 Solution

The fundamental group of the figure-eight is the free group on its two circle loops $a,b$. A two-sheeted covering space is specified by their permutations of the two-point fiber. It is connected exactly when this action is transitive, equivalently when at least one permutation exchanges the sheets. Since $S_2$ is abelian, relabeling the two sheets does not identify any further cases. There are therefore three connected double covers, given by $(a,b)=(\tau ,1),(1,\tau ),(\tau ,\tau )$, where $\tau$ exchanges the sheets.

In either of the first two covers there are two vertices joined by two parallel edges, with a loop at each vertex. The two covers differ as covers of the fixed base, but their total spaces are homeomorphic by exchanging the base-circle labels. The third cover has two vertices joined by four parallel edges and no loops.

These last two graph types are not homeomorphic. Their two degree-four branch points are intrinsically identifiable topologically. Removing either branch point in the graph with loops leaves two components: the punctured loop at that point, and the remaining connected graph. Removing either branch point in the four-edge graph leaves a connected four-armed graph. Thus the three covers give two homeomorphism types of total space. This is the distinction between equivalence of covers and homeomorphism of covering spaces.