The Lifting Criterion

A central question in the theory of covering spaces is determining when a map can be “lifted” to a map such that . The answer is purely algebraic.

Lifting Criterion

Let be a covering space. If is a path-connected and locally path-connected space, a continuous map can be lifted to a map if and only if the image of its induced homomorphism is a subgroup of the image of :

Proof Sketch Necessity (): If a lift exists, then . Applying the fundamental group functor gives , which directly implies that . Sufficiency (): Assuming the subgroup condition holds, we construct . For any point , we choose a path from to . The path in is lifted to a unique path in starting at . We define as the endpoint of this lift.

  • Well-definedness: If we choose a different path from to , the loop is mapped by into . This condition guarantees that the lifts of and have the same endpoint.
  • Continuity: The local path-connectedness of is essential to prove that the constructed map is continuous.

Proposition

Suppose is a connected space and we have two lifts of a map . If there is a point such that , then the lifts are identical, i.e., .

Proof Let and . The sets and are disjoint, their union is , and they can be shown to be open. By assumption, is non-empty. Since is a connected space, it cannot be the union of two disjoint non-empty open sets. Therefore, we must have , which implies for all .

Universal Property of Universal Cover

If a universal cover of exists and is locally path-connected, then for any other connected covering space of , there exists a unique lift (covering map) . universal_property_universal_cover

Proof Note that a universal cover is automatically path-connected and locally path-connected. So that we can apply the lifting criterion. And by the above proposition, this lifting is unique.

Morphisms of Covering Spaces

A morphism of covering spaces (covering space map) from to is a continuous map such that . map_of_covering_spaces An isomorphism of covering spaces is a covering space map for which there exists an inverse map such that and .

Uniqueness of the Universal Cover

If a locally path-connected space has two universal covers, they are isomorphic.

Proof The following diagram tells the story: universal_cover_uniqueness

Existence of the Universal Cover

Semi-Locally Simply-Connected

A space is semi-locally simply-connected if for any point and any open neighborhood of , there exists a smaller open neighborhood with such that the map induced by the inclusion is trivial.

Fundamental Theorem of Covering Spaces

If a space is path-connected, locally path-connected, and semi-locally simply-connected, then a universal cover of exists.

Proof Sketch We can construct as the set of homotopy classes of paths in starting at a fixed basepoint . An element in is denoted as , where with . The basepoint of is , the homotopy class of the constant path at . The projection map is defined by mapping a path class to its endpoint: . The topology on is defined using a basis of open sets. For any point and any open neighborhood of that satisfies the SLSC condition (i.e., is trivial), and for any point , we define a basis open set: These sets form a basis for the topology on . With this topology, the projection restricted to is a homeomorphism onto , and the full map is the desired universal covering map. (The detailed point-set topology proof is omitted.)

Categorical Perspective

We can see immediately from the above that, the covering spaces of a given topological space form a category . It is a subcategory of the slice category . Then the fundamental theorem of covering spaces can be interpreted as follows: If a space is path-connected, locally path-connected, and semi-locally simply-connected, then the category is equivalent to the category of (or functors from the fundamental groupoid to ).

Also, the universal property of the universal cover can be interpreted as the universal cover is the initial object of the category of connected covering spaces of a “nice” space .