$\pi_1(M)=\langle a,b|b^{-1} b^{-1},a \rangle \cong \Bbb Z?$ Why?
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2026-03-25 14:22:56.1774448576
$\pi_1(M)=\langle a,b|b^{-1} b^{-1},a \rangle \cong \Bbb Z?$
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No. It's $\Bbb Z_2$, since $a=e$.