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Yes, the result follows from a theorem of Livesay that if $M$ is a compact connected non-orientable 3-manifold with $\pi_1(M)$ finite, then $M$ is homeomorphic to $P^2\times I$ minus a collection of disjoint open 3-balls.

The first answer to the original question that I wrote here was flawed. Here is a corrected version, thanks to the comments below.

First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.

If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$.

By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $q:N\to N$ be the covering transformation, an orientation reversing involution. Now $H_2(N;\mathbb{Q})\cong\mathbb{Q}^{k-1}$ and $H_2(N;\mathbb{Q})$ is generated by any $k-1$ components of $\partial N$. Since $q$ acts on each component of $\partial N$ as an orientation reversing homeomorphism, it follows that the Lefschetz number $\Lambda_q$ is given by $$\Lambda_q=1-0+(-(k-1))=2-k.$$ But $q$ has no fixed points, so $\Lambda_q=0$ and $k=2$. See Epstein for the original (and more general) proof of this step.

Therefore $N$ is homeomorphic to $S^2\times I$, and it follows by Livesay or Rubinstein that $M$ is homeomorphic to $P^2\times I$. The argument I had in mind here involves doubling $N$ along its boundary to get $S^2\times S^1$. The covering map $p$ also extends to the double. Then the classification of Seifert fibered spaces covered by $S^2\times\mathbb{R}$ tells us that the only compact quotient of $S^2\times S^1$ which contains a 2-sided projective plane is $P^2\times S^1$. Hence $M$ is homeomorphic to $P^2\times I$.

Yes, the result follows from a theorem of Livesay and the Poincaré Conjecture that if $M$ is a compact connected non-orientable 3-manifold with $\pi_1(M)$ finite, then $M$ is homeomorphic to $P^2\times I$ minus a collection of disjoint open 3-balls.

The first answer to the original question that I wrote here was flawed. Here is a corrected version, thanks to the comments below.

First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.

If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$.

By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $q:N\to N$ be the covering transformation, an orientation reversing involution. Now $H_2(N;\mathbb{Q})\cong\mathbb{Q}^{k-1}$ and $H_2(N;\mathbb{Q})$ is generated by any $k-1$ components of $\partial N$. Since $q$ acts on each component of $\partial N$ as an orientation reversing homeomorphism, it follows that the Lefschetz number $\Lambda_q$ is given by $$\Lambda_q=1-0+(-(k-1))=2-k.$$ But $q$ has no fixed points, so $\Lambda_q=0$ and $k=2$. See Epstein for the original (and more general) proof of this step.

Therefore $N$ is homeomorphic to $S^2\times I$, and it follows by Livesay or Rubinstein that $M$ is homeomorphic to $P^2\times I$. The argument I had in mind here involves doubling $N$ along its boundary to get $S^2\times S^1$. The covering map $p$ also extends to the double. Then the classification of Seifert fibered spaces covered by $S^2\times\mathbb{R}$ tells us that the only compact quotient of $S^2\times S^1$ which contains a 2-sided projective plane is $P^2\times S^1$. Hence $M$ is homeomorphic to $P^2\times I$.

Yes.

First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.

If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$. Then $\chi(\partial M)=k$. But $\chi(\partial M)=2\chi(M)$ for any compact 3-manifold so $k$ is even.

By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $S$ be a 2-sphere embedded in $N$ which separates two components of $\partial N$ from the other $k-2$ components. Let $V$ and $W$ be the closures of the components of $N\setminus S$ such that $\partial V$ consists of $k-1$ 2-spheres and $\partial W$ consists of three 2-spheres.

Suppose $p(S)$ is a projective plane. Then $p(W)$ is a compact 3-manifold with $\chi(\partial p(W))=3$, a contradiction. Hence $p(S)$ is a 2-sphere, and $p(S)$ is separating in $M$.

Note that $M$ is prime. Otherwise $\pi_1(M)$ would split as a free product. But non-trivial free products are infinite (and non-abelian). So we again use the Poincaré Conjecture to infer that any simply-connected summand must be the 3-sphere.

Thus $p(S)$ must bound a 3-ball in $M$. Hence $p(V)$ is a 3-ball. Then $V$ is also a 3-ball, which implies that $k=2$. Therefore $N$ is homeomorphic to $S^2\times I$, and it follows that $M$ is homeomorphic to $F\times I$.

Yes, the result follows from a theorem of Livesay that if $M$ is a compact connected non-orientable 3-manifold with $\pi_1(M)$ finite, then $M$ is homeomorphic to $P^2\times I$ minus a collection of disjoint open 3-balls.

The first answer to the original question that I wrote here was flawed. Here is a corrected version, thanks to the comments below.

First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.

If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$.

By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $q:N\to N$ be the covering transformation, an orientation reversing involution. Now $H_2(N;\mathbb{Q})\cong\mathbb{Q}^{k-1}$ and $H_2(N;\mathbb{Q})$ is generated by any $k-1$ components of $\partial N$. Since $q$ acts on each component of $\partial N$ as an orientation reversing homeomorphism, it follows that the Lefschetz number $\Lambda_q$ is given by $$\Lambda_q=1-0+(-(k-1))=2-k.$$ But $q$ has no fixed points, so $\Lambda_q=0$ and $k=2$. See Epstein for the original (and more general) proof of this step.

Therefore $N$ is homeomorphic to $S^2\times I$, and it follows by Livesay or Rubinstein that $M$ is homeomorphic to $P^2\times I$. The argument I had in mind here involves doubling $N$ along its boundary to get $S^2\times S^1$. The covering map $p$ also extends to the double. Then the classification of Seifert fibered spaces covered by $S^2\times\mathbb{R}$ tells us that the only compact quotient of $S^2\times S^1$ which contains a 2-sided projective plane is $P^2\times S^1$. Hence $M$ is homeomorphic to $P^2\times I$.

First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.

If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$. Then $\chi(\partial M)=k$. But $\chi(\partial M)=2\chi(M)$ for any compact 3-manifold so $k$ is even.

By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $S$ be a 2-sphere embedded in $N$ which separates two components of $\partial N$ from the other $k-2$ components. Let $V$ and $W$ be the closures of the components of $N\setminus S$ such that $\partial V$ consists of $k-1$ 2-spheres and $\partial W$ consists of three 2-spheres.

Suppose $p(S)$ is a projective plane. Then $p(W)$ is a compact 3-manifold with $\chi(\partial p(W))=3$, a contradiction. Hence $p(S)$ is a 2-sphere, and $p(S)$ is separating in $M$.

Note that $M$ is prime. Otherwise $\pi_1(M)$ would split as a free product. But non-trivial free products are infinite (and non-abelian). So we again use the Poincaré Conjecture to infer that any simply-connected summand must be the 3-sphere.

Thus $p(S)$ must bound a 3-ball in $M$. Hence $p(V)$ is a 3-ball. Then $V$ is also a 3-ball, which implies that $k=2$. Therefore $N$ is homeomorphic to $S^2\times I$, and it follows that $M$ is homeomorphic to $F\times I$.

Yes.

First note that $M$ is non-orientable since it contains a 2-sided projective plane. Let $N$ be the orientation double cover of $M$, and let $p:N\to M$ be the covering map. Then $N$ is simply-connected since $\pi_1(M)\cong\mathbb{Z}/2\mathbb{Z}$.

If some component of $\partial M$ had non-positive Euler characteristic, then the same would be true of some component of $\partial N$, but this cannot happen since $N$ is simply-connected. By assumption, no component of $\partial M$ is a 2-sphere. Hence $\partial M$ is a collection of projective planes. Let $k$ be the number of components of $\partial M$. Then $\chi(\partial M)=k$. But $\chi(\partial M)=2\chi(M)$ for any compact 3-manifold so $k$ is even.

By the Poincaré Conjecture, $N$ is homeomorphic to a 3-sphere with the interiors of $k$ disjoint 3-balls removed. Let $S$ be a 2-sphere embedded in $N$ which separates two components of $\partial N$ from the other $k-2$ components. Let $V$ and $W$ be the closures of the components of $N\setminus S$ such that $\partial V$ consists of $k-1$ 2-spheres and $\partial W$ consists of three 2-spheres.

Suppose $p(S)$ is a projective plane. Then $p(W)$ is a compact 3-manifold with $\chi(\partial p(W))=3$, a contradiction. Hence $p(S)$ is a 2-sphere, and $p(S)$ is separating in $M$.

Note that $M$ is prime. Otherwise $\pi_1(M)$ would split as a free product. But non-trivial free products are infinite (and non-abelian). So we again use the Poincaré Conjecture to infer that any simply-connected summand must be the 3-sphere.

Thus $p(S)$ must bound a 3-ball in $M$. Hence $p(V)$ is a 3-ball. Then $V$ is also a 3-ball, which implies that $k=2$. Therefore $N$ is homeomorphic to $S^2\times I$, and it follows that $M$ is homeomorphic to $F\times I$.