Graph covers and semi-covers: Who is stronger?

The notion of graph cover, also known as locally bijective homomorphism, is a discretization of covering spaces known from general topology. It is a pair of incidence-preserving vertex- and edge-mappings between two graphs, the edge-component being bijective on the edge-neighborhoods of every vertex and its image. In line with the current trends in topological graph theory and its applications in mathematical physics, graphs are considered in the most relaxed form and as such they may contain multiple edges, loops and semi-edges. Nevertheless, simple graphs (binary structures without multiple edges, loops, or semi-edges) play an important role. It has been conjectured in [Bok et al.: List covering of regular multigraphs, Proceedings IWOCA 2022, LNCS 13270, pp. 228--242] that for every fixed graph $H$, deciding if a graph covers $H$ is either polynomial time solvable for arbitrary input graphs, or NP-complete for simple ones. A graph $A$ is called stronger than a graph $B$ if every simple graph that covers $A$ also covers $B$. This notion was defined and found useful for NP-hardness reductions for disconnected graphs in [Bok et al.: Computational complexity of covering disconnected multigraphs, Proceedings FCT 2022, LNCS 12867, pp. 85--99]. It was conjectured in [Kratochv\'{\i}l: Towards strong dichotomy of graphs covers, GROW 2022 - Book of open problems, p. 10, {\tt https://grow.famnit.upr.si/GROW-BOP.pdf}] that if $A$ has no semi-edges, then $A$ is stronger than $B$ if and only if $A$ covers $B$. We prove this conjecture for cubic one-vertex graphs, and we also justify it for all cubic graphs $A$ with at most 4 vertices.