 Solution

The based lifting criterion isUnder the stated path-connected and locally path-connected domain hypotheses, it is necessary and sufficient for a continuous based lift. Two based coverings are equivalent if a homeomorphism between their total spaces takes one chosen basepoint to the other and commutes with their projections.

For the usual classification by subgroups, total spaces are taken to be connected. They are then path-connected and locally path-connected because they cover a locally path-connected base. If two such coverings have the same subgroup $H$, apply the lifting criterion to each projection to obtain based maps $F:\tilde{X}\to {\tilde{X}}^{\prime }$ and $G:{\tilde{X}}^{\prime }\to \tilde{X}$ over $X$. Their composites and the identity maps are based lifts of the same projections. Uniqueness of lifts on a connected domain gives $GF=\mathrm{id}$ and $FG=\mathrm{id}$. Thus $F$ is an equivalence, proving at most one connected based covering realizes a given subgroup.

The connected-total-space qualification is necessary. If arbitrary disconnected coverings are allowed, the identity cover of a circle and the disjoint union of two identity covers have the same subgroup at a basepoint in the first component, yet are inequivalent. The subgroup only records the chosen component in that setting.

The degree is the cardinality of a fibre. Over a path-connected base this is independent of the basepoint, by path lifting. For a connected total space it equals the subgroup index $[{\pi }_1(X,b):H]$, and it may be infinite.