Let $G$ be a simple graph and $F$ be a closed surface. I will say that two embeddings $f_1 : G \to F$ and $f_2 : G \to F$ are equivalent if there is a homeomorphism $h$ of $F$ and an automorphism $\sigma$ of $G$ such that $h \circ f_1 \circ \sigma = f_2$. Let $q : S^2 \to \mathbb{R}P^2$ be the canonical 2-sheeted covering.
I was reading the paper "Enumertion of projective-planar embeddings of graphs" by Negami and he states the following is clear: if $f_1$ and $f_2$ are equivalent embeddings of a (3-connected, but I don't believe that is being used for what I will now say) graph into $\mathbb{R}P^2$, then the graphs $q^{-1}(f_1(G))$ and $q^{-1}(f_2(G))$ are isomorphic.
Why is this true? Maybe I am missing some very basic fact about covering spaces.
The basic fact is that $q$ is a universal covering map, since $S^2$ is path-connected and simply connected.
Suppose $f_1$ and $f_2$ are equivalent embeddings of $G$ into $\mathbb{R}P^2$, i.e., there is a homeomorphism $h$ of $\Bbb RP^2$ and an automorphism $\sigma$ of $G$ such that $h \circ f_1 \circ \sigma = f_2$.
Since $q:S^2 \to \Bbb{R}P^2$ is a covering map and $h$ is a homeomorphism of $\Bbb RP^2$, $h\circ q$ is also a covering map. Sincd $q$ is universal, $h$ can be lifted to an auto-homeomorphism $\alpha$ of $S^2$ such that $$q\circ\alpha=h\circ q$$
Let $s\in q^{-1}(f_1(G))$. So $q(s)=f_1(g)$ for some $g\in G$. Then $$f_2(\sigma^{-1}(g))=(h\circ f_1\circ \sigma)(\sigma^{-1}(g))\\=h\circ f_1(g)=h\circ q(s)=q\circ \alpha(s),$$ which means $\alpha(s)\in q^{-1}(f_2(G))$.
Symmetrically, $\alpha^{-1}$ maps $q^{-1}(f_2(G))$ to $q^{-1}(f_1(G))$.
We can check that $\alpha$ induces a homeomorphism between $q^{-1}(f_1(G))$ and $q^{-1}(f_2(G))$ as topological subspaces of $S^2$.