Every compact manifold with fundamental group $ \mathbb{Z} $ is a mapping torus Suppose that $ M $ is the mapping torus of some homeomorphism of a manifold $ F $. Then $ M $ is a fiber bundle with fiber $ F $ and base space $ S^1 $
$$
F \to M \to S^1 
$$
So if $ F $ is connected and simply connected then by LES homotopy we have
$$
\pi_1(M) \cong \pi_1(S^1) \cong \mathbb{Z}
$$
Let $ M $ be a compact $ n $ dimensional manifold with fundamental group $ \pi_1(M) \cong \mathbb{Z} $. Is it true that $ M $ must be the mapping torus of a compact connected simply connected $ n-1 $ dimensional manifold?
My thoughts so far:

*

*True for $ n=1 $ because $ S^1 $ is a mapping torus for the point.

*True for $ n=2 $ because there are no compact simply connected 1 manifolds so there are no compact 2 manifolds with fundamental group $ \mathbb{Z} $

*True for $ n=3 $ since the only compact simply connected 2 manifold is the sphere $ S^2 $ and indeed the only 3 manifolds with fundamental group $ \mathbb{Z} $ are the trivial mapping torus $ S^2 \times S^1 $ and the mapping torus of the antipodal map (these are the only two possible mapping tori for $ S^2 $). For example see What closed 3-manifolds have fundamental group $\Bbb Z$?
I expect this already fails for $ n=4 $ and almost certainly fails for generic $ n $. But I figured it couldn't hurt to ask.
 A: Let $F$ be a path-connected topological space and $\phi : F \to F$ a homeomorphism. Denote the mapping torus of $\phi$ by $M$ so that we have a fiber bundle $F \to M \to S^1$. Explicitly, $M = F\times[0, 1]/\sim$ where $(x, 0) \sim (\phi(x), 1)$. From the long exact sequence in homotopy, there is a short exact sequence $$0 \to \pi_1(F) \to \pi_1(M) \to \pi_1(S^1) \to 0.$$ The covering space of $M$ corresponding to the image of $\pi_1(F)$ in $\pi_1(M)$ is homeomorphic to $F\times\mathbb{R}$. This follows from the fact that $M$ can be constructed as the quotient of $F\times\mathbb{R}$ by the $\mathbb{Z}$-action generated by $(x, t) \mapsto (\phi(x), t + 1)$. In particular, the universal cover of $M$ is homeomorphic to $\widetilde{F}\times\mathbb{R}$ where $\widetilde{F}$ is the universal cover of $F$.
Suppose now that $F$ is a closed $(n-1)$-manifold, so that $M$ is a closed $n$-manifold. If $\pi_1(M) \cong \mathbb{Z}$, then $\pi_1(F) = 0$ by the short exact sequence above. As suggested by Moishe Kohan, let $N = M \# X$ where $X$ is a closed simply connected $n$-manifold, then $\widetilde{N}$, the universal cover of $N$, is homeomorphic to the connected sum of $\widetilde{F}\times\mathbb{R} = F\times\mathbb{R}$ and countably many copies of $X$. If $H_k(X; \mathbb{Z}) \neq 0$ for some $0 < k < n$, then it follows from Mayer-Vietoris that $H_k(\widetilde{N}; \mathbb{Z}) \cong H_k(\widetilde{F}; \mathbb{Z})\oplus\bigoplus_{i\in\mathbb{Z}}H_k(X)$; in particular, $H_k(\widetilde{N}; \mathbb{Z})$ is not finitely generated. As $\pi_1(N) \cong \pi_1(M) \cong \mathbb{Z}$, if $N$ were a mapping torus, the fiber $F'$ would be a closed simply connected manifold, so $\widetilde{N}$ would be homeomorphic to $\widetilde{F'}\times\mathbb{R} = F'\times\mathbb{R}$ which has finitely generated homology. Therefore $N$ is not a mapping torus. Note, if $H_k(X; \mathbb{Z}) = 0$ for all $0 < k < n$, then $X$ is a simply connected homology sphere and hence homeomorphic to $S^n$, in which case $N$ is homeomorphic to $M$ (which is a mapping torus).
I believe that the hypothesis $\pi_1(M) \cong \mathbb{Z}$ is unnecessary, but the proof above only extends to the case where $\pi_1(M)$ is an extension of $\mathbb{Z}$ by a finite group, i.e. $F$ has finite fundamental group.
A: Just posting this edit as an answer instead since there is nothing else to say here. All credit to Moishe Kohan:
I see. Thanks Moishe Kohan. I guess this is a good general lesson that too much of my intuition about topology is from dimension $ n \leq 3 $ in those dimensions (almost) everything can be determined from fundamental group. In dimension 2 and 1 manifolds are homeomorphic if and only if they have same fundamental group. This is almost true for dimension 3 as well. For example given $ \pi_1 $ for an orientable 3 manifold we can find all the homology and cohomology up to 3. Moreover, in dimensions 3 and under the only compact connected simply connected manifold is the sphere. So we can expect information about the fundamental group to be able to tell us very strong things about a manifold. But in dimension 4 and up there are lots of things other than the sphere that are compact connected simply connected (for example $ K_3 $ surface) and taking connected sum with any of those will give a manifold with same fundamental group but otherwise a totally different topology. So knowing only that a manifold is compact and has $ \pi_1 \cong \mathbb{Z} $ we should not expect to be able to show anything as strong as that it must be a mapping torus (at least not in dimension 4 and up).
