No, we cannot, as it is not true in general.

One way to build examples is via "stabilisation". Take $N_0 = M \times \{1/2\}$. Suppose that $\alpha$ is an arc in $M \times [0,1]$ with the following properties.

• $\alpha$ is simple (does not self-intersect).
• $\alpha \cap N_0 = \partial \alpha$.
• $\alpha$ is isotopic, relative to its boundary, to a simple arc embedded in $N_0$.

We take a regular neighbourhood of $N_0 \cup \alpha$. Let $N$ be the boundary component of the neighbourhood that is on the same side of $N_0$ as $\alpha$ is. The genus of $N$ is one higher than that of $M$. Also, $N$ separates $M \times \{0\}$ from $M \times \{1\}$.

Of course the arc $\alpha$ could also be "knotted". Also, we could stabilise more than once. Finally, we could combine these operations in various ways. It is a pretty theorem, appearing early in the theory of three-manifolds that fibre over the circle, that the above operations are all that can happen.

This is not true in general.

One way to build examples is via "stabilisation". Take $N_0 = M \times \{1/2\}$. Suppose that $\alpha$ is an arc in $M \times [0,1]$ with the following properties.

• $\alpha$ is simple (does not self-intersect).
• $\alpha \cap N_0 = \partial \alpha$.
• $\alpha$ is isotopic, relative to its boundary, to a simple arc embedded in $N_0$.

We take a regular neighbourhood of $N_0 \cup \alpha$. Let $N$ be the boundary component of the neighbourhood that is on the same side of $N_0$ as $\alpha$ is. The genus of $N$ is one higher than that of $M$. Also, $N$ separates $M \times \{0\}$ from $M \times \{1\}$.

Of course the arc $\alpha$ could also be "knotted". Also, we could stabilise more than once. Finally, we could combine these operations in various ways. It is a pretty theorem, appearing early in the theory of three-manifolds that fibre over the circle, that the above operations are all that can happen.