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The Borel Conjecture asserts that homotopy equivalent aspherical closed manifolds are homeomorphic, which is still open in general. But, for three-dimensional manifolds, this conjecture holds (I read this in Bessieres-Besson-Boileau), whose proof depends on the geometrization theorem (Perelman).

Question: Does the relative version of the Borel conjecture also hold for compact 3-manifolds with boundary (by the geometrization)? The relative version: If there is a homotopy equivalence between two compact aspherical manifolds that is a homeomorphism between their boundaries, are those manifolds homeomorphic?

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Yes.

When the manifolds are Haken this is a theorem of Waldhausen. See Ian Agol's answer here.

Since your manifolds are aspherical, they are irreducible by the Poincaré conjecture. Since they have boundary and are irreducible, they are Haken.

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  • $\begingroup$ Could you please give a brife proof that geometrization conjecture implies Borel conjecutre in dimension 3? I read [BBB], it seems they didn't explictely proved Borel Conjecture, just said it's a corollary of GC, which I don't know why. $\endgroup$
    – user16750
    Commented Oct 7, 2011 at 2:52
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    $\begingroup$ @unknown: The Poincaré Conjecture implies that a $3$--manifold is irreducibile if it is aspherical. Say your manifolds are $M$ and $N$. If they are homotopy equivalent, they have isomorphic fundamental groups. If this group does not contain a $\mathbb{Z} \oplus \mathbb{Z}$ subgroup, then $M$ and $N$ are hyperbolic, by Geometrization, and so, by Mostow Rigidity, they are homeomorphic. $\endgroup$
    – Autumn Kent
    Commented Oct 7, 2011 at 3:10
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    $\begingroup$ @unkown: Continued: If this group contains a $\mathbb{Z} \oplus \mathbb{Z}$ subgroup, then, by the (strong) Torus Theorem, $M$ and $N$ are either Haken or small Seifert fibered spaces. If both are Haken, then Waldhausen's theorem comes to the rescue. I think you do something like show that $\pi_1$ of a small SFS isn't an amalgamated product to rule out $M$ Haken and $N$ a small SFS. If both are small SFSs, then the euler class tells you they're homeomorphic. $\endgroup$
    – Autumn Kent
    Commented Oct 7, 2011 at 3:15
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    $\begingroup$ @unkown: Actually, to rule out $M$ Haken and $N$ small SFS, argue like this: $\pi_1$ has infinite cyclic center, and so $M$ is a SFS too (this is part of the proof of the Strong Torus Theorem). The center is a completely algebraic thing, and so the quotient of $\pi_1(M)$ by the center is the orbifold fundamental group of the underlying orbifold of the Seifert fibered structure on $M$. This must be isomorphic to the orbifold group for the underlying orbifold for $N$, and so $M$ is a small SFS. Euler class to the rescue. $\endgroup$
    – Autumn Kent
    Commented Oct 7, 2011 at 3:28

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