Let \(Y=(I\times I)/\sim\) be the quotient of the square using the equivalence relation generated by \((0,s)\sim(1,1-s)\) for all \(s\in I\) and \((0,0)\sim(0,1)\). That is, \(Y\) is a Möbius band with two distinct points on its boundary glued together.
Compute \(\pi_1(Y)\).
Describe a \(\Delta\)-complex structure on \(Y\), and use your complex to compute the simplicial homology groups of \(Y\).
Solution
Label the vertices of the square counterclockwise by \[ A=(0,0),\qquad B=(1,0),\qquad C=(1,1),\qquad D=(0,1), \] and triangulate the square by the diagonal \(AC\).
Orient the following edges: \[ a:A\to B, \qquad b:D\to C, \qquad x:B\to C, \qquad d:A\to C. \]
1. All four vertices have one common image \(v\) in \(Y\).
The Möbius identification \[ (0,s)\sim(1,1-s) \] gives \[ A\sim C, \qquad D\sim B. \] The additional identification gives \[ A\sim D. \] By transitivity, \[ A\sim B\sim C\sim D. \]
2. The left edge \(A\to D\) is identified with the right edge \(C\to B\), hence with \(x^{-1}\).
Parametrize the left edge by \[ s\longmapsto(0,s), \qquad 0\le s\le1. \] Under the defining side identification it maps to \[ s\longmapsto(1,1-s), \] which runs down the right edge from \(C\) to \(B\). Since \(x\) is oriented from \(B\) to \(C\), this is the reverse orientation \(x^{-1}\).
3. The quotient of the chosen triangulation is a \(\Delta\)-complex with one \(0\)-simplex \(v\), four \(1\)-simplices \(a,b,x,d\), and two \(2\)-simplices\[ \sigma=[A,B,C], \qquad \tau=[A,C,D]. \]
By 1 all original vertices become the single \(0\)-simplex \(v\). The bottom edge, top edge, and diagonal have distinct interiors and give the edge classes \(a,b,d\). The two vertical edges are identified with opposite orientations by 2 and give the single edge class \(x\).
The quotient relation acts only on the boundary of the square, so the interiors of the two triangles remain disjoint and inject into \(Y\). Their face maps land in the four stated edge classes. Thus the quotient characteristic maps satisfy the defining conditions of a \(\Delta\)-complex.
4. The two \(2\)-simplex attaching words are\[ a x d^{-1} \qquad\text{and}\qquad d b^{-1}x. \]
Traverse the oriented boundary of \(\sigma=[A,B,C]\): \[ A\to B\to C\to A. \] These three directed edges are respectively \[ a,\quad x,\quad d^{-1}, \] giving \(a x d^{-1}\).
For \(\tau=[A,C,D]\), traverse \[ A\to C\to D\to A. \] The first edge is \(d\). Since \(b\) is oriented \(D\to C\), the edge \(C\to D\) is \(b^{-1}\). Finally \(A\to D\) is \(x^{-1}\) by 2, so its reverse \(D\to A\) is \(x\). Thus the word is \[ d b^{-1}x. \]
5. The fundamental group has presentation\[ \pi_1(Y,v) \cong \left\langle a,b,x,d \ \middle|\ a x d^{-1}=1,\ d b^{-1}x=1 \right\rangle. \]
4. This is the cellular form of the Seifert–van Kampen theorem and gives the displayed presentation.
6. We have\[ \boxed{\pi_1(Y)\cong F_2.} \]
From the first relator in 5, \[ d=ax. \] From the second, \[ d b^{-1}x=1 \quad\Longleftrightarrow\quad b=xd. \] Thus both \(d\) and \(b\) can be eliminated by Tietze transformations, leaving no relation on \(a\) and \(x\): \[ \pi_1(Y) \cong \langle a,x\mid\ \rangle =F(a,x)\cong F_2. \] This proves part (a).
7. The simplicial chain complex of the \(\Delta\)-complex is\[ 0\longrightarrow \mathbb Z^2 \xrightarrow{\partial_2} \mathbb Z^4 \xrightarrow{0} \mathbb Z \longrightarrow0, \]where, in the ordered edge basis \((a,b,x,d)\),\[ \partial_2(\sigma)=a+x-d, \qquad \partial_2(\tau)=x+d-b. \]
3. Every edge begins and ends at \(v\), so \(\partial_1=0\).
For \(\sigma=[A,B,C]\), the simplicial boundary formula gives \[ \partial_2\sigma =[B,C]-[A,C]+[A,B] =x-d+a. \] For \(\tau=[A,C,D]\), \[ \partial_2\tau =[C,D]-[A,D]+[A,C]. \] Here \[ [C,D]=-b, \qquad [A,D]=-x \] by the chosen orientation of \(b\) and 2, so \[ \partial_2\tau=-b+x+d=x+d-b. \]
8. The homomorphism \(\partial_2:\mathbb Z^2\to\mathbb Z^4\) is injective.
Suppose \[ m(a+x-d)+n(x+d-b)=0. \] The coefficient of \(a\) is \(m\), hence \(m=0\). The coefficient of \(b\) is \(-n\), hence \(n=0\). Therefore the kernel is zero.
9. We have\[ H_2(Y)=0 \qquad\text{and}\qquad H_i(Y)=0\quad(i\ge3). \]
By 8, \[ H_2(Y)=\ker\partial_2=0. \] There are no simplices of dimension at least \(3\), so all higher simplicial chain groups and homology groups vanish.
10. We have\[ H_1(Y)\cong\mathbb Z^2. \]
Since \(\partial_1=0\), \[ H_1(Y) \cong \mathbb Z\langle a,b,x,d\rangle \big/ \mathbb Z\langle a+x-d,\ x+d-b\rangle. \] The first relation gives \[ d=a+x, \] and the second then gives \[ b=x+d=a+2x. \] Hence every class is generated by the classes of \(a\) and \(x\).
To see that there is no relation between these two generators, define \[ \psi:\mathbb Z^4\longrightarrow\mathbb Z^2 \] on the edge basis by \[ \psi(a)=(1,0), \quad \psi(x)=(0,1), \quad \psi(d)=(1,1), \quad \psi(b)=(1,2). \] Then \[ \psi(a+x-d)=0, \qquad \psi(x+d-b)=0, \] so \(\psi\) descends to a homomorphism \[ \overline\psi:H_1(Y)\to\mathbb Z^2 \] sending the classes of \(a,x\) to the standard basis. The map \[ \mathbb Z^2\to H_1(Y), \qquad (m,n)\mapsto m[a]+n[x] \] is therefore both surjective and inverse to \(\overline\psi\). Thus \(H_1(Y)\cong\mathbb Z^2\).
11. We have\[ H_0(Y)\cong\mathbb Z. \]
The \(\Delta\)-complex has one vertex and is nonempty, hence connected. Equivalently, from 7, \[ H_0(Y)=C_0(Y)/\operatorname{im}\partial_1 \cong \mathbb Z/0 \cong\mathbb Z. \]
12. Therefore the simplicial homology groups are\[ \boxed{ H_i(Y)\cong \begin{cases} \mathbb Z, & i=0,\\ \mathbb Z^2, & i=1,\\ 0, & i\ge2. \end{cases}} \]
Combine <1>9–<1>11. This proves part (b).