Let $K$ be a knot (embedded circle) in $S^3$ and let $M$ be obtained from $S^3$ by $0$-surgery on $K$. A meridian of $K$ in $S^3$ can be viewed as a circle $C$ in $M$. Consider the product manifold $X=S^1\times M$. $X$ contains an embedded torus $T:=S^1\times C$. Let $\nu T$ denote the closed tubular neighborhood of $T$ in $X$. How can we show that the inclusion $\partial \nu T\to X-\text{int}(\nu T)$ induces a surjection on $\pi_1$?
First, we have $\nu T=S^1\times \nu C$ where $\nu C$ is the closed tubular neighborhood of $C$ in $M$. According to this question Perform 0-framed surgery, then remove neighbourhood of meridian. Is this the knot complement?, the complement of $\text{int}(\nu C)$ in $M$ is the knot complement $S^3-\text{int}(\nu K)$, where $\nu K$ is the closed tubular neighbrohood of $K$ in $S^3$. So we have $X-\text{int}(\nu T)\cong S^1\times (S^3-\text{int}(\nu K))$. Thus we can view the inclusion $\partial \nu T\to X-\text{int}(\nu T)$ as $S^1\times \partial \nu K\to S^1\times (S^3-\text{int}(\nu K))$. This implies that we only need to show that the inclusion $\partial \nu K\to (S^3-\text{int}(\nu K))$ induces a surjection on $\pi_1$, which I can't show. Any hints for this?