# Non-nullhomologous incompressible tori in finite covers of 3-manifolds

I'm interested in proving the following claim:

Lemma: Let $$M$$ be an aspherical, closed 3-manifold which is not hyperbolic. Then, there exists a finite cover $$M'$$ of $$M$$ and an incompressible embedded torus $$T$$ in $$M'$$, such that $$T$$ is non-nullhomologous in $$M'$$.

I'm aware that if $$M$$ is Seifert fibered, then there exists a finite cover $$M'$$ of $$M$$ such that $$M'$$ is a circle bundle over a surface. I would like to say that we can take a further finite cover $$M''$$ which is a circle bundle over a surface $$F$$, where $$F$$ contains an embedded non-separating curve $$\gamma$$. Then the torus $$T$$ in $$M''$$ corresponding to the curve $$\gamma$$ is non-separating and thus non-nullhomologous.

If the above is correct, then it suffices to prove the Lemma in the case where the JSJ-decomposition of $$M$$ does not contain any Seifert fibered pieces. However, here I am stuck.

• Since the tori in the JSJ decomposition are by definition incompressible, it seems we can reduce to the case where the $3$-manifold is geometric, no? Aug 5, 2021 at 20:44
• It’s possible, however, that the tori in the JSJ decomposition are all nullhomologous Aug 5, 2021 at 21:11
• Can you do something like Van Kampen? (as long as the pieces have non-trivial $\pi_{1}$) it seems it will work alot of the time (although I can't prove that). Is there some example where the JSJ tori are null-homologous? Aug 6, 2021 at 10:06
• The tori are nullhomologous exactly when they are separating in the 3-manifold. How do you suggest I should apply Van Kampen in this situation? Aug 7, 2021 at 9:31
• I am now not so sure anymore sorry. Btw, It seems that there is a proof also for Sol $3$-manifolds. Take a finite cover w,hich is a $T^2$-bundle over $S^1$. Then, a $T^{2}$-fibre of the bundle is not seperating and incompressible. Aug 7, 2021 at 11:52