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 ?

closedmanifolds, I guess the generalization to compact manifolds is straightforward. $\endgroup$ – H1ghfiv3 Dec 11 '15 at 13:02