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?



Surface separating the boundary of a cylinder
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.)



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> 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 $F \times 0$.
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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.




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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.


