Covering Spaces, Covers and π sets are Equivalent

From One Category to Another

Let S be a path connected, semilocally simply connected space. As you recall, the covers of S form a category, which I will call C₁. The morphisms are continuous maps from one cover into another, such that the diagram commutes. In this section we're going to develop a covariant functor from C₁ into C₂, a category of groups and sets. This is what you would expect from algebraic topology - a correspondence between topological spaces and algebraic structures. If you're not familiar with category theory, you might want to check it out. If you enjoy category theory then you know the drill. We're going to define C₂, map the objects of C₁ (i.e. the covers of S) into the objects of C₂, map the morphisms of C₁ into the morphisms of C₂, and show that the composition of morphisms in C₁ leads to the composition of morphisms in C₂. (Since morphisms are actually functions on sets, we get this last part for free.)

Defining C₂

If G is a group, the category of G sets is the category of sets such that G acts on each set, and morphisms respect the action of G. For example, let f be a morphism from A into B. Let x be an element of A, and let q be an element of G that acts on A. If q takes x to y, then q, acting on B, takes f(x) to f(y). We can apply f, or the action of G, in either order. Verify that this is indeed a category. The composition of morphisms gives another morphism, and the identity map is a morphism.

Does G act in the forward or backwards direction? It doesn't matter, because the two categories are equivalent. Invert the permutations across the board, and the objects of one category become the objects of the other. As functions on sets, the morphisms remain the same. So, for our convenience, consider the category of G sets where the action runs forward, i.e. the permutation of e*f is the permutation of e followed by the permutation of f.

In the following, G is the fundamental group π(S), and the G sets are called π sets. In other words, C₂ is the category of π sets, where π is the fundamental group of S at some base point x. Since S is path connected, selecting another base point would produce a group isomorphic to π. This will come into play later, when we compress each equivalence class down to a single object. For now, assume x is a fixed base point in S, and define π(S) at x.

Mapping Covers to π sets

Let S~ be a covering space for S. The corresponding set is now the fiber of x. If x~ is a point in the fiber of x, and e is a loop in π(S), then e acts on x~ by the path lifting theorem, carrying x~ to a new point in the fiber of x. Thus x~ is at the beginning of e~, and the action of e caries x~ forward to the end of e~. Now, applying the loops e and f in sequence is the same as applying the composite loop e*f. Thus the loops act on the fiber of x.

Since S is evenly covered, the points in the fiber of x can be housed in disjoint open sets. Thus the fiber of x attains the discrete topology. Suppose e and f are homotoppic loops that move x~ to two different points. Lift the homotopy, and the end of the path, as it evolves from e~ to f~, traces a continuous path that jumps from one point in the fiber of x to another. Since the fiber of x is discrete, this is a contradiction. The action of e and f is the same, and the action is indeed a function of the group π(S).

The action is transitive iff S~ is connected. In fact, the components of S~ become disjoint orbits under the action of π.

Mapping the Morphisms

A morphism from S~ into T~ carries the fiber of x in S~ into the fiber of x in T~. Apply the action of e, then the morphism; or apply the morphism first. Either way you have the lift of e in T~, and the final destination of x~ is found at the end of this lifted path. The induced map is a morphism in C₂. this completes the construction of the functor from C₁ into C₂.

Change the Base Point

Let y be another base point in S and let l be a fixed path from x to y. Lift l to map the fiber of x to the fiber of y. The action of e on a point in the fiber of y can be found by running along l~ in reverse, then e~, then l~. Thus π acts on the fiber of y, and a bijection, determined by l, connects the fiber of x with the fiber of y, in a manner that respects the action of π. Finally, the fundamental group at y is isomorphic to the fundamental group at x. Had we chosen y instead of x, we would find the same group, and the same π set, up to isomorphism.

Equivalence Classes become Objects

Compress each equivalence class in C₂ down to a representative object. (This can be done for any category.) If two π sets are equivalent, they are really the same object in C₂.

Combine this witht the previous result and find base point invariance. For any base point x, the map from C₁ into C₂ is the same up to isomorphism, and determines the same (compressed) object in C₂.

Next, compress the equivalence classes of C₁ down to objects. Two covers that are isomorphic are essentially the same cover. Fortunately these covers induce a bijection on their associated π sets, so the two covers imply the same (up to isomorphism) π set in C₂. In other words, the map from C₁ into C₂ is well defined.

Surjective

We will show that C₁ maps onto C₂. In other words, every possible π set has a corresponding cover.

Start with a π set and separate it into orbits. If we can find a cover for each orbit, the disjoint union of these covers becomes a covering space for the entire π set. So we may assume the π set consists of one orbit. In other words, π acts transitively.

Select an element d in the π set and let H be the stabilizing subgroup that fixes d. If e is a member of H, and f moves d somewhere else, then e*f moves d to the same place. The right cosets of H carry d to all the points in the π set.

Now we need S to be semilocally simply connected, in order to apply the previous theorem. Create a cover whose fundamental group is H, with a fiber defined by the right cosets of H. Using our map between categories, this cover creates the desired π set. Each π set has a corresponding cover.

Let U~ be the universal cover. Remember there is just one in C₁, because equivalent covers have been compressed down to a single object. This cover was described in the previous section. It is simply connected, and the cosets are all of π. In other words, the set is the group itself. Verify that the action of π on itself is right translation. Conversely, such a π set leads to a cover with a trivial fundamental group - a connected and simply connected cover - a universal cover.

Injective

If the objects of C₁ remain distinct in C₂, then the two categories are equivalent. Suppose two distinct covers S~ and T~ lead to equivalent π sets. The map between π sets carries orbits to orbits. These determine connected components in S~ and T~ respectively. Two covering spaces are equivalent iff they are equivalent per component. It is enough to look at each component. In other words, we may assume S~ and T~ are connected spaces, and π acts transitively on the corresponding (equivalent) π sets in C₂.

When S~ over S is mapped into C₂, a base point is assumed. Move this base point to x in S. The result is a π set that is equivalent to the original. Do the same for T~, and the two π sets remain equivalent. In other words, we may assume both S~ and T~ use the same base point x in S to create their π sets in C₂.

Let H be the subgroup of π that fixes the π set, in other words, the stabilizing subgroup. Since the bijection respects group action, the same stabilizing subbgroup H applies to both π sets, implied by S~ and T~.

Start at x and run through any loop in H. The result fixes the fiber of S~, and the fiber of T~. Furthermore, any loop outside of H moves the fiber of S~ and of T~. The fundamental group of S~ embeds in π, and maps onto H, and is equal to H. The same is true of T~. That's a good sign, since we want S~ and T~ to be homeomorphic.

Start with x~ in the fiber of x in S~, and let the function f map x~ to the point in T~, in the fiber of x, that is indicated by the map on π sets. In other words, equivalence in C₂ already tells us how to map the fiber of x in S~ to the fiber of x in T~. This will seed the homeomorphism f between S~ and T~.

Since the fundamental groups of S~ and T~ both become H in S, and S~ and T~ are connected, and f is defined on one point, the associated isomorphism f can be created. In other words, S~ and T~ are equivalent, and they represent the same object in C₁. The two categories, C₁ and C₂, are equivalent.

This is a bit surprising. When we first defined a covering space, the possibilities seemed endless. However, the condition of locally homeomorphic is a substantial constraint. There is in fact one cover for each π set, and if we restrict attention to connected covers, there is one cover for each transitive π set. This includes every quotient group.

Apply this to the circle S¹, whence π = Z. Let 1, the generator of Z, move an element through its orbit. If this orbit is infinite then the π set is Z, acting on itself. This is the universal cover R¹.

If the orbit is trivial then the cover is S¹.

Finally let the orbit have size n, whence H, generated by n, is the stabilizing subgroup. The fundamental group of the cover is still Z. Geometrically, the cover looks like a circle wrapped around S¹ n times. Since S¹ is well behaved, i.e. path connected and locally simply connected, these are the only connected covers up to isomorphism.