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This is a simple version of the question asked here.

Let $M$ be a compact 3-manifold without boundary, $\mathbb{T}^2$ be the standard 2-torus, and $i:\mathbb{T}^2\to M$ be an embedding.

(*) Assume the induced homomorphism $i_*:\pi_1(\mathbb{T}^2)\to\pi_1(M)$ is trivial: $i_*(\pi_1(\mathbb{T}^2))=0$.

Does $i(\mathbb{T}^2)$ bound a solid torus $\mathbb{D}^2\times S^1$?

If it does, could we weaken the (*) assumption?

Thanks!

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There are embedded tori in $\mathbb{R}^3$ which do not bound solid tori. –  user641 Nov 28 '11 at 14:04
    
Really? I have no idea what it is. –  Pengfei Nov 29 '11 at 10:55
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2 Answers

up vote 3 down vote accepted

The condition that $i_*$ is trivial is neither necessary nor sufficient to to guarantee that $i(\mathbb{T}^2)$ bounds a solid torus.

To show that it is not necessary, consider the manifold $M = S^2 \times S^2$. If $E$ is the equator of $S^2$, then $E\times S^1$ is a solid torus in $M$. However, $\pi_1(M) = \mathbb{Z}$, and the homomorphism $i_*\colon \pi_1(E\times S^1)\to\pi_1(M)$ is onto.

To show that it is not sufficient, consider a standard torus in $S^3$, whose removal separates $S^3$ into two solid tori. Now suppose that we modify one of these solid tori by removing two solid balls and gluing in a copy of $S^2\times[0,1]$ (i.e. by attaching a 3-dimensional "handle"). This will not affect the fact that $i_*$ is trivial, but the inside is no longer a solid torus. If we do the same to the outside, we obtain a torus in a 3-manifold with trivial $i_*$ that does not bound a solid torus.

Indeed, the requirement that $i_*$ is trivial does not even guarantee that the torus is a boundary. For example, if we start with a standard torus in $S^3$ and attach an $S^2\times[0,1]$ handle between the inside and the outside, then $i_*$ will still be trivial, but the torus is no longer the boundary of any region.

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Thank you! I know how wrong I was. –  Pengfei Nov 29 '11 at 10:56
    
(I think your $M$ should be $S^2\times S^1$, not $S^2\times S^2$) –  Jason DeVito Feb 17 at 19:07
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I think that this is false. Take a connect sum of two nontrivial 3-manifolds along a 2-sphere $M \#_S N$. Cut two disks out of $S$, and tube them together inside $M$, such that the meridian of the tube together with a curve on $S$ is contractible. I don't think that the resulting torus is going to bound a solid torus in $M \#_S N$, even though it satisfies your hypothesis- it's going to bound $N$ plus a "handle" on one side, and $M$ minus a "handle" on the other.

It looks to me like you need to require that the torus contains an embedded contractible loop, and that after cutting along the embedded disc which it bounds, the sphere that remains is compressible. Something like that.

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Thank you! Your example is very clear. –  Pengfei Nov 29 '11 at 11:01
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