Torus theorem for simply connected 3-manifolds

Question

I'm trying to prove the following statement (ignoring that it follows easily from the Poincare conjecture), which is exercise 4C6 in Rolfsen's Knots and Links:

If $M$ is a closed and simply connected 3-manifold and $T\subset M$ is an embedded torus, then $T$ separates $M$ and at least one component of $M\setminus T$ has fundamental group $\mathbb{Z}$.

I'm taking everything in the above statement to be PL. I have successfully shown that $T$ separates $M$. I have also used Dehn's lemma and the loop theorem to conclude that there is a simple closed curve $J$ in $T$ such that $J$ bounds a disk in one of the closed components $A$ of $M\setminus T$. Similarly to the proof of the torus theorem, we can then take a bicollared neighborhood $N$ of $D$ and show that $A\setminus N$ is bounded by a 2-sphere $S$.

At this point the fact that we are not necessarily in $S^3$ becomes an issue. In the proof of the torus theorem we can conclude that the sphere $S$ necessarily bounds a 3-ball. In this instance I believe that we cannot conclude this. My main idea for continuing is to show that every non-trivial curve in $A$ must intersect $D$. This would show that $A\setminus N$ is simply connected and from here I believe we can conclude that $\pi_1(A)\cong\mathbb{Z}$. This is obvious if $A$ happens to be a solid torus but I'm not sure how one would show this (or if it's even true) for a general 3-manifold with torus boundary.

Is this the right approach? Any hints/tips would be appreciated.

Answer 1

You're very close, and it seems to follow from only homotopy theory once you have constructed $S$. First, note that $S$ separates $M$ into two simply connected regions (since otherwise the van Kampen theorem would imply $M$ has a nontrivial fundamental group). Consider the fact that $A$ is $A\setminus N$ with the closure of $N$ attached. Thinking of the closure of $N$ as $D\times [0,1]$, let $I=0\times[0,1]$, which is one of the fibers of the neighborhood, a path connecting one side of $N$ to the other. By collapsing $N$ onto $0\times(0,1)$, we see that $A$ is homotopy equivalent to $(A\setminus N)\cup I$ (we homotope the attachment to $A$ through a deformation retract of $N$ onto the fiber). Since $\partial(A\setminus N)=S$ is path connected, we can drag one end of $I$ to the other, giving the homotopy equivalent space $(A\setminus N)\vee S^1$, the wedge product of $A\setminus N$ with $S^1$. Hence, $$\pi_1(A) \approx \pi_1((A\setminus N)\vee S^1) \approx \pi_1(A\setminus N)*\pi_1(S^1) \approx 1*\mathbb{Z}\approx\mathbb{Z}.$$