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 ?
 A: 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. 
A: 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.
