Show that all 3 presentations of \(\pi_1(K)\) are isomorphic
Square with sides glued
Two mobius strips glues along boundary
Multiplication rule
Solution
Let \[ G_1=\langle a,b\mid bab^{-1}=a^{-1}\rangle, \qquad G_2=\langle c,d\mid c^2=d^2\rangle, \] and let \[ G_3=\ZZ\rtimes\ZZ, \qquad (m,n)(p,q)=(m+(-1)^n p,n+q). \]
1. The square model of the Klein bottle has fundamental group \(G_1\).
Its CW structure has one vertex, two oriented \(1\)-cells \(a,b\), and one \(2\)-cell attached by the Klein-bottle word \(aba^{-1}b\) (equivalently, after replacing a generator by its inverse, \(bab^{-1}a\)). Seifert–van Kampen therefore gives the presentation \(bab^{-1}=a^{-1}\).
2. The decomposition into two Möbius bands has fundamental group \(G_2\).
If \(c\) and \(d\) are the core-circle generators of the two Möbius bands, each boundary circle represents twice the corresponding core generator, hence \(c^2\) and \(d^2\). Van Kampen for the union along the common boundary circle gives \[ \pi_1(K)\cong\langle c,d\mid c^2=d^2\rangle=G_2. \]
3. Define\[ \Phi:G_1\to G_2,\qquad \Phi(a)=cd^{-1},\quad \Phi(b)=d. \]Then \(\Phi\) is a well-defined homomorphism.
In \(G_2\) the relation \(d^2=c^2\) gives \(d^{-2}=c^{-2}\). Hence \[ \Phi(bab^{-1}) =d(cd^{-1})d^{-1} =dcd^{-2} =dcc^{-2} =dc^{-1} =(cd^{-1})^{-1} =\Phi(a^{-1}). \] Thus the defining relation of \(G_1\) is respected.
4. Define\[ \Psi:G_2\to G_1,\qquad \Psi(c)=ab,\quad \Psi(d)=b. \]Then \(\Psi\) is a well-defined homomorphism.
From \(bab^{-1}=a^{-1}\) one obtains \(ba=a^{-1}b\). Therefore \[ \Psi(c^2)=(ab)^2=a(ba)b=a(a^{-1}b)b=b^2=\Psi(d^2), \] so the defining relation of \(G_2\) is respected.
5. The maps \(\Phi\) and \(\Psi\) are inverse isomorphisms, so \(G_1\cong G_2\).
On generators, \[ \Psi\Phi(a)=\Psi(cd^{-1})=(ab)b^{-1}=a, \qquad \Psi\Phi(b)=b, \] and \[ \Phi\Psi(c)=\Phi(ab)=(cd^{-1})d=c, \qquad \Phi\Psi(d)=d. \] Hence both composites are the identity.
6. Define\[ \Theta:G_1\to G_3,\qquad \Theta(a)=(1,0),\quad \Theta(b)=(0,1). \]Then \(\Theta\) is a well-defined homomorphism.
Using the multiplication in \(G_3\), \[ (0,1)(1,0)(0,-1)=(-1,0)=(1,0)^{-1}, \] so the relation \(bab^{-1}=a^{-1}\) is satisfied.
7. For all \(n,p\in\ZZ\) one has\[ b^n a^p=a^{(-1)^n p}b^n \]in \(G_1\).
The defining relation gives \(ba^p=a^{-p}b\) for every \(p\in\ZZ\). Iterating for positive \(n\) gives the formula; replacing \(n\) by \(-n\) gives the negative cases as well.
8. The map\[ \Lambda:G_3\to G_1,\qquad \Lambda(m,n)=a^m b^n, \]is a homomorphism.
By 7, \[ \Lambda(m,n)\Lambda(p,q) =a^m b^n a^p b^q =a^{m+(-1)^n p}b^{n+q} =\Lambda(m+(-1)^n p,n+q), \] which is exactly the multiplication law of \(G_3\).
9. The maps \(\Theta\) and \(\Lambda\) are inverse isomorphisms, so \(G_1\cong G_3\).
For \((m,n)\in G_3\), \[ \Theta\Lambda(m,n)=\Theta(a)^m\Theta(b)^n=(m,n). \] Conversely, \(\Lambda\Theta\) fixes the generators \(a,b\) of \(G_1\), hence is the identity.
10. Consequently\[\boxed{G_1\cong G_2\cong G_3\cong\pi_1(K).}\]
Combine <1>1–<1>2, 5, and 9.