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Let $n=(0,0,1)$ and use the open coverThe cap $V$ is open in $Y$: it equals $Y\cap \{z>1/2\}$ because the attached chord has $z=0$. The cap is homeomorphic to an open disk, and $U\cap V=V\setminus \{n\}$ is a punctured open disk, hence homotopy equivalent to a circle. Both sets and their intersection are path-connected. Choose the base point $b$ in that intersection.

We first calculate ${\pi }_1(U,b)$. A stereographic projection identifies $S^2\setminus \{n\}$ with ${\mathbb{R}}^2$ and sends the two chord endpoints to $(-1,0),(1,0)$. Thus $U$ becomes the Euclidean plane with an extra closed interval attached at those two points. This description is a homeomorphism: gluing the two closed pieces along their endpoints gives exactly the original subspace topology.

The Euclidean plane has a deformation retraction onto $L=[-1,1]\times \{0\}$ while fixing its endpoints. An explicit homotopy isExtend it by the identity map on the extra chord. The pasting lemma for closed subspaces gives a deformation retraction of $U$ onto two distinct closed intervals sharing just their endpoints, a circle. Consequently ${\pi }_1(U,b)\cong \mathbb{Z}$, with the routine base point transport if the retraction moves $b$.

The inclusion-induced group homomorphism ${\pi }_1(U\cap V,b)\to {\pi }_1(U,b)$ is trivial: it factors through $S^2\setminus \{n\}\cong {\mathbb{R}}^2$, which is contractible. Also ${\pi }_1(V,b)=1$. All hypotheses of the Seifert-van Kampen theorem are now checked, and it gives the fundamental group of a sphere with an attached chord:where both maps from the amalgamating infinite cyclic group are trivial. A group generator follows the chord from one endpoint to the other and returns along any path on the sphere, with a connecting path from the chosen base point.