The space \(Y\) in the preceding exercise can be obtained from a disk with two holes by identifying its three boundary circles. There are only two essentially different ways of identifying the three boundary circles. Show that the other way yields a space \(Z\) with \(\pi_1(Z)\) not isomorphic to \(\pi_1(Y)\). [Abelianize the fundamental groups to show they are not isomorphic.]
Solution
1. \(Y\) has \(\pi_1(Y) = \langle a, b, c \mid aba^{-1}b^{-1}cb^\varepsilon c^{-1}\rangle\) (from the preceding exercise).
given.
2. \(H_1(Y) = \ZZ^2\).
2.1. Abelianizing the relation \(aba^{-1}b^{-1}cb^\varepsilon c^{-1} = 1\) gives \(b^\varepsilon = 1\).
in the abelianization, \(aba^{-1}b^{-1} = 1\) and \(cb^\varepsilon c^{-1} = b^\varepsilon\), so the relation becomes \(b^\varepsilon = 1\).
2.2. Hence \(b = 1\) in \(H_1(Y)\) (since \(\varepsilon = \pm 1\)).
2.1.
2.3. Therefore \(H_1(Y) = \langle a, b, c \mid b = 1\rangle = \ZZ^2\) (free abelian on \(a\) and \(c\)).
2.
3. The other identification yields \(Z\), the nonorientable surface of genus \(3\) (the connected sum of three projective planes), with \(\pi_1(Z) = \langle a, b, c \mid a^2 b^2 c^2\rangle\).
identifying the three boundary circles of a disk with two holes in the other (nonorientable, boundary-closing) way produces the closed nonorientable surface \(N_3\), whose fundamental group has the standard presentation \(\langle a,b,c \mid a^2b^2c^2 = 1\rangle\).
4. \(H_1(Z) = \ZZ^2 \oplus \ZZ/2\).
4.1. Abelianizing \(a^2 b^2 c^2 = 1\) gives \(2a + 2b + 2c = 0\), i.e. \(2(a+b+c) = 0\).
additive notation in the abelianization.
4.2. Hence \(H_1(Z) = \langle a, b, c \mid 2(a+b+c) = 0\rangle \cong \ZZ^2 \oplus \ZZ/2\).
the single relation \(2(a+b+c) = 0\) introduces one \(\ZZ/2\) torsion summand, leaving free rank \(2\).
5. \(H_1(Y) = \ZZ^2 \not\cong \ZZ^2 \oplus \ZZ/2 = H_1(Z)\).
2 and 4; one is torsion-free, the other has a \(\ZZ/2\) summand.
6. Hence \(\pi_1(Y) \not\cong \pi_1(Z)\).
isomorphic groups have isomorphic abelianizations, but 5 shows the abelianizations differ.
7. Q.E.D.
6.