Let $\mathbb{Z}_2=\{0,1\}$ act on $S^1\times S^2$ via $$1\cdot (w,z)=(\overline{w},-z),$$ with $S^1\subset \mathbb{C}$ and $\overline{w}$ is the conjugate of $w$. I'm trying to understand topologically who is $(S^1\times S^2)/\mathbb{Z}_2$.
At a first glance I was tempted to say that it was a product of a semi circle with a projective plane. But this action is identifying the antipodal points in distinct $2$-spheres. It seems that only when $w=\pm 1$ the associated spheres are projected onto a projective plane.
It is $\mathbb{R}P^3\#\mathbb{R}P^3$, since $\mathbb{R}P^3$ can be seen as a half hemisfere of $S^3$ after the identification given by the antipodal map, in which each meridian is a $S^2$ and the equator is $\mathbb{R}P^2$. That is, $\mathbb{R}P^3$ is the mapping cone of the canonical projection $\pi:S^2\to \mathbb{R}P^2$.
If one takes the pole of $\mathbb{R}P^3$ as the base point of the connected sum $\mathbb{R}P^3\#\mathbb{R}P^3$, we have exactly the double mapping cylinder given by $\pi:S^2\to \mathbb{R}P^2$, which is topologically the same thing that I have said in a the comment, $]0,1[\times S^2\cup \{0,1\}\times \mathbb{R}P^2$.