A Dehn twist of Heegaard gluing is integral surgery

Push $c$ a small distance into $H_{-}$ to obtain an embedded knot $K$ and choose a closed tubular neighbourhood $N(K)$ disjoint from $\mathrm{\Sigma }$. The collar between $c$ and the boundary of $N(K)$ is an embedded annulus $A$; its parallel boundary curve on $\mathrm{\partial }N(K)$ is ${\lambda }_A$. Choose the product coordinates on $N(K)$ so that the meridian $\mu$ bounds a normal disk.

Cut $H_{-}\setminus \mathrm{int}N(K)$ along $A$ and reglue the two annular faces after one full turn. In annular coordinates the turn adds $2\pi$ to the angular coordinate between the two ends and is stationary near the seams. Thus on $\mathrm{\Sigma }$ the induced map is ${\tau }_c$ (or its inverse if the turn is reversed). On $\mathrm{\partial }N(K)$, a meridian crossing the annulus once returns with one extra longitude traverse, so the induced homology shear is $[\mu ]\mapsto [\mu ]+[{\lambda }_A]$ or $[\mu ]\mapsto [\mu ]-[{\lambda }_A]$, respectively; the longitude is fixed. The operation fixes the rest of $H_{-}$ and is smooth after rounding the product corners.

Glue the identity on $H_{+}$ to the cut-and-reglued map on the complement of $N(K)$. Along $\mathrm{\Sigma }$ the maps agree because the new gluing is $\phi \circ {\tau }_c^{\pm 1}$; hence they give an orientation-preserving diffeomorphism between the two knot exteriors. The meridian of the replacement solid torus is identified with $[\mu ]\pm [{\lambda }_A]$, exactly the framed surgery slope from step 2.1. By [F3], filling this boundary is the stated integral surgery, so its result is the Heegaard-glued manifold with the twisted gluing map.

The annular parallel and its product framing are part of the supplied data, and the only auxiliary collar/tubular choices are within the ${\mathrm{A}\mathrm{C}}_{\omega }$ interface [A1]. The construction therefore uses an actual framed knot and introduces no rational coefficient.