Perform 0-framed surgery, then remove neighbourhood of meridian. Is this the knot complement?

Question

Let $K$ be a knot in $S^3$ and let $m$ be a meridian of $K$. Let $M_K$ be the 3-manifold obtained by performing 0-framed surgery on $K$. The meridian $m$ can also be viewed as a circle in $M_K$. Is $M_K\setminus (m\times D^2)$, i.e. the 3-manifold minus a regular neighbourhood of the meridian the original knot complement?

Answer 1

Yes. Briefly, this is because the meridian $m$ becomes a longitude of the solid torus glued in during 0-surgery and is therefore isotopic to its core.

More precisely (and generally): Given a solid torus $V$ in a 3-manifold $Y$, a framing of $V$ is equivalent to an embedding $S^1 \times D^2 \to V \subset Y$ such that $ \partial(S^1 \times D^2) \to \partial V$ takes $\{x\} \times \partial D^2$ to a meridian $m$ and $S^1 \times \{y\}$ to a longitude $\ell$. Performing 0-surgery on $V$ gives you a manifold $$Y_0(V)= (Y \setminus \mathring{V}) \cup (S^1 \times D^2)$$ where the gluing map instead takes $\{x\} \times \partial D^2$ to $\ell$ and $S^1 \times \{y\}$ to $m$. Since $m=S^1 \times \{y\}$ is isotopic to the core of the new $S^1 \times D^2$, its complement in $Y_0(V)$ is homeomorphic to the complement of $S^1 \times D^2$ in $Y_0(V)$, i.e. the original solid torus exterior $Y \setminus \mathring{V}$.

Edit - I'm implicitly using the following standard style of argument: An isotopy of embeddings $f:S^1 \times [0,1] \to Y_0(V)$ between $f_0(S^1)=m \subset Y_0(V)$ and the core of the glued-in solid torus $f_1(S^1)=S^1 \times \{0\}\subset Y_0(V)$ extends to an ambient isotopy $F: Y_0(V) \times [0,1] \to Y_0(V)$ by the isotopy extension theorem. Then $F_1$ restricts to a homeomorphism between the complement of $m$ and the complement of the core of the solid torus in $Y_0(V)$. This latter space is clearly homeomorphic to the complement of the entire solid torus in $Y_0(V)$, and is thus homeomorphic to $Y \setminus V$.