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For a framed embedding $S^{\lambda -1}\times D^{m+1-\lambda }\hookrightarrow M^m$, index-$\lambda$ surgery on a smooth manifold removes the interior of that region and replaces it by $D^{\lambda }\times S^{m-\lambda }$. Their common boundary is $S^{\lambda -1}\times S^{m-\lambda }$, along which the framing specifies the gluing. This is a $(\lambda ,m+1-\lambda )$ surgery. Its trace is an $(m+1)$-dimensional cobordism obtained by attaching the handle $D^{\lambda }\times D^{m+1-\lambda }$ to $M\times [0,1]$ along the outgoing boundary.

In the present surface case, $\lambda =2$ and $m=2$. The operation removes an annulus around the embedded curve $\gamma$ and caps its two boundary circles by closed disks. Since $[\gamma ]=pa+qb\ne 0$, the curve cannot separate the oriented torus: a separating curve is the oriented boundary of one of the complementary subsurfaces and is therefore null-homologous. Thus it is a nonseparating curve, and the cut-open surface $F$ is connected with two boundary components. Its Euler characteristic is zero, since both the removed annulus and the gluing circles have Euler characteristic zero. Capping adds two, so the resulting oriented closed connected surface has $\chi (M)=2$. By the allowed surface classification it is a sphere. ThereforeEquivalently the classification shows that $F$ is a cylinder. The argument is conditional on the asserted embedding existing; it does not assign a surgery manifold to a nonexistent embedded curve.

Choose an arc in this connected cylinder joining its two boundary circles and close it by an arc crossing the removed annulus. After smoothing the joints, its closed curve $\delta$ meets $\gamma$ transversely exactly once. If $[\delta ]=ra+sb$, then their algebraic intersection number of curves on an oriented surface isThis proves that $[\gamma ]$ is a primitive homology class and forces $\gcd (p,q)=1$.

Conversely, if $\gcd (p,q)=1$, the mapis an immersion with the required homology class. If its values at $t,s$ coincide, then $p(t-s)$ and $q(t-s)$ are integers; Bézout's identity gives $t-s\in \mathbb{Z}$. It is therefore injective and, by compactness, an embedded circle. An explicit annular thickening isThe determinant of its coordinate derivatives is $\delta (p^2+q^2)\ne 0$. If two image points coincide, solving for the difference in the normal coordinate shows that $\delta (u-u^{\prime })$ is an integer multiple of $1/(p^2+q^2)$. The strict choice of $\delta$ forces that integer to be zero, so $u=u^{\prime }$ and then $t=t^{\prime }$ modulo one. This supplies the actual required embedding, not just a homology class. Hence