Covering Spaces, Universal Covering Space

Morphisms Between Covering Spaces

Let S be a base space with two different covering spaces T and U. A continuous function h from T into U acts as a morphism between covering spaces if the projection from T onto S is the same as h followed by the projection from U onto S. In other words, the diagram commutes.

Verify that the identity map from T onto T is a morphism, and the composition of morphisms gives another morphism.

Covering spaces with base points also support morphisms. The morphism h must map the base point of T to the base point of U, where the respective base points are in the fiber of the base point of S. Again, the identity map is a morphism, and the composition of morphisms gives another morphism.

A Bijective Morphism is Invertible

We know that h and its inverse are compatible with the projections onto S; we only need show h inverse is continuous.

Pull any point c in U back to b in T. Let a = T(b), which is the same as U(c). Let W be an open set about b whose image evenly coveres a. In other words, W is homeomorphic to an open set V in S, such that V contains a. Restrict V to a possibly smaller open set that is evenly covered under U. Lift this to an open set about c, which lies in the image of h(W). This can be done for every point in an open set in T, hence the image of an open set is open, and h is bicontinuous. Its inverse is a morphism from U to T.

Let g and h be inverse morphisms, so that their composition implements the identity map on T. Both h and g must be 1-1, and onto, hence h is a bijection, and g is the inverse of this bijection.

Morphisms as Lifts

The previous theorem can sometimes be used to create a morphism between covering spaces. Let S~ and D be covering spaces for S, such that D is connected, and let π(D) embedded in S be a subgroup of π(S~) embedded in S. Map any point in the fiber of x in D to any point in the fiber of x in S~ and lift the projection of D up to a compatible function from D into S~. This function implements a morphism from D into S~.

Conversely, a morphism between covering spaces is a lift of the projection of the first up to the second. Thus, if a morphism maps D into S~, and D is connected, it is possible to embed π(D) into π(S~).

Category

Given a space S, or a space S with a designated base point a, the covering spaces form a category. verify that the morphisms described above satisfy the conditions of a category.

Equivalent objects in this category are equivalent covering spaces, with inverse morphisms between them. We showed above that these morphisms implement a homemomorphism. Conversely, a morphism that implements a homeomorphism is invertible, and makes the covering spaces equivalent.

Lift Becoms Equivalence

Let S~ and T~ be connected covering spaces of S, presenting the same fundamental group G. For some base point x, π(S~) and π(T~) both map onto G, based at x. Let x~ be a point in the fiber of x in S~, and let y~ be a point in the fiber of x in T~. The lifting criteria are satisfied, hence a morphism from S~ into T~ carries x~ to y~. At the same time, a morphism from T~ into S~ carries y~ back to x~. The composition produces a third morphism from S~ into itself that fixes x~. This morphism is a lift of the projection up to S~, and two lifts are equal or everywhere distinct. This lift agrees with the identity map on x~, hence it is the identity map. The original morphisms are inverses of each other, and the covering spaces are homeomorphic, or isomorphic, or equivalent. Such an equivalence can be constructed for any x~ and any y~.

Universal Covering Space

Since pointed covering spaces form a category, we can say that U is a universal covering space if it is a universal object within its category. Now, if you don't know what categories are, here is the translation. For any covering space T and base point c, there is a unique morphism from U to T, mapping b to c.

Notice that a morphism from U to T is actually the projection of U onto S, lifted up to T. In the previous theorem we showed this lift is unique, hence if the morphism exists it is unique.

There is but one morphism from U into itself, fixing b, namely the identity map.

Two universal spaces are homeomorphic. This is an immediate consequence of category theory, or you can prove it manually. Show that the two universal spaces lift each other, and the lifts compose to a lift that must be the unique lift from U into U, namely the identity map. The lifts are inverses of each other, and the two universal spaces are homeomorphic.

Topologically, there is but one universal covering space.

A Simply Connected Covering Space is Universal

If U is simply conneccted, the function from U onto S can be lifted up to any other covering space T. We proved this in the previous theorem. Therefore U is universal.

For example, R1 is universal over S1, and Sn is universal over Pn for n > 1.

A Universal Cover for the Figure 8

Start at the origin and draw four segments of length 1, heading north south east and west.

Head east from the origin, out to a distance of ¾, and draw a verticle segment of height ½ across this line. In other words, it goes up ¼ and down ¼. Then move west from the origin, through a distance of ¾, and draw another vertical segment of height ½, ¼ up and ¼ down. Then draw horizontal segments of length ½ across the north and south arms, at hights ±¾.

There are now 12 dangling ends to this shape. Select any of these 12 tails and let it have length t. (In this case the length is always ¼.) Move out along the tail through a distance of ¾t, and draw a segment of length ½t across the tail. This creates two more tails of length ¼t, and it cuts the original tail to a length of ¼t. Do this for all twelve tails and build a shape that has 36 tails. Then do the same thing for all 36 tails to produce 108 tails, and so on, creating a fractal.

Verify that this shape is simply connected.

Map the origin to the point of tangency between the two circles. As you move east through a distance of ¾, to the next intersection, a projection wraps this segment around the right circle counterclockwise. Moving west to the next intersection wraps the segment around the right circle clockwise. Moving north to the next intersection wraps the segment around the left circle counterclockwise, while moving south implements a clockwise rotation. Extend this map to the next intersection, and the next, until it is defined across the entire fractal.

show that this projection is continuous, and that it evenly covers every point on the figure 8. Thus our fractal is a universal covering for the figure 8.

If S consists of three circles, tahgent at a point, build a fractal in 3 dimensions. Start with 6 arms pointing up down left right front and back. At every iteration, two mutually perpendicular lines are added to each tail, thus multiplying the number of tails by 5. Repeat this indefinitely, creating a fractal in R3. Map the six directions to loops, clockwise and counterclockwise, around the three circles. This is a universal cover for S.

The above can be generalized to n circles, tangent at a common point.