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If $\gamma \cdot u$ uses the ordered-tangent orientation convention for algebraic intersection number of curves on an oriented surface, a right-handed Dehn twist takes $u$ to $u+([\gamma ]\cdot u)[\gamma ]$ on first homology. In particular a separating-curve twist acts trivially there.

The exterior of a link is the complement of the interior of a closed tubular neighborhood of the entire link. Its boundary consists of one torus per component; its first homology has the component meridians as a free basis.

Take two solid tori $V_1,V_2$, each with a meridian-longitude basis $({\mu }_i,{\lambda }_i)$ on its boundary. Use opposite boundary-basis orientations on the two sides, so that a determinant-one coefficient matrix describes an orientation-reversing gluing map. For $A\in SL(2,\mathbb{Z})$ choose a torus homeomorphism $f:\partial V_2\to \partial V_1$ inducingand define $M_A=V_1{\cup }_f{V}_2$. The linear map on ${\mathbb{R}}^2/{\mathbb{Z}}^2$ realizes the matrix. Torus homeomorphisms with the same action on first homology are isotopic, and an isotopy extends over a boundary collar, so the resulting three-manifold depends only on $A$. Specifying opposite boundary conventions is necessary: gluing two identically oriented boundary bases by a determinant-one map would not describe the usual oriented gluing.

We now identify this gluing directly in the quotient from part (a). Split $S^3$ into the two invariant solid tori $W_1=\{|z_2|\le 1/\sqrt{2}\}$ and $W_2=\{|z_1|\le 1/\sqrt{2}\}$. On $W_1$ write $z_1=\sqrt{1-|w{|}^2}{e}^{2\pi ix}$, $z_2=w$. The quotient coordinatesidentify $W_1/⟨g⟩$ with $S^1\times D^2$. They are invariant under $g$; choosing a $p$th root of $u$ reconstructs $(x,w)$, and different choices are exactly the same cyclic orbit. Thus this is a genuine solid-torus coordinate model.

On the common boundary, measure arguments in turns as $(x,y)$. Its quotient lattice isThe meridian and longitude of the first quotient torus can be chosen as ${\mu }_1=(0,1)$ and ${\lambda }_1=(1/p,q/p)$. They form a lattice basis, since $(1,0)=p{\lambda }_1-q{\mu }_1$. The meridian of the second torus is precisely that $(1,0)$ loop. Consequently

The determinant condition is $qr+ps=-1$. Put $k=-r$, so $kq\equiv 1(\mathrm{mod}p)$. On $W_2$, the analogous quotient coordinates use $e^{2\pi ipy}$ and $e^{-2\pi iky}{z}_1$, and a longitude is $(k/p,1/p)$. Choose its negative for the second boundary-basis convention. Thensince $ps+qr=-1$. Thus the two quotient tori are glued by exactly $A$. This proves the genus-one gluing model of a lens space andDifferent choices of $r,s$ satisfying the determinant condition differ by $(s,r)\mapsto (s,r)+j(-q,p)$. This changes $f$ by a meridional Dehn twist on $V_2$, which extends over $V_2$: in solid-torus coordinates it is $(z,e^{i\theta })\mapsto (e^{ij\theta }z,e^{i\theta })$. Therefore there is no hidden dependence on that choice of longitude.

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.

Fix the algebraic intersection number of curves on an oriented surface by declaring $u\cdot v=+1$ when the ordered tangent pair $(u,v)$ agrees with the surface orientation. Orient an annular neighbourhood of $\gamma$ with coordinates $(\theta ,r)$ and orientation $d\theta \wedge dr$. A right-handed Dehn twist is represented there by $(\theta ,r)\mapsto (\theta +2\pi f(r),r)$, where $f$ increases from zero to one and is constant near the two boundary circles; it is the identity outside this annulus. A transverse arc gains one oriented copy of $\gamma$ per signed crossing. Consequently its homology action of a Dehn twist isReversing the orientation of $\gamma$ changes both factors' signs and leaves this expression unchanged. A separating curve has $[\gamma ]=0$, so its Dehn twist acts trivially on first homology.

The first assertion is true. Choose a basis $(a,b)$ for the first homology group of the torus with $a\cdot b=1$. The positive Dehn twists about these curves have matricesThe mapping class group of the oriented closed torus is $S{L}_2(\mathbb{Z})$, generated by $A,B$ and their inverses; the Euclidean algorithm on a primitive column gives this generation. The displayed relation rewrites those inverses as positive words:Replacing every inverse in a generating word proves the assertion. The closed torus hypothesis matters: a boundary twist is retained when a boundary circle must be fixed pointwise.

Give the three Hopf fibration components coherent orientations, so their pairwise linking numbers are one. In the link exterior, their longitudes satisfy $[{\lambda }_i]={\sum }_{j\ne i}{\mu }_j$. Filling along $p_i{\mu }_i+q_i{\lambda }_i$ imposes that relation on first homology. For the three rational coefficients the presentation matrix isIts determinant is $30$, and the greatest common divisor of its two-by-two minors is one (for example, $-5$ and $-3$ occur). Its Smith normal form is $\mathrm{diag}(1,1,30)$, soA nonzero rational coefficient with numerator one does not by itself make a multi-component surgery an integral homology sphere: the nonzero linking numbers must be included.

For the integer filling, use the Seifert fibered space structure $P\times S^1$. Its central regular fiber is $h=abc$, and each preferred longitude is $h{\mu }_i^{-1}$. The three filling relations are consequentlySubstituting $b=h$ into $abc=h$ gives $c=a^{-1}$. The remaining equations say $a^2=a^4$, hence $a^2=1$ and $h=1$. Thus its fundamental group is cyclic of order two.

Geometrically, a filling coefficient $r$ attaches a Seifert fibered space solid torus with multiplicity $|r-1|$, the distance of $r\mu +\lambda$ from the regular fiber $\mu +\lambda$. The multiplicities are $2,1,4$; the middle filling creates no exceptional fiber. The result is a Seifert fibered space over the sphere with at most two exceptional fibers, hence a union of two solid tori, or a lens space. A lens space with fundamental group of order two is $L(2,1)$. ThereforeAs a separate arithmetic check, the integer surgery linking matrix has determinant $-2$, in agreement with its first homology of order two.

On a genus-two surface, let $\alpha$ be a separating simple closed curve that cuts the surface into two once-punctured tori. Choose a simple closed curve $\beta$ that passes from one side to the other and back, with the two crossings arranged in minimal position. The crossings have opposite signs, sobut the bigon criterion shows that they cannot be removed and henceEquivalently, $\alpha$ is separating and therefore represents zero in first homology, forcing its algebraic intersection with every curve to vanish even though its geometric intersection need not vanish.