= Solution
Each <Möbius strip> deformation retracts onto its core circle. Its boundary travels twice around that circle. Thicken the two strips slightly across their common boundary to obtain an open cover whose intersection deformation retracts onto that boundary circle. The <Seifert-van Kampen theorem> therefore gives
$$
\boxed{\pi_1(X)=\langle a,b\mid a^2=b^2\rangle.}
$$
Depending on the boundary orientation the relation may first read $a^2=b^{-2}$; replacing $b$ by its inverse gives the displayed presentation.
The map $a,b\mapsto1$ respects the relation and surjects onto the infinite <cyclic group> $\mathbb Z$, so the <fundamental group> is infinite. It also surjects onto the <infinite dihedral group>: map $a$ and $b$ to the reflections $s(x)=-x$ and $t(x)=2-x$ of the real line. Both squares are the identity, but $st(x)=x-2$ and $ts(x)=x+2$ differ. An abelian group cannot have this nonabelian quotient. \b[The <fundamental group> is infinite and nonabelian.]
Back to article page
