$P^2$-irreducibility of a $3$-manifold

Question

A $3$-manifold $M$ is called $P^2$-irreducible if it is irreducible and there is no $2$-sided $P^2$ contained in $M$.

Can we show $M$ is $P^2$-irreducible iff $\pi_2(M)=0$?

Notice that one direction follows directly from the Sphere theorem.

Answer 1

Yes.

One direction is immediate by the Sphere theorem (projective plane theorem) as pointed out by the OP.

Assume $M$ satisfies $\pi_2(M)=0$. Note that this condition with the Poincare conjecture means any sphere in $M$ bounds a $3$-ball (see here). Therefore, we conclude that $M$ is irreducible.

Suppose $M$ contains a $2$-sided $P^2$. One considers the orientation double cover $M'\overset{p}{\longrightarrow} M$ and assume $p^{-1}(P^2)=S$ which is a sphere (not $P^2$s by our assumption). As above, $S$ bounds a $3$-ball $B$ and we consider $B\overset{p}{\longrightarrow} p(B)$. It is not hard to show this is a double covering map. However, the involution on $B$ contains a fixed point by Brouwer fixed-point theorem. We obtain a contradiction.

Answer 2

Suppose that $P \subset M$ is a two-sided real projective plane. Let $M'$ be the orientation double cover of $M$. Let $S$ be the two-sphere in $M'$ that double covers $P$.

Exercise A: $P$ is non-separating in $M$ if and only if $S$ is non-separating in $M'$.

Exercise B: If $S$ separates $M'$, then neither component of $M' - S$ is a three-ball.

We deduce, in both the separating and non-separating cases, that $\pi_2(M')$ is non-trivial; thus $\pi_2(M)$ is non-trivial.