# Is the following 3-manifold always a trivial I-bundle over a surface?

Let $M$ be a compact, orientable and irreducible 3-manifold with with boundary consisting of two incompressible components $N_0,N_1$, with $N_i \stackrel{f_i}{\cong} S_g$ for some diffeomorphism $f_i: S_g \to N_i$, where $S_g$ is the closed orientable surface of genus $g$. I wonder if the following is true:

If $f_0 \simeq f_1$, i.e, if the $f_i$ are homotopic as maps from $S_g$ to $M$, then $M$ is homeomorphic to $S_g \times [0,1]$.

I have the following idea for a proof: Since $f_0 \simeq f_1$, there is a map $f: S_g \times [0,1] \to M$ with $f(x,i) = f_i(x)$ for $i=0,1$. In particular, we have $f_0 = f \circ i_0$ with $i_0: S_g \to S_g \times [0,1]$ the natural inclusion of $S_g$. Since $f_0$ is $\pi_1$-injective and $i_0$ is a homotopy equivalence, $f$ must be $\pi_1$ injective. I would be done once I've shown that $f$ is also $\pi_1$-surjective, since this implies that $f$ is indeed homotopic to a homeomorphism.

Intuitively, this should be the case, but i cannot come up with a formal proof of this. Does anybody have an idea, or a counterexample ?

• Perhaps a possible approach: using Morse theory your bordism can be decomposed into handles. If $M$ is not a product, then it should be obtained by adding handles to $S_g\times I$. And then your homotopy can be made transversal to the handles of that decomposition. – ThiKu Dec 11 '15 at 10:04
• Show that $f$ is a degree 1 map. Degree 1 maps must be surjective on the fundamental group (since they can't lift to a non-trivial cover). – user83633 Dec 11 '15 at 11:50
• $f$ has degree 1 as a map of manifolds with boundary, ie the degree of the map on $H_2(M,\partial M) \cong Z$. This is because the degree of the map on the boundary is 1. That the degree is the same as the covering degree is a basic fact that you can find in textbooks. – Danny Ruberman Dec 11 '15 at 12:50
• You probably mean $H_3(M,\partial M)$, don't you ? I was aware of the equality of homological degree and covering degree for closed manifolds, I guess the generalization to compact manifolds is straightforward. – Berni Waterman Dec 11 '15 at 13:02
• It is a remarkable skill, being able to even write coherent setences while fast asleep. A small mistake does not matter – Berni Waterman Dec 11 '15 at 20:11

This follows, fairly easily, from the hypothesis of irreducibility and from the "annulus theorem" (see page 130 of Jaco-Shalen's book "Seifert Fibered Spaces in 3-Manifolds"). You can remove the hypothesis of irreducibility if you are willing to use the Poincaré conjecture.

• I thought the Poincare conjecture was a theorem these days? – Igor Rivin Dec 11 '15 at 20:53
• @IgorRivin - Sure. But the original poster might want to avoid using a giant hammer of a theorem, when a tiny spatula of an assumption will do the same job. – Sam Nead Dec 11 '15 at 21:28
• Actually, my understanding has been that the whole "irreducibility" hypothesis was invented to weasel out of the PC, so now that we don't need to weasel any more... – Igor Rivin Dec 11 '15 at 22:00
• You mean only in context of this question? Because in general, it seems to me that even to this day, irreducible 3 manifolds are far better understood than general 3 manifolds. – Berni Waterman Dec 12 '15 at 1:43
• @IgorRivin - Yes, correct. But we wouldn't want to waste a good weasel! – Sam Nead Dec 13 '15 at 9:59

Although this question has already been answered, I would like to shortly explain the idea on how to prove that $f$ is $\pi_1$-surjective, as it has been established in the comments below my original question. The result then follows from the classification of Haken manifolds (See Hempel's "3 Manifolds", for example).

I will use the notation from the origianl question:

Let $X:= S_g \times [0,1]$, and $x$ some basepoint in $X$, let $G := f_*(\pi_1(X,x)) \subset \pi_1(M,f(x))$ and let $g: N \to M$ be a covering map corresponding to the subgroup $G$. The map $f$ then lifts to a map $h:X \to N$, so that we have $g \circ h = f$. Moreover, $N$ is also a (possibly noncompact) manifold-with-boundary; the homological degree $deg(g)$ is therefore well-defined and its absolute value coincides with the covering degree of $g$. In particular, since $deg(f) = deg(g)*deg(h)$, we must have $deg(g)= \pm 1$. That is because $f$ restricts to a homeomorphism on the boundary, implying $deg(f)= \pm1$. But then $g$ must be the trivial covering map, so $G = \pi_1(M)$. Therefore, $f$ is indeed $\pi_1$-surjective.