When are mapping tori isomorphic as bundles over the circle? Suppose $\Sigma$ is an orientable genus-$g$ surface (possibly with boundary). The mapping torus corresponding to an orientation-preserving diffeomorphism $\phi: \Sigma \to \Sigma$ is the quotient $M_\phi = \Sigma \times [0,1] / (x,0) \sim (\phi(x),1)$, which has a natural bundle structure $\pi_\phi : M_\phi \to S^1$ with fiber $\Sigma$. If two mapping tori are isomorphic as bundles over $S^1$, then they are certainly homeomorphic. But the converse is false. I'm curious about a stronger condition:

If a homeomorphism of mapping tori $f: M_\phi \to M_{\phi'}$ can be homotoped to carry at least one fiber of $\pi_\phi$ to one of $\pi_{\phi'}$, is it homotopic to a bundle isomorphism? 

 A: Yes, this is true. We may as well suppose $f$ is a homeomorphism that already preserves one of the fibers. Now cut the homeomorphism open along that fiber. Then both manifolds are now homeomorphic to $\Sigma \times [0,1]$, and we may identify your homeomorphism with a homeomorphism $\Sigma \times [0,1]$ to itself; assume it preserves $\Sigma \times \{0\}$ or just compose with a reflection so that this is true; then, if $f_0$ is the restriction of the homeomorphism to $\Sigma \times \{0\}$, compose with $f_0^{-1} \times \text{id}$. Now you have a homeomorphism of $\sigma \times [0,1]$ to itself that's the identity on one side. To be precise, this is the same thing as an pseudoisotopy of $\Sigma$. Your question is: is this pseudoisotopy homotopic (rel the boundary) to a fiberwise homeomorphism (that is, an isotopy)? Even better, is it isotopic to an isotopy?
The answer to this question says: yes! (Here they allow themselves to modify the map on the $\times \{1\}$ side, but you can easily change the arguments so that they don't do this.) So your homeomorphism is isotopic to a fiber-preserving one; gluing the sides back together, we get that your homeomorphism $M_\phi \to M_{\phi'}$ is isotopic to a bundle isomorphism.
