A continuous map of 3-dimensional manifolds $f \colon M^3 \to N^3$ is called a branched covering if there is a link $L \subset N^3$, such that the restriction $f \colon M \setminus f^{-1}(L) \to N \setminus L$ is a covering map in the usual sense. In that case we say that $f$ is branched along $L$, and the degree of $f$ is the degree of the aforementioned restriction. A classical result by Hilden and Montesinos states that each closed orientable 3-manifold $M$ can be represented as a branched covering $M \to S^3$ over the 3-sphere branched along a knot.
In the non-orientable case Berstein and Edmonds showed that each compact non-orientable manifold $N$ without boundary can be represented as a branched covering over $N_1=\mathbb{R}P^2 \times S^1$. Moreover, a non-orientable manifold $M$ branched covers $N_2=$ the non-trivial $S^2$-bundle over $S^1$ iff the Bockstein $\in H^2(M,\mathbb{Z})$ of its first tangent Stiefel-Whitney class is zero. The question is: is there a branched covering map from $S^3$ to at least one of the manifolds $N_1,N_2$? If this turns out to be true, then every compact 3-manifold without boundary, orientable or not, would be a branched covering of $\mathbb{R}P^2 \times S^1$.
Best Answer
A branched covering must induce a surjection on rational homology, which one can see, for example, by the existence of a transfer homomorphism. So there is no branched covering $S^3 \to \mathbb{R}P^2 \times S^1$.