Dehn surgery on knots and homology

Since $\gcd (p,q)=1$, Bézout's identity gives integers $r,s$ with $ps-qr=1$. The integer matrix with columns $(p,q)$ and $(r,s)$ is unimodular, so its linear map on ${\mathbb{R}}^2$ descends to a torus homeomorphism. The image of a coordinate circle is embedded and has class $p[\mu ]+q[\lambda ]$; reversing its parameter changes both coefficients' signs and leaves the unoriented curve unchanged. With the product orientation on $V=D^2\times S^1$, the standard meridian–longitude basis on $\mathrm{\partial }V$ has intersection $+1$. The matrix has determinant $+1$, whereas $\mu \cdot \lambda =-1$ on $T_K$, so the induced boundary map reverses orientation and gives the required gluing map. Thus every primitive pair defines a slope and an oriented filling.

Let $C_0,C_1$ be essential simple closed curves whose primitive classes agree up to sign. Reverse one parameter if needed so their classes agree. By [F1], the curves are isotopic. Thus the unoriented slope depends only on the primitive class modulo simultaneous sign.

Let $V=D^2\times S^1$ and let ${\varphi }_0,{\varphi }_1:\mathrm{\partial }V\to T_K$ be orientation-reversing gluing maps whose meridians map to the same slope. The transition map $g={\varphi }_0^{-1}{\varphi }_1$ preserves boundary orientation and sends the meridian class to $ϵ$ times itself for some $ϵ\in \{1,-1\}$. Its matrix in the meridian–longitude basis is therefore $\left(\begin{array}{cc}ϵ& k\\ 0& ϵ\end{array}\right)$ for some $k\in \mathbb{Z}$. By [F2], $g$ is isotopic to $T^k$ if $ϵ=1$, and to $T^{-k}R$ if $ϵ=-1$, where $T(z,w)=(wz,w)$ and $R(z,w)=(\bar{z},\bar{w})$. These representatives extend over $V$: $T^j$ extends by $(z,w)\mapsto (w^{j}z,w)$, and $R$ extends by simultaneous complex conjugation. Both extensions preserve the orientation of $V$. To extend the isotopy from a representative $h$ to $g$, use a collar $\mathrm{\partial }V\times [0,1]$ and the map $(x,s)\mapsto (I_{\rho (s)}(h^{-1}(x)),s)$, where $I_t$ is the isotopy from $h$ to $g$, $\rho (0)=1$, and $\rho$ is zero near the inner collar edge; composing this collar map with the extension of $h$ extends $g$ over $V$. Since ${\varphi }_0\circ g={\varphi }_1$, gluing this extension to the identity on $E_K$ gives an oriented homeomorphism of the fillings. Therefore the oriented filling type depends only on the meridian slope.

Choose a bicollar of $T_K$ in $S^3$ and thicken $E_K$ and $N(K)$ slightly to an open cover. The two sets deformation-retract to $E_K$ and $N(K)$, and their intersection deformation-retracts to $T_K$. By [F1] and [F2], the segment $H_2(S^3;\mathbb{Z})\to H_1(T_K;\mathbb{Z})\to H_1(E_K;\mathbb{Z})\oplus H_1(N(K);\mathbb{Z})\to H_1(S^3;\mathbb{Z})$ has zero groups at both ends, so the middle map is an isomorphism.

Let $M=E_K(p\mathrm{/}q)$ and choose a bicollar of the gluing torus. Thickening the exterior and attached solid torus across this collar gives an open cover of $M$ whose two sets retract to the pieces and whose intersection retracts to $T_K$. Write $\alpha =p\mu +q\lambda$, and choose $\beta =r\mu +s\lambda$ with $ps-qr=1$, so $\alpha ,\beta$ form a basis of $H_1(T_K;\mathbb{Z})$. In the Mayer–Vietoris map, $\alpha$ maps to $(p[\mu ],0)$ because it bounds the attached solid-torus meridian disk, and $\beta$ maps to $(r[\mu ],\pm v)$ where $v$ generates the first homology of the attached solid torus; the coefficient is $\pm 1$ because $\beta$ intersects $\alpha$ once. The pieces and their overlap are path-connected: the exterior is a connected manifold and hence path-connected, and the solid torus and boundary torus are path-connected. By [F6], the following $H_0$ map sends $1$ to $(1,-1)$ and is injective. Exactness therefore identifies $H_1(M;\mathbb{Z})$ with the cokernel of the $H_1$ map.

By [F3], $H_1(T_K;\mathbb{Z})\cong {\mathbb{Z}}^2$, and by [F4] and [F2] the solid torus $N(K)\simeq S^1$ has $H_1(N(K);\mathbb{Z})\cong \mathbb{Z}$. Since $H_1(E_K;\mathbb{Z})\oplus \mathbb{Z}\cong {\mathbb{Z}}^2$, the direct summand $H_1(E_K;\mathbb{Z})$ is free of rank one. Under the isomorphism in step 1.1, $\mu$ maps to $([\mu ],0)$ because it bounds a disk in $N(K)$. The longitude maps to $(0,\pm 1)$: [A1] gives its zero exterior image, and degree $+1$ on the core gives a generator of $H_1(N(K);\mathbb{Z})$ under the deformation-retraction identification. Since the map is an isomorphism, $[\mu ]$ generates $H_1(E_K;\mathbb{Z})$.

Thus $H_1(M;\mathbb{Z})$ has generators $[\mu ],v$ and relations $p[\mu ]=0$ and $r[\mu ]\pm v=0$. Eliminating $v$ leaves $\mathbb{Z}⟨[\mu ]⟩\mathrm{/}⟨p[\mu ]⟩$. If $p\mathrm{\ne }0$, this is $\mathbb{Z}\mathrm{/}\mathrm{\mid }p\mathrm{\mid }\mathbb{Z}$; if $p=0$, coprimality gives $q=\pm 1$ and the group is $\mathbb{Z}$.

The attaching region of the $2$-handle is $S^1\times D^2\subset \mathrm{\partial }(D^2\times D^2)$. It is glued to the tubular neighborhood $N(K)\subset S^3$, so the portion of $\mathrm{\partial }{B}^4$ left in the boundary after attachment is exactly $E_K=S^3\setminus \mathrm{int}N(K)$.

The free face of the handle boundary is $D^2\times S^1$, a solid torus. Its meridian disk $D^2\times \{z_0\}$ has boundary $\mathrm{\partial }{D}^2\times \{z_0\}$; the specified product framing sends this curve to the attaching parallel of slope $n[\mu ]+[\lambda ]$ on $\mathrm{\partial }{E}_K$. Thus the outgoing boundary is the union of $E_K$ and a solid torus with precisely the meridian image in [F1].

The orientation-compatible handle attachment identifies the oriented attaching region with $N(K)$ by a boundary-orientation-reversing map. The free face $D^2\times S^1$ inherits its boundary orientation from the oriented handle, so its gluing to $E_K$ reverses the induced orientations on the common torus; this is exactly the oriented filling convention in [F1]. Rounding corners changes the boundary only inside a collar and preserves its orientation. Thus $\mathrm{\partial }{X}_n(K)$ is the stated oriented filling.

In [F1], for $0<t<1\mathrm{/}2$ the left-starting motion is $(2ht-h,-2ht)$ and the right-starting motion its negative. At the crossing the latter has greater $y$, hence is over when viewed along $+y$. Their projected tangents in coordinates $(t,x)$ are $(1,-2h)$ and $(1,2h)$, with determinant $4h>0$. Rounding the diamond corners and flattening endpoints preserves this strict sign. The downward three-crossing diagram therefore has three positive crossings. The diagram agrees with the fixed geometric closure: the change from stacked diamonds to rounded diagonal strands is a height-preserving interpolation with nonzero pair difference, followed by the fixed closure map. In particular it is not the mirror braid.

Label the initial left and right meridians $a,b$. At a positive crossing the right input is over, and the left input follows the positive over-normal to become the right output. Successive rows of the braid have input/output pairs $\begin{array}{cccc}\text{row}& \text{input }(L,R)& \text{output }(L,R)& \text{underpass relation}\\ 1& (a,b)& (b,c)& c=b^{-1}ab\\ 2& (b,c)& (c,a)& a=c^{-1}bc\\ 3& (c,a)& (a,b)& b=a^{-1}ca\end{array}$ Here the last two output labels use the closure identifications of bottom and top positions. By [F8], this is the finite three-crossing code to which the Wirtinger correctness theorem [F3] applies. These are all crossing relations by [F3]. Substituting $c=b^{-1}ab$ into $b=a^{-1}ca$ gives $aba=bab$; the remaining relation follows because $b^{-1}ab=ab{a}^{-1}$ and then $c^{-1}bc=a$. Thus ${\pi }_1(S^3\setminus T)=⟨a,b\mid aba=bab⟩$, with the stated positive meridians. In the chosen tube, the radial normal homotopy from positive radius $r<ϵ$ to $(1-s)r+sϵ$ identifies this group with ${\pi }_1(E_T)$.

Trace a small parallel starting at the initial left arc. Along the oriented component its underpasses occur in row order $1,3,2$, with over-meridians $b,a,c$; it passes over the intervening crossings. This order follows directly from the input/output table: $a$ becomes $c$ on the right in row 1, passes over in row 2, becomes $b$ in row 3, closes to the initial right position, passes over in row 1, becomes $a$ in row 2 and closes after passing over in row 3. At each underpass the access-path calculation in [F3] is ${\alpha }_{\mathrm{i}\mathrm{n}}\beta =g_o{\alpha }_{\mathrm{o}\mathrm{u}\mathrm{t}}$, where $\beta$ is the path along the small parallel; thus the three successive factors are $b,a,c$, with positive exponents. At an overpass the access path slides through its upper product chart and contributes no factor. On the crossing-free exterior closure arcs use the same product charts, again contributing no factor. Concatenating gives the based parallel word $w=bac$. These path identities can be made on the upper side of the normal circles. The upper normal section and the blackboard section are homotopic without zero: in the normal quotient the blackboard vector and viewing vector form a positively oriented basis, so the interpolation rotating the former through a fixed quarter-turn to the latter is everywhere nonzero. This proves that the traced parallel has the blackboard framing and oriented degree $+1$, rather than identifying a peripheral word only from its exponent sum. By [F4] and step 1.1 its self-linking is $+3$.

The boundary torus has product coordinates from this actual parallel: its degree-one section is $w$ and its positive fibre is $a$. A section making three negative fibre turns is an embedded degree-one curve with based word $w{a}^{-3}$; the torus rectangle homotopy makes the fibre and section commute. In the abelianization of step 2.1 the relation $aba=bab$ gives $[a]=[b]$, and sending each to $1$ supplies an inverse to the cyclic generator map. Thus $H_1(E_T)=\mathbb{Z}$ on $[a]$ by [F6], and $[w]=[b]+[a]+[c]=3[a]$. Therefore $\lambda =w{a}^{-3}$ has zero exterior homology and degree $+1$, precisely [F5]. Moreover $c=ab{a}^{-1}$, so $w{a}^3=b{a}^{2}b{a}^2=ba(bab)a=(ba{)}^3=(ab{)}^3\mathrm{.}$ Both last cubes equal $(aba{)}^2$, which commutes with $a,b$ because conjugation by $aba$ exchanges them. This proves $\lambda =(ab{)}^3{a}^{-6}$ with the fixed positive meridian and the correct sign.

The $+1$ slope is the primitive boundary curve $\lambda \mu$, since its coefficients in the oriented meridian/longitude basis are $(1,1)$. In [F7] the solid torus has cyclic fundamental group; a complementary boundary generator identifies its core generator and eliminates it, while its meridian kills precisely $\lambda \mu$. Thus filling adds $w{a}^{-2}=1$. Substituting $w=bac$ and $c=b^{-1}ab$ gives $ba{b}^{-1}ab{a}^{-2}=1$. Equivalently step 4.1 gives $(ab{)}^3=a^5$. These are the asserted presentations of the same positive-trefoil $+1$ filling.