 Solution

For the intended standard Hopf link, identify $S^3$ with the unit sphere in ${\mathbb{C}}^2$ and its two components with the coordinate circles $z_1=0$ and $z_2=0$. Outside small neighborhoods of those circles, both coordinates are nonzero, and their arguments together with $|z_1|$ giveOn this product the longitude of either component is the meridian of the other. Therefore the two attaching meridians become the two coordinate curves of $T^2$, intersecting exactly once. Adding the product collar does not change a gluing of solid tori. The meridian-intersection-one argument in Question 2(c) identifies the result with the standard genus-one splitting of the sphere. Thus zero surgery on the Hopf link gives

There is a genuine link-type qualification in the wording: merely specifying two linked but individually unknotted components does not determine the Hopf link. For example the two components of the torus link $T(2,4)$ are unknotted but have linking number $2$. Filling their preferred longitudes is zero surgery with linking matrixso the resulting first homology is its cokernel, $\mathbb{Z}/2\oplus \mathbb{Z}/2$, rather than the zero group of $S^3$. The cokernel formula follows because zero longitudes impose twice the other meridian as the homology relations. Thus the sphere answer applies to the standard once-linked Hopf interpretation; the literal unspecified link description alone is insufficient to force a single manifold.