= 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 \b[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 \b[two homeomorphism types of total space]. This is the distinction between equivalence of covers and homeomorphism of covering spaces.
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