2\pi-\theta\). Conversely, those two points occur only in these same generating pairs. Thus no two distinct interior points of \(e_0\) are identified with one another, and the three open arcs descend to one open \(1\)-cell \(a\).
2. The quotient \(X\) has a CW structure with one \(0\)-cell, one loop \(1\)-cell \(a\), and one \(2\)-cell whose attaching word is\[ aa a^{-1}=a. \]
The quotient relation acts only on the boundary of \(D^2\), so the open disk is unchanged and descends homeomorphically to one open \(2\)-cell. By 1, the quotient of the boundary is the single loop \(a\).
Traverse \(\partial D^2\) once counterclockwise. Along \(e_0\) the quotient map traverses \(a\) positively. Along \(e_1\) it again traverses \(a\) positively, because the \(e_0\)–\(e_1\) identification preserves orientation. Along \(e_2\) it traverses \(a\) negatively, because the \(e_0\)–\(e_2\) identification reverses orientation. Therefore the attaching loop represents \[ aa a^{-1}=a. \]
3. We have the presentation\[ \pi_1(X)\cong\langle a\mid a=1\rangle\cong 1. \]
The \(1\)-skeleton is a circle, so its fundamental group is the free group on the generator \(a\). Attaching the unique \(2\)-cell quotients this free group by the normal closure of its attaching word. By 2 that word is \(a\), so \[ \pi_1(X) \cong \langle a\mid a=1\rangle, \] the trivial group. This proves part (a).
4. The space \(X\) is not homotopy equivalent to \(S^1\).
Homotopy equivalent path-connected spaces have isomorphic fundamental groups. By 3, \[ \pi_1(X)=1, \] whereas \[ \pi_1(S^1)\cong\mathbb Z. \] These groups are not isomorphic. Therefore \(X\) is not homotopy equivalent to \(S^1\).