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Dehn surgery removes tubular neighborhoods of a link in a three-dimensional manifold and performs Dehn filling on their boundary tori with specified primitive meridian slopes. Dehn surgery on a framed link is the integral-slope special case.

Use the meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

Take a Hopf link with its two unknot components labeled by integer surgery coefficients $2,3$.

The slam-dunk move eliminates a meridional component with coefficient $b$ and replaces the other coefficient $a$ by $a-1/b$. Eliminating the component with coefficient three givesThus this Dehn surgery on a framed link becomes $5/3$ Dehn filling on the unknot, with the required orientation. As a check, its surgery linking matrix iswhose determinant is five.

Write $V=M_{S^2}(p^{\ast }/q,\ast )$. Filling one end of $S^1$ times an annulus gives a solid torus, so $V$ is a solid torus. The prescribed algebraic intersection number of curves on an oriented surface convention gives ${\tilde{f}}_2\cdot {\tilde{h}}_2=1$. HenceIn $H_1(V;\mathbb{Z})$, the first Dehn filling imposes $p^{\ast }f-q{h}_1=0$, while $h_1+h_2=0$. ThereforeThe class ${\lambda }_2$ is primitive because $\gcd (p^{\ast },q)=1$, a consequence of $p{p}^{\ast }-q{q}^{\ast }=1$. It is therefore the disk-bounding meridian of a solid torus of $V$. Since ${\mu }_2$ meets it once, Dehn filling $V$ along ${\mu }_2$ glues two solid torus pieces with disk boundaries meeting once. This is the standard genus-one Heegaard splitting of $S^3$, soTo identify the regular fiber geometrically, put it on their common boundary torus. Invert the displayed basis change:Here ${\mu }_2$ bounds a disk in the newly attached solid torus and ${\lambda }_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary solid torus pieces. Both coefficients are positive and $\gcd (p,q)=1$, so it is the positive $(p,q)$ torus knot. This identifies the torus knot directly from the genus-one splitting, without invoking a theorem specifically about torus knots.

All classes in the nontrivial homological equalities are taken in $H_1(X_{(p,q)};\mathbb{Z})$, where $X_{(p,q)}=M_{S^2}(p^{\ast }/q,-q^{\ast }/p,\ast )$. The two internal gluing torus components still have inclusion maps into this knot exterior. Interpreting the maps instead as maps into the closed $S^3$ would make every displayed class zero.

Before the two Dehn fillings, $H_1(S^1\times \text{pair of pants};\mathbb{Z})$ has generators $f,h_1,h_2,h_3$ and relation $h_1+h_2+h_3=0$. The Dehn fillings add $p^{\ast }f-q{h}_1=0$ and $-q^{\ast }f-p{h}_2=0$. Put $m=-h_3$. Eliminating the relations using $p{p}^{\ast }-q{q}^{\ast }=1$ givesFor completeness, the three relation rows in generators $(f,h_1,h_2)$ are $(p^{\ast },-q,0)$, $(-q^{\ast },0,-p)$, $(0,1,1)$ when computing the quotient by $m$. Their determinant is $p{p}^{\ast }-q{q}^{\ast }=1$, so $m$ really generates the entire first homology group, and no finite torsion or index is hidden in the elimination.

Orient each exceptional-fiber longitude by ${\mu }_i\cdot {\lambda }_i=1$. In the boundary basis $(m_i,{\tilde{f}}_i)=(-{\tilde{h}}_i,{\tilde{f}}_i)$, write ${\mu }_i={\alpha }_i{m}_i+{\beta }_i{\tilde{f}}_i$ and ${\lambda }_i=u_i{m}_i+v_i{\tilde{f}}_i$, where ${\alpha }_i{v}_i-{\beta }_i{u}_i=1$. ThenThe filled meridian of a solid torus maps to zero, and $({\alpha }_1,{\alpha }_2)=(q,p)$, yieldingAdding multiples of ${\mu }_i$ to ${\lambda }_i$ has no effect. The word “any” therefore means any longitude with this compatible orientation; negating a longitude would negate the corresponding equality.

To compute Turaev torsion, use the general product formula $\tau (S^1\times F)\doteq (1-[f]{)}^{-\chi (F)}$ and multiplicativity under gluing along torus components. For a pair of pants, $\chi (F)=-1$, so the unfilled piece contributes $1-[f]$. The two filling solid torus pieces contribute $(1-[{\lambda }_1]{)}^{-1}$ and $(1-[{\lambda }_2]{)}^{-1}$. With $t=[m]$, the equalities above give $[f]=t^{pq}$, $[{\lambda }_1]=t^p$, and $[{\lambda }_2]=t^q$. ThereforeHere $\doteq$ allows the unit $\pm t^n$: a fully refined Turaev torsion requires an Euler structure and a homology orientation, neither of which the statement specifies. The displayed rational function is the representative whose expansion at $t=0$ starts with $1$, equivalently the standard nonnegative-exponent normalization.

Choose a fiber monodromy representative fixing the boundary of the fiber Seifert surface pointwise. The boundary of each fiber is the Seifert longitude, precisely the slope used by zero Dehn filling. Attach the filling solid torus as $D^2\times S^1$, matching $\partial D^2\times \{\theta \}$ with the boundary of the fiber at $\theta$. The fibers cap off to closed topological surfaces $\hat{\Sigma }$ of the same genus $g$; extend the fiber monodromy across the disk by the identity. ThusThis is a foliation by fibers of a bundle over $S^1$, and every leaf is compact with genus $(p-1)(q-1)/2$.

Choose a point in the interior of the capping disk, which $\hat{\varphi }$ fixes. Its suspension is an embedded closed curve $\{x\}\times S^1$ in the mapping torus. It is transverse to the foliation and meets every fiber once. Hence every leaf meets a closed transversal to a foliation, which is the defining criterion for a taut foliation. Therefore the fiber foliation is taut.

The curve $\mu$ crosses the alpha disk ${\alpha }_2$ once and is disjoint from ${\alpha }_1$, so in the presentation it represents $y$ up to inversion and conjugacy. The Dehn filling adds the relation $y=1$, after which the beta relator becomes $x^2$. Thus the filled genus-one Heegaard diagram has two alpha–beta intersections andCompressing the displayed genus-two diagram along $\mu$ visibly cancels the annular handle and leaves that genus-one Heegaard diagram. It is therefore the lens space

If a meridian of a knot component in a surgery link links only one other component, filling it with coefficient $b\ne 0$ removes it and replaces the other coefficient $a$ by $a-1/b$. This gives the negative continued-fraction chain description of rational Dehn filling.