Surface separating the boundary of a cylinder Let $M^2$ be a connected closed surface. Suppose there exists an smooth embedding from a connected closed surface $N$ into the interior of $M \times [0,1]$ such that $N$ separates $M \times \{0\}$ and $M \times \{1\}$.
If $N$ is homeomorphic to $M$, can we prove that the region bounded by $M \times \{0\}$ and $N$ is homeomorphic to $M \times [0,1]$?
If the conclusion is true, can we generalize it to the higher dimensional case?
 A: I am going to focus on the oriented case here. A similar argument should hold in the unoriented case but the degree argument ought to use twisted coefficients, which I don't want to go through here.
Write $W_0$ and $W_1$ for the closure of the two components of $M \times [0,1] \setminus N$, with $M \times \{i\} \subset W_i$. Then $W_0 \cap W_1 = N$ and $W_0 \cup W_1 = M \times [0,1]$.
Write $i: N \to M \times [0,1]$ for the inclusion and $p: M \times [0,1] \to M$ for projection onto the first component. Write $j_i: W_i \to M \times [0,1]$ for the inclusion of the two components.
Because $N$ is homologous to $M \times \{i\}$ via $W_i$, the map $pi$ is degree 1; degree 1 maps between oriented surfaces are homotopy equivalences. It follows that $pi$ induces an isomorphism on $\pi_1$.
Goal: We should argue that $i: N \to W_0$ is an isomorphism on $\pi_1$. (The argument is unchanged for $W_1$.) Then we can cite Theorem 10.2 of Hempel, "3-manifolds", which is as follows. (I have changed some notation to avoid conflicting with yours.)

Theorem 10.2: Let $Y$ be a compact 3-manifold and $F$ a compact connected surface in $\partial Y$ so that $i_*: \pi_1 F \to \pi_1 Y$ is an isomorphism. Then the Poincare associate $\mathcal P(Y)$ is homeomorphic to $F \times [0,1]$ by a homeomorphism which takes $F$ to $F \times 0$.

Here, the Poincare associate is the unique manifold so that $Y$ is the connected sum of $\mathcal P(Y)$ and some number of homotopy spheres. Since Perelman, we now know that $\mathcal P(Y) = Y$. But in this particular case --- because $M \times [0,1]$ is irreducible --- it follows that $\mathcal P(W_i) = W_i$ without any knowledge about homotopy spheres.
Then the desired result follows, without any need to cite Perelman.

Let's prove that $i_*: \pi_1` N \to \pi_1 W_0$ is an isomorphism. (An identical argument works for $W_1$.)
Because the composite $$\pi_1 N \xrightarrow{i_*} \pi_1 W_0 \xrightarrow{(pj_0)_*} \pi_1 M$$ is an isomorphism (and in particular injective), $i_*$ is injective. The hard part is showing that $i_*$ is surjective. This is equivalent to showing that $K := \text{ker}(pj_0)_* = \{1\}$.
Pick $\gamma \in K$ a representative contained in the interior $W_0^\circ$. Because $p_*: \pi_1(M \times [0,1]) \to \pi_1 M$ is an isomorphism, it follows that $(j_0)_* \gamma = 1$; rephrased, $\gamma$ is null-homotopic in $M \times [0,1]$.
Choose a null-homotopy $H: D^2 \to M \times (0,1)$ which is transverse to $N$, and so that $H^{-1}(N)$ has the fewest possible connected components. I claim that $H^{-1}(N) = \varnothing$, and hence $H$ is a null-homotopy inside $W_0^\circ$, so that $\gamma = 1 \in K$; since $\gamma$ was arbitrary, this implies $K = 1$ as desired.
Because $\gamma$ does not meet $N$, $H^{-1}(N)$ does not meet the boundary of $D^2$, and hence consists of a finite number of disjoint circles. If $H^{-1}(N)$ is nonempty, choose an innermost circle of $H^{-1}(N)$. This represents a loop $\xi$ in $N$ which is null-homotopic in either $W_0$ or $W_1$. Because the inclusion of $N$ into both $W_i$ induce an injection on $\pi_1$, it follows that $\xi$ is null-homotopic in $N$. Now redefine $H$ by replacing the given innermost disc by a null-homotopy inside $N$, and then push this disc inside the respective component that the next most innermost area maps to. The new null-homotopy meets $N$ in exactly one fewer circle, contradicting minimality. It follows that $H^{-1}(N)$ is empty, as desired.
