Dehn surgery on knots and homology — Examples and Counterexamples

A standard unknot is the coordinate circle in $S^3\subset {\mathbb{C}}^2$ given by $w=0$. Its tubular neighborhood and its exterior are both solid tori; on the exterior, ${\lambda }_U$ bounds a meridional disk and ${\mu }_U$ is a longitude. Therefore each $p\mathrm{/}q$ filling is a union of two solid tori, and for $p>0$ it is $L(p,q)$ by [F3].

When $p=0$, primitivity forces $q=\pm 1$, so the attaching meridian matches the meridian of the exterior solid torus. By the gluing-map independence in the slope definition [F4], use the product gluing; then $S^1\times D^2$ and $S^1\times D^2$ glue along $S^1\times S^1$ as $S^1\times (D^2{\cup }_{S^1}{D}^2)=S^1\times S^2$.

For $p=1,q=1$, the attaching slope is one meridian plus one longitude of the exterior solid torus. A Dehn twist along its meridian disk extends over that torus and changes this slope to its longitude. By [F4] the gluing may then be chosen to be the standard decomposition $S^3=(D^2\times S^1)\cup (S^1\times D^2)=\mathrm{\partial }(D^2\times D^2)$, so the filled orientation is that of $S^3$.

For all coprime $p,q$, [F1] gives the stated $H_1$ group; the cases $p=0$ and $p=1$ agree with steps 2.1 and 2.2.

For every integer $n$, [F2] identifies the $n\mathrm{/}1$ filling with the outgoing boundary of $X_n(U)$. This adds no rational-framing claim.

By [F3], $M$ is the connected closed oriented manifold obtained by filling the connected knot exterior along its boundary torus. Apply [F1] with $p=q=1$. Then $H_1(M;\mathbb{Z})\cong \mathbb{Z}\mathrm{/}1\mathbb{Z}=0$.

Let $A_5$ denote the subgroup of even permutations of $\{1,2,3,4,5\}$; every $5$-cycle is even because it is a product of four transpositions. Set $a=(12345)$ and $b=(13542)$, composing right to left. Applying the factors of $aba$ and $bab$ from right to left gives the same image tuple $(2,1,3,5,4)$ on inputs $(1,2,3,4,5)$, so both words equal $(12)(45)$. Thus the braid relator in [F2] is satisfied.

Applying the factors of $ba{b}^{-1}ab{a}^{-2}$ from right to left, the six successive images for inputs $1,2,3,4,5$ are respectively $(4,2,3,1,2,1)$, $(5,4,5,3,4,2)$, $(1,3,4,5,1,3)$, $(2,1,2,4,5,4)$, and $(3,5,1,2,3,5)$. Their final entries are $(1,2,3,4,5)$, so the filling relator is the identity. The assignments therefore satisfy both relators in [F2] and induce a homomorphism ${\pi }_1(M)\to A_5$.

The image of $a$ is a nonidentity $5$-cycle, so this homomorphism has nontrivial image and ${\pi }_1(M)\mathrm{\ne }1$. Together with step 1.1, this refutes the statement.