A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES

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1 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES YAN SHUO TAN Abstract. In this aer, we try to give as comrehensive an account of covering saces as ossible. We cover the usual material on classification and deck transformations, and also show how to erceive the subject from a more abstract categorical view oint. The reader is assumed to ossess a working knowledge of basic toology and category theory. Contents 1. Introduction 1 2. Preliminaries 2 3. Lifting Proerties 3 4. Classification of Covering Saces - Isomorhism Conditions 5 5. Classification of Covering Saces - Existence of Covers 7 6. Classification of Covering Saces - A Lattice Isomorhism Interlude - Homotoy Relationshis with the Base Sace Covering of Grouoids Grou Actions and Grouoid Actions Some Equivalences of Categories Orbit Saces and a Galois Theory of Covering Saces Postlude - Some Philosohical Remarks 25 Acknowledgments 26 References Introduction The theory of covering saces is an old subject that is canonical material for most introductions to algebraic toology. It is, however, sometimes seen to be not terribly interesting in its own right, but instead useful mostly as a tool for roving a motley of results, such as that the fundamental grou of the circle is isomorhic to the integers and that every subgrou of a free grou is free. Part of the oint of this aer is to argue otherwise. I seek to show that there is value in attaining a dee understanding of covering saces not just so that one can better use it as a tool, but also because the theory embodies the following recurrent theme in algebraic toology: The same subject material can often be resented at different levels of abstraction, and it is shuttling between these that hels us to understand and contextualize the subject by revealing its similarities and differences with other mathematical theories. 1

2 2 YAN SHUO TAN To resent this icture, I have organized this aer into two arts. In the first, I set down the foundations of the subject as they are traditionally resented as in [7] and [3]. After a brief interlude, I take u the more abstract aroach of [6] in the second art. This chronology of eeling layers off the subject to reveal an inner skeleton can be taken as a way of interreting the title to this aer. There is an element of covering sace theory that has the air of folklore, and that is the oft-cited analogy between the classification of covering saces and the corresondence between subfields and subgrous in Galois theory. I have not managed to find a single source that deals rigorously with what this analogy actually is, and it turns out that there are in fact two distinct ways in which authors have drawn analogies between the two subjects. As such, the second aim of this aer is to demystify this confusion by discussing and distinguishing these two threads of argument. This rather unfortunate state of affairs is also the second way of interreting the title. There is, in fact, a third way, which we shall reveal at the end of this aer. 2. Preliminaries Definition 2.1. Let : E B be a surjective ma. The oen set U of B is said to be evenly covered by if the inverse image 1 (U) can be written as the union of disjoint oen sets V α in E such that for each α, the restriction of to V α is a homeomorhism of V α onto U. The collection {V α } is called a artition of 1 (U) into slices. Definition 2.2. Let : E B be a surjective ma. If every oint b of B has a neighborhood U that is evenly covered by, then is called a covering ma, and E is said to be a covering sace of B. We shall follow the convention of the extant literature and sometimes shorten these two terms to covering and cover resectively. Given a oint b B, we denote its reimage by F b and call it the fiber of above b. Note that a cover is insearable from its associated covering. We hence often consider both as a single object and simly call it a covering. Examle 2.3. Let : R S 1 be defined by (t) = (cos 2πit, sin 2πit). Then every oen arc sanned by an angle of radian less than π is an evenly covered by, so is a covering ma. Examle 2.4. Now define : S 1 S 1 to be the n-th ower ma z z n,where this time we view S 1 as being embedded in C. Once can check that neighborhoods of the form S 1 {z} for some z are evenly covered. Letting n run over the natural numbers, these two examles exhaust all ossibilities for covering saces of S 1, u to some suitable notion of isomorhism. This fact may seem surrising at first, but will be made aarent after we classify covering saces, which gives a corresondence between the baseoint-reserving isomorhism classes of covering saces of a given sace, and the subgrous of the fundamental grou. The following two easy observations allow us to enlarge our ool of examles for covering saces.

3 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 3 Proosition 2.5. Let : E B be a covering ma, let B 0 be a subsace of B, and write E 0 = 1 (B 0 ). Then 0 : E 0 B 0 obtained by restricting is a covering ma. Proosition 2.6. Let : E 1 B 1, q : E 2 B 2 be covering mas, then their roduct q : E 1 E 2 B 1 B 2 is also covering ma. Corollary 2.7. The usual rojection : R 2 T is a covering ma for the torus. 3. Lifting Proerties Given a covering sace : E B, we know that aths and homotoies in E roject down to aths and homotoies in B. More generally, any ma f : X E from another sace X into E induces a ma f : X B. But what can we say when we start from the oosite direction, i.e. when we are given a ma g : X B instead? It turns out that aths and homotoies can always be lifted, but general mas require additional conditions on the domain sace. Lemma 3.1. Let : E B be a covering ma and γ : I B aathstartingata oint b 0. Then for every oint e 0 1 (b 0 ), there is a unique lift γ : I E such that γ(0) = e 0. Proof. Pick an oen cover U for B consisting of evenly covered neighborhoods. Using the Lebesgue number lemma, we can ick a subdivision of [0, 1], s 0,...,s n such that each subinterval [s i,s i+1 ] is maed by γ into a single oen set U i U. We now roceed by induction: Suose that we have defined a lift for γ restricted to [0,s j ], for 0 j n 1, we claim that we can extend this to a lift for γ defined over [0,s j+1 ]. Consider the set U j in which [s j,s j+1 ] lies. Its reimage consists of a collection of disjoint oen sets. Now, we have already defined γ(s j ) to lie in one of these, call it V. Since V : V U i is a homeomorhism, we can then define a continuous function g :[s j,s j+1 ] E by setting g(s) =( Vα ) 1 (f(s)). By the asting lemma, combining g and γ yields a continuous function on [0,s j+1 ]. Relabelling the entire function γ, we clearly have γ = γ. We now rove uniqueness via the same ste by ste argument. Suose we have two lifts of γ, call them γ and γ that agree on [0,s j ], with V the unique slice of 1 (U j ) containing γ(s j )= γ (s j ). Then since γ([s j,s j+1 ]) and γ ([s j,s j+1 ]) are each connected, they must both lie in V. By the local homeomorhism roerty of, each oint γ(t) for s j t s j+1 must have a unique reimage in V,which imlies that γ and γ agree over [0,s j+1 ]. Lemma 3.2. Let : E B be a covering ma, and H : I I B amawith H(0, 0) = b 0. Then for every oint e 0 1 (b 0 ), there is a unique lift H : I I E such that H(0, 0) = e 0. Proof. We use the exact same argument as in the revious roosition, noting this time that we can subdivide the unit square into subsquares such that the image of each square under H is contained in an evenly covered neighborhood of B. Now, note that we have roved the roosition for general mas from the unit square into the base sace B. In the secial case of ath homotoies, H {0} I is a constant ath in B, which lifts to the constant ath in E at e 0. Since H {0} I

4 4 YAN SHUO TAN is also a lift, by the uniqueness of ath liftings, it must be the constant ath. By the same argument, H {1} I is the constant ath at e 0. Hence, H is a homotoy of aths. This fact has imortant imlications. In articular, it says that if two aths are homotoic in the base sace, their lifts at a given starting oint are also homotoic and in articular must have the same end oint. As a secial case, if a loo in the base sace lifts to a loo in the covering sace, then its entire homotoy class does as well. We can now rove the following imortant theorem, which I shall follow [3] in calling the Lifting Criterion. Theorem 3.3 (Lifting Criterion). Let :(E,e 0 ) (B,b 0 ) be a covering ma, and f :(X, x 0 ) (B,b 0 ) any ma with X ath-connected and locally ath-connected. Then a lift f :(X, x 0 ) (E,e 0 ) exists iff f (π 1 (X, x 0 )) (π 1 (E,e 0 )). Furthermore, if such a lift exists, it is unique. Proof. Suose there is a lift f such that f = f. Then alying the fundamental grou functor gives f = f, so the only if statement follows. Given x X, choose a ath γ in X from x 0 to x. Then fγ is a ath in B beginning at b 0, and we can lift it to a ath fγ in E beginning at e 0. We also have that fγ is a ath in E beginning at e 0, but since lifts of aths are unique, we must have fγ = fγ, and in articular f(x) = fγ(1) = fγ(1), which is indeendent of the choice of f, which tells us that f is unique. We now rove the if statement. Given x X, choose a ath γ in X from x 0 to x and define f(x) tobefγ(1) where fγ is the unique lift of fγ to E beginning at e 0. To show that this is well-defined, let γ be another ath from x 0 to x in X. Then γ γ is a loo in X based at x 0,sofγ fγ = fγ fγ = f(γ γ ) is a loo in B based at b 0. In articular, its homotoy class is in the image of f, and hence the image of by assumtion. This imlies that fγ fγ is homotoic to a loo in B that lifts to a loo in E based at e 0, but since ath homotoies lift to ath homotoies, we see that fγ fγ itself lifts to a loo α in E based at e 0. Note that we can write α = β β where β and β are lifts of fγ and fγ resectively. By the uniqueness of lifted aths, we need to have β = fγ, and similarly, since β is a lift of fγ starting at e 0,wemust have β = fγ. In articular, this means that fγ and fγ have the same end oint. To rove continuity of f at x X, choose an evenly covered neighborhood U of f(x) inb, and let V be the slice of 1 (U) containing f(x). It suffices to show that we can ick a neighborhood W of x that is maed by f into V. Let 0 : V U be the homeomorhism obtained by restricting. Sincef is continuous, we can ick W such that f(w ) U. Then given x W,wehave f(x) = 1 0 f(x) V by unique lifting, which comletes the roof. Local ath-connectedness is a necessary condition in the Lifting Criterion. To see why, consider the following counterexamle. Examle 3.4. Let W be the quasi-circle. It is the subsace of R 2 comrising the closed toologist s sine curve (the grah of y =sin(1/x) defined for 0 x 1with the line segment [ 1, 1] in the y-axis attached) with its two ends connected via an arc. This sace is ath-connected but not locally ath-connected. Now let q be the quotient ma that collases [ 1, 1] to a oint. Note that q can be viewed as

5 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 5 a ma from W to S 1 because it takes the comactification of R with [ 1, 1] to its one-oint comactification. But while π 1 (W ) = 0, one can show that q does not lift to the covering sace : R S 1 by arguing that any lift must ma [ 1, 1] to a coy of the integers in R, which causes roblems with injectivity on the rest of W. 4. Classification of Covering Saces - Isomorhism Conditions Given a base sace B, we want to classify all its covering saces and the mas between them. To do this, we need to establish a suitable notion for a ma between covering saces. Since classification in mathematics is always done u to isomorhism tye, we also need to define what it means for two covering saces to be isomorhic. Definition 4.1. Let : E B and : E B be covering mas, and let g : E E be a ma such that the following diagram commutes: E g E B We then say that g is a ma of coverings over B. Ifg is in addition a homeomorhism, we say that it is an isomorhism of coverings. Remark 4.2. Many oular texts, in articular [3] and [7], omit the notion of a ma of coverings, focusing solely on the cases where there is an isomorhism. This is unfortunate because doing so loses a lot of the richness of covering sace theory. It turns out that the existence and uniqueness statements in the Lifting Criterion are recisely the tools we need to determine when mas of coverings exist: Taking the domain sace X to be not just any sace, but another covering sace for the base sace B, we can classify mas between covering saces, and taking X to be E allows us to classify endomorhisms. Examle 3.4 tells us, however, that this analysis is only ossible when the covering sace E is locally ath-connected. By the local homeomorhism roerty, this means that we also require B to be locally ath-connected. If this is the case, B can be decomosed into disjoint oen sets B α that are its ath comonents, and Proosition 2.5 tells us that the mas 1 (B α ) B α obtained by restricting are also covering mas. This means that, for a given base sace, we lose no information by considering its ath comonents individually, so we may go ahead and assume that B is ath-connected. Finally, we make the following easy observation: Proosition 4.3. Let B be ath-connected and locally ath-connected. Let : E B be a covering ma. If E 0 is a ath-comonent of E, then 0 : E 0 B obtained by restricting is still a covering ma. Once again, we lose no information by considering the ath-comonents of the covering sace individually. With these observations in mind, therefore, we shall make the following: Convention 4.4. Unless stated otherwise, all saces are assumed to be both athconnected and locally ath-connected.

6 6 YAN SHUO TAN For the remainder of this section, we will answer the question of when two covering saces are isomorhic. As one might susect from the Lifting Criterion, it turns out that this has everything to do with the fundamental grous of the covering saces, and their images under the homomorhisms induced by covering mas. Theorem 4.5. Let :(E,e 0 ) (B,b 0 ) and :(E,e 0 ) (B,b 0 ) be covering mas. There is an isomorhism of coverings h :(E,e 0 ) (E,e 0 ) if and only if the subgrous H = (π 1 (E,e 0 )) and H = (π 1 (E,e 0)) of π 1 (B,b 0 ) are equal. If h exists, it is unique. Proof. Suose such an isomorhism h exists. Then we have = h and by taking inverses, = h 1. These resectively imly that H H and H H, so the two grous are equal. To rove the converse statement, assume that H = H. We shall aly the Lifting Criterion four times. First, lifts to : (E,e 0 ) (E,e 0), and lifts to :(E,e 0) (E,e 0 ), which gives the following two commutative diagrams: E E E E B This shows that and are mas of coverings. We also have = =, so is an isomorhism of E with itself taking e 0 to e 0. Since the identity is also such an isomorhism, the uniqueness art of the Lifting Criterion tells us that = id E. Analogously, we get = id E. Hence, and are inverse isomorhisms. This theorem tells us that two based covering saces have a base-oint reserving isomorhism between them if and only if they are associated with the same subgrou of the π 1 (B,b 0 ). It is imortant to note, however, that we have not determined when two covering saces are isomorhic in general, i.e. when we do not secify base oints. We consider this roblem now. Lemma 4.6. Let : E B be a covering ma. Let e 0 and e 1 be oints of 1 (b 0 ), and let H i = (π 1 (E,e i )) for i =0, 1. Then H 0 and H 1 are conjugate. Conversely, given e 0, and given a subgrou H of π 1 (B,b 0 ) conjugate to H 0, there exists a oint e 1 of 1 (b 0 ) such that H 1 = H. Proof. We rove the first statement. Recall that we have assumed E to be athconnected, so we can find a loo γ in B based at b 0 lifting to a ath γ going from e 0 to e 1. Let [α] be an element of H 1 and take a reresentative α that lifts to a loo α based at e 1. Then γ α γ is a loo based at e 0. Projecting down gives [γ][α][γ] 1 H 0,sowehave[γ]H 1 [γ] 1 H 0. By swaing the laces of H 0 and H 1 in the above argument, we also get [γ] 1 H 0 [γ] H 1, so equality follows. To rove the converse, suose H is a conjugate subgrou of H 0. We can write H 0 =[γ]h[γ] 1 for some element [γ] π 1 (B,b 0 ). Lift γ to a ath γ in E starting at e 0, and let e 1 be its end-oint. By the above argument, we have that H 1 = [γ] 1 H 0 [γ] =H Theorem 4.7. Let : E B and : E B be covering mas, and let (e 0 )= (e 0 )=b 0. There is an isomorhism of coverings h : E E if and only if the subgrous H = (π 1 (E,e 0 )) and H = (π 1 (E,e 0)) of π 1 (B,b 0 ) are conjugate. B

7 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 7 Proof. Given an isomorhism h : E E, let e 1 = h(e 0 ), and denote H1 = (π 1 (E,e 1)). Then by Theorem 4.5, we have H 0 = H1, and by the Lemma 4.6, we have that H1 is conjugate to H0. Conversely, suose H 0 and H0 are conjugate. Then by Lemma 4.6, we can ick a oint e 1 E such that H 1 = (π 1 (E,e 1 )) is equal to H0. By Theorem 4.5, there is an isomorhism h that takes e 1 to e Classification of Covering Saces - Existence of Covers Before discussing the main theoretical framework in this section, let us first consider an illustrative examle. Examle 5.1. Consider the circle S 1. We stated earlier that its only covering saces are R and S 1 with covering mas the usual rojection and f n : z z n resectively. One question that naturally arises is: What does the induced homomorhism of a covering ma do to the fundamental grou of a covering sace? Let γ : I S 1 be a generator for π 1 (S 1,s 0 )(wheres 0 is the canonical base oint.) Then f n γ wras around the circle n-times, so (f n ) is clearly the n-th ower ma. There are three things to notice about this. First, (f n ) is injective so it embeds the fundamental grou of the covering sace as a subgrou of the fundamental grou of the base sace. Moreover, under the isomorhism that takes π 1 (S 1,s 0 ) to Z, (f n ) (π 1 (S 1,s 0 )) is taken to nz. Finally, there are n reimages of any given oint x under f n, and this number is recisely the index of (f n ) (π 1 (S 1,s 0 )) in π 1 (S 1 ). Letting n run over all the natural numbers, we see that every subgrou of π 1 (S 1,s 0 ) is the image of the fundamental grou of a covering sace under its induced homomorhism. Furthermore, in accordance with the results of Section 4, for each subgrou of π 1 (S 1,s 0 ), there is only one isomorhism tye of covering sace associated to it. (Here, isomorhism tye and base-oint reserving isomorhism tye are equivalent notions since π 1 (S 1,s 0 ) is abelian.) In other words, there is a bijection between isomorhism tyes of covering saces of S 1 and the subgrous of its fundamental grou. The various facets of the above henomenon haen much more generally. In articular, we have: Proosition 5.2. Let : (E,e 0 ) (B,b 0 ) be a covering ma, and let : π 1 (E,e 0 ) π 1 (B,b 0 ) be the induced homomorhism. This ma is injective. Furthermore, the image subgrou (π 1 (E,e 0 )) π 1 (B,b 0 ) consists of the homotoy classes of loos in B based at b 0 whose lifts to E starting at e 0 are loos. Proof. This follows immediately from the ath and homotoy lifting lemmas. We now see the results of the revious section in a new light: There we comared subgrous of π 1 (B,b 0 ) to solve the roblem of when two covering saces were isomorhic. But the observation can be turned on its head to say that each subgrou of π 1 (B,b 0 ) has at most one baseoint reserving isomorhism tye of covering sace realizing it, and each conjugacy class of subgrous, at most one isomorhism tye. In fact, given a certain local niceness condition on the base sace that is in articular satisfied by S 1, we can establish the existence of a covering sace realizing each subgrou of π 1 (B,b 0 ), and the association then becomes a corresondence that

8 8 YAN SHUO TAN is reminiscent of that between subfields of a given field and the subgrous of its Galois grou in Galois theory. The true icture is trickier than it seems. We shall deal with it gradually over the rest of this aer, and here content ourselves to rove the setwise bijection. The first ste is to find a covering sace realizing the trivial subgrou. We define it as follows: Definition 5.3. Let : E B be a covering ma. We say that E is a universal covering sace or simly a universal cover if it is simly connected. Now, not every sace affords a universal cover. We mentioned earlier that some local niceness condition needs to be imosed. We exlain what this condition is now. Definition 5.4. A sace B is said to be semilocally simly connected if for each b B, there is a neighborhood U of b such that the homomorhism i : π 1 (U, b) π 1 (B,b) induced by inclusion is trivial. Lemma 5.5. If the sace B has a universal cover, then it is semilocally simly connected. Proof. Let : E B be a covering ma from a universal cover. Let b be a oint in B, and ick a neighborhood U of b that is evenly covered by. Fix a oint e 1 (b), and let V be the slice of 1 (U) containing e. Let γ be a loo in U based at b. Using the local homeomorhism roerty, we can lift it to a loo γ in V based at e. ButsinceE is simly connected, there is a homotoy H in E between γ and the constant loo, and this rojects down to a homotoy between γ and the constant loo in B. In other words, i ([γ]) is trivial. It turns out that this condition characterizes the class of saces that ossess universal covers. In other words, being semilocally simly connected is not only necessary but also sufficient, and we shall rove this by constructing a universal cover for an arbitrary semilocally simly connected sace. To motivate the construction, we first make the following observation, whose roof we leave as an easy exercise. Proosition 5.6. AsaceE is simly connected if and only if there is a unique homotoy class of aths connecting any two oints in E. Suose we have a simly connected sace E and ick a base oint e 0. Then this roosition tells us recisely that there is a bijection between the homotoy classes of aths starting at e 0 and the oints in E. On the other hand, if E were a covering sace for a base sace B under a ma that sends e 0 to b 0 B, then the ath and homotoy lifting roerties tell us that there is a bijection between homotoy classes of aths in E starting at e 0 and homotoy classes of aths in B starting at b 0. This means that we could try to construct a universal E 0 for a base sace (B,b 0 ) by thinking of homotoy classes of aths starting at b 0 as the oints of E 0, which has the advantage of allowing us to describe E 0 urely in terms of B. We shall use this idea to rove: Theorem 5.7. The sace B has a universal cover if and only if it is semilocally simly connected.

9 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 9 Proof. Fix a baseoint b 0 of B, and define a sace E 0 by setting its underlying oint set to be E 0 = {[γ] γ is a ath in B starting at b 0 } where, [γ] denotes the (end-oint reserving) homotoy class of γ. Let : E 0 B send [γ] toγ(1). Then is surjective, and we shall see that defining the right toology on E 0 will make into a covering ma. Let U be the collection of ath-connected oen sets U B such that the homomorhism induced by inclusion, π 1 (U) π 1 (B) is trivial. Note that this does not deend on our choice of baseoint in U. We can find such a neighborhood for every oint in B because it is both simly connected and locally ath-connected. Furthermore, if U U and V is a ath-connected oen subset of U, then the homomorhism induced by inclusion of V in B factors as π 1 (V ) π 1 (U) π 1 (B), and is thus also trivial. This imlies that U is a basis for the toology on B. We now define a toology on E 0.GivenU U, and a ath γ in B from b 0 to a oint in U, let U [γ] = {[γ η] η is a ath in U with η(0) = γ(1)}. Since the oeration is well defined on homotoy classes, we see that the definition is indeendent of our choice of reresentative γ. We have that : U [γ] U is surjective because U is ath-connected, and it is injective because different choices of η with the same endoint are homotoic in B by assumtion. Moreover, if [γ ] U [γ], then we can write γ = γ η for some η in U, and elements of U [γ ] are of the form [γ η µ], where µ is a ath in U. Since η µ is a ath in U, wehaveu [γ ] U [γ]. Analogously, we get the oosite inclusion, so U [γ ] = U [γ]. We claim that the sets U [γ] form a basis for a toology on E 0. To rove this, suose we ick an element [γ ] in the intersection of two sets U [γ] and V [γ ].Pick a neighborhood W of γ (1) that is contained in U V. Then we clearly have W [γ ] U [γ ] = U [γ], and W [γ ] V [γ ] = V [γ ],sow [γ ] is a neighborhood of [γ ] that lies in the intersection. This roves the claim, and we let the toology on E 0 be that generated by this basis. To show that is a covering ma, we first verify that each neighborhood U U is evenly covered by. It is clear that 1 (U) is the union of the sets U [γ]. Our remarks earlier tell us that these sets are disjoint, and if we fix one of these, we have already roved that the restriction U[γ] gives a bijection with U. Furthermore, one can check that U[γ] induces a bijection between the oen subsets contained in U [γ] and those contained in U, and is thus a homeomorhism. The fact that is a local homeomorhism automatically imlies that it is continuous. Finally, we show that E 0 is simly connected. Let [b 0 ] denote the constant ath at b 0 and choose it as our baseoint. Observe that given [γ] E 0, we can define a family of aths, {γ s s I} by γ s (t) =γ(st). The ma I E 0 given by s [γ s ] can easily be checked to be continious, and thus defines a ath between [b 0 ] and [γ]. This roves that E 0 is ath-connected. Let f : I E 0 be a loo based at [b 0 ]. Then f is a loo γ based at b 0. Define the ma f : s [γ s ] as in the above aragrah. Then f is a lift of γ based at [b 0 ], so by the unique lifting of aths, we have f = f. In articular, [b 0 ]=f(1) = f (1) = [γ 1 ]=[γ], so γ is nullhomotoic in B, and by the homotoy lifting roerty, f must also be nullhomotoic. In other words, π 1 (E 0, [b 0 ]) = 0.

10 10 YAN SHUO TAN Since universal covers are unique u to isomorhism, we often talk about the universal cover of a given base sace. The reason for the adjective universal is because it is also a covering sace for the covering saces of B realizing the intermediate subgrous of π 1 (B,b 0 ). We shall consider this in a later section, but we note here that the existence of a universal cover is sufficient to guarantee the existence of covering saces reresenting all of the subgrous of π 1 (B,b 0 ). Proosition 5.8. Let B be semilocally simly connected. Then for every subgrou H π 1 (B,b 0 ), there is a covering sace : E H B such that (π 1 (E H,e 0 )) = H for a suitably chosen baseoint e 0 E H. Proof. Construct E 0, the universal cover of B with resect to the baseoint b 0, and define an equivalence relation on the set of oints as follows: We set [γ] [γ ]if γ(1) = γ (1) and [γ γ ] H. One can check that this relation is well-defined. We then let E H be the quotient sace E 0 / generated by this equivalence relation. Now, recall that the ma : E 0 B sends [γ] toγ(1). Our definition of tells us that factors through the quotient ma and induces a continuous ma : E H B. We want to show that is a covering ma, so let q denote the quotient ma and let U be a neighborhood that is evenly covered by. First, observe that since U [γ] is locally homeomorhic with U, q is injective on each U [γ]. Now, suose we have γ(1) = γ (1) U, and [γ] [γ ]. Then for any ath η with η(0) = γ(1), we have [γ η] [γ η], so q(u [γ] )=q(u [γ ]). Furthermore, for each oen subset V U, ick[µ], [µ ] such that µ(1) = µ (1) V, and V [µ] U [γ],v [µ ] U [γ ]. It is clear that we have q(v [µ] )=q(v [µ ]). As such, the restriction q U[γ] : U [γ] q(u [γ] ) is in fact a homeomorhism. We can then write q(u[γ] ) = U[γ] (q U[γ] ) 1 so is a local homeomorhism on each set q(u [γ] ). This roves that E H is a covering sace for B, and it remains to show that (π 1 (E H, [b 0 ])) = H. But given a loo γ in B based at b 0, we see that [γ] isin the image of if and only if γ lifts to a loo in E H. But using the same argument from the roof of Theorem 5.7, we see that the lift of γ starts at [b 0 ] and ends at [γ], and so is a loo in E H if and only if [γ] [b 0 ], or equivalently, [γ] H. The end result of the revious two sections is the following classification theorem: Theorem 5.9 (Classification of connected coverings). Let B be semilocally simly connected. Fix a baseoint b 0. Then (1) There is a bijection between the set of isomorhism classes of based covering saces : (E,e 0 ) (B,b 0 ) and the set of subgrous of π 1 (B,b 0 ) obtained by associating the covering sace : E B with the subgrou (π 1 (E,e 0 )). (2) This association also induces a bijection between the set of isomorhism classes of covering saces : E B and conjugacy classes of subgrous of π 1 (B,b 0 ). 6. Classification of Covering Saces - A Lattice Isomorhism It is common in the literature (c.f. [3], [7]) to sto an exosition into the classification of covering saces with Theorem 5.9. We realize, however, that we can easily rove something deeer based on the following observation:

11 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 11 Proosition 6.1. Let X, Y and Z be saces. Let, q and r be continuous mas such that the following the diagram commutes: X q Y Z r (1) If and r are covering mas, so is q. (2) If and q are covering mas, so is r. (3) Suose Z has a universal cover, then if q and r are covering mas, so is. Proof. The roof for (1) and (2) is a rather tedious and unenlightening affair in oint-set toology. We hence refer the reader to [7] for a comlete roof. To rove (3), let f : Z Z be the universal cover for Z. By the Lifting Criterion, there is a lift of f to g : Z Y.Sincer : Y Z and f : Z Z are both coverings of Z, by (1), g is also a covering. We then reeat the argument one more: Alying the Lifting Criterion, we obtain a lift of g to h: Z X, and observe that since q : X Y and g : Z Y are both coverings of Y, h is also a covering by (1). One can then check that we have the following string of equalities: h = rqh = rg = f, so we can aly (2) to conclude that is a covering. The argument is encasulated in the following diagram: q X Y g h f Z r Z This roosition roves our assertion in the last section that all intermediate covering saces of a given base sace are themselves covered by the universal cover. In fact, more is true: It tells us that in the context of a sace with a universal cover, mas of coverings and covering mas are one and the same thing, so the erstwhile confusing similarity between the two terms may be excused on this account. Furthermore, the roosition tells us that the following category embeds in To, the category of based saces, as a full subcategory. Definition 6.2. Let (B,b 0 ) have a universal cover. We define a category Cov(B,b 0 ) by setting its objects to be based covers of (B,b 0 ) and its morhisms to be the based coverings : (E,e 0 ) (E,e 0). Our goal now is to state art (1) of the classification theorem of the last section in terms of an isomorhism of categories. Unfortunately, Cov(B,b 0 ) does not quite fit our uroses because the objects described in the classification theorem are

12 12 YAN SHUO TAN isomorhism tyes of based covers rather than individual saces. We get around this by constructing a second category Cov(B,b 0 ) as follows: Pick one reresentative from each isomorhism tye of based cover. We define Cov(B,b 0 ) to be the full subcategory sanned by these objects in Cov(B,b 0 ). Notice that this actually defines a skeleton of Cov(B,b 0 ), and as with skeletons in general, our choice of Cov(B,b 0 ) is non-unique. Now, a oset (P, ) can be interreted as a category, with objects the set elements of P and morhisms a b if a b in P. We can then turn this observation on its head and define a oset to be a category with at most one morhism between any two objects. Proosition 6.3. The category Cov(B,b 0 ) is a oset. Proof. Let (E,e 0 ) and (E,e 0) be objects of Cov(B,b 0 ). If there is a covering : (E,e 0 ) (E,e 0), then by our discussion above, is a based ma of coverings of the base sace (B,b 0 ), and so is unique by the Lifting Criterion. Furthermore, if there is a covering q :(E,e 0) (E,e 0 ), then the two based saces are isomorhic and hence must be the same object by construction. The set of subgrous of a grou G can be ordered under inclusion to form a oset, which we shall denote P(G). If we let Λ: P(π 1 (B,b 0 )) Cov(B,b 0 )bethe set bijection taking a subgrou of π 1 (B,b 0 ) to the based sace realizing it, we can ask whether Λ is a functor. Theorem 6.4. The ma Λ: P(π 1 (B,b 0 )) Cov(B,b 0 ) is an isomorhism of categories. In fact, it is a lattice isomorhism. Proof. We have already shown that it is a bijection on objects. To rove Λ is a functor, we just need to show it is order-reserving in both directions, i.e. that it is bijective on morhisms. Now, suose we have a morhism H K in P(π 1 (B,b 0 )), so that H is a subgrou of K. Write (E H,e 0 )=Λ(H), (E K,e 0)=Λ(K), and let : (E H,e 0 ) (B,b 0 ), and :(E K,e 0) (B,b 0 ) denote the resective covering mas. Then by the Lifting Criterion, there is lift of to a covering ma q :(E H,e 0 ) (E K,e 0), and so a morhism Λ(H) Λ(K). The converse is roved by reversing the above argument. Since P(π 1 (B,b 0 )) is a lattice, we automatically get that Cov(B,b 0 ) is a lattice, and that Λ is a lattice isomorhism. Remark 6.5. Many texts and online sources on covering saces call attention to the analogy between the classification of covering saces and the following fact from Galois Theory: If K/F is a Galois extension, then there is a lattice isomorhism between the oset of subfields of K containing F and the oset of subgrous of Gal(K/F ) given by the inverse functors Aut(K/ ) and Fix( ). This is not entirely helful, because the analogy breaks down when we take a look at the bigger icture. If K/F is an arbitrary field extension, Aut(K/ ) and Fix( ) are no longer isomorhisms, and it is ossible to have Fix(Gal(K/L)) = L. On the other hand, they are still order-reversing mas of osets, and the corresondence they give satisfies the conditions for what is known in order theory as a Galois connection. There is no such generalization for Cov(B,b 0 ). Moreover, there is still another analogy between covering sace theory and this fact from Galois theory that arises when we consider the automorhism grous of

13 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 13 covers. In this case, we actually have a full analogue of the Fundamental Theorem of Galois Theory. We ursue this observation in a later section. 7. Interlude - Homotoy Relationshis with the Base Sace Let us take a brief detour from our main exosition to examine how we can use covering saces as a tool for studying the homotoy roerties of saces. Proosition 7.1. A covering ma : (E,e 0 ) (B,b 0 ) induces isomorhisms : π n (E,e 0 ) π n (B,b 0 ) for n 2. Proof. Since S n is simly connected for n 2, any ma f :(S n,s 0 ) (B,b 0 ) satisfies the Lifting Criterion trivially, and so has a lift f :(S n,s 0 ) (E,e 0 ). This roves surjectivity. Injectivity follows from the fact that a homotoy in the base sace lifts to a homotoy in the covering sace. Proosition 7.2. Let E 0 and E 0 be universal covers for B and B resectively. If B and B are homotoically equivalent, then so are E 0 and E 0. Proof. Let f : B B and g : B B be homotoy inverses, and let : E 0 B and : E 0 B denote the covering mas. By the Lifting Criterion, we have lifts f: E 0 E 0 and g : E 0 E 0, giving the following commutative diagram: E f g E B f B g We claim that the two lifts are homotoy inverses. To see this, observe that f = f,sowehaveg f = gf. Now let H be the homotoy between gf and id B. Then H ( id B ): E 0 I B is a ma from a simly connected sace into B. The Lifting Criterion alies, and we obtain a lift that one can check to be a homotoy between gf and. But is an automorhism of covers that fixes a oint, so by the uniqueness ortion of Theorem 4.5, it must be the identity. Arguing similarly for fg comletes the roof of our claim. The converse is not true as seen from the fact that R 2 is a covering sace for both the torus and the Klein bottle, but these two base saces are not homotoically equivalent. Proosition 7.3. Let X be ath-connected and locally ath-connected, and fix a base oint x 0.Ifπ 1 (X, x 0 ) is finite, then every ma X S 1 is nullhomotoic. Proof. Let f :(X, x 0 ) (S 1,s 0 ) be a ma. Then f (π 1 (X, x 0 )) is isomorhic to a quotient of π(x, x 0 ), and is hence finite. In articular, it is torsion. On the other hand, f (π 1 (X, x 0 )) π(s 1,s 0 ) = Z, and as a subgrou of a free grou, must itself be free. So the only ossibility is that f (π 1 (X, x 0 )) = 0, so by the Lifting Criterion, there is a lift f :(X, x 0 ) (R,e 0 ), which is nullhomotoic. This nullhomotoy rojects down to give a nullhomotoy for f. Corollary 7.4. Every ma f : S n S 1 is nullhomotoic.

14 14 YAN SHUO TAN Observe that the roof of the roosition still works when we relace S 1 by any other sace with a free (or free abelian) fundamental grou and a contractible universal cover, and if we allow π 1 (X, x 0 ) to be infinite so long as it remains torsion. We thus also have: Corollary 7.5. Let X be ath-connected and locally ath-connected, and let W be homotoically equivalent to a wedge of circles or to a torus of any dimension. If every element of π 1 (X, x 0 ) has finite order, then every ma X W is nullhomotoic. 8. Covering of Grouoids Fundamental grous serve as natural algebraic models for connected based toological saces. They are not, however, entirely convenient when the sace we consider is disconnected, in which case the fundamental grou deends on the choice of ath comonent, or when the sace is not based, so that we have to make an arbitrary choice of a base oint. We hence introduce something far more natural. Definition 8.1. Let X be a sace. Then the fundamental grouoid of X, denoted Π(X), is the category with objects the oints of X, and whose morhisms x y are the homotoy classes of aths between x and y. A ma f : X Y takes homotoy classes of aths to homotoy classes of aths. In other words, it induces a ma of grouoids f : Π(X) Π(Y ). Letting Π act on morhisms by f f then defines a functor Π( ): To Gd from the category of toological saces to the category of grouoids. Now, recall that a skeleton Sk(C) of a category C is defined as a full subcategory with one object from each isomorhism class of objects of C. It is an imortant notion because the inclusion functor J : Sk(C) Cgives an equivalence of categories. Given a oint x 0 X, the fundamental grou based at that oint, π 1 (X, x 0 ), comrises all homotoy classes of aths starting and ending at x 0, and so embeds as a full subcategory of Π(X). As such, a skeleton for Π(X) comrises the disjoint union of a collection of fundamental grous, one for each ath-comonent of X. In articular, we have Proosition 8.2. Let X be a ath-connected sace. Then for each oint x 0 X, π 1 (X, x 0 ) is a skeleton for Π(X), and the inclusion functor π 1 (X, x 0 ) Π(X) is an equivalence of categories. It turns out that much of the theory of covering saces has an analogue for grouoids. In fact, both theories have a beautiful confluence, which we shall resent in the next section. In this section, we first develo the fundamentals of a theory for covering grouoids, loosely following the aroach of [6]. We say that a grouoid is connected if any two objects have at least one morhism between them. It is easy to see that a sace X is ath-connected if and only if Π(X) is connected. With the same rationale as with saces, we restrict our attention to connected grouoids. Let us first develo an analogue of a fundamental neighborhood for grouoids. Definition 8.3. Let C be a small grouoid, and x an object of C. The star of x, denoted St C (x) is the set of morhisms of C with domain x. We also use the suggestive notation π(c,x) to denote C(x, x) St C (x), the grou of automorhisms of the object x.

15 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 15 Remark 8.4. If X is a sace and x 0 a oint of X, wehaveπ 1 (X, x 0 )=π(π(x),x 0 ). Definition 8.5. Let E and B be small grouoids. A covering of grouoids : E B is a functor that is surjective on objects and restricts to a bijection : St(e) St((e)) for each object e of E. We immediately get Proosition 8.6. Let :(E,e 0 ) (B,b 0 ) be a covering of grouoids, then the restriction : π(e,e 0 ) π(b,b 0 ) is injective. Observe that if we let F b denote the set of objects of E such that (e) =b, then 1 (St(b)) is the disjoint union over e F b of St(e). This is the analogue of the artition of the reimage of a fundamental neighborhood into slices. Recall that with covering saces, the first ste was to use this roerty of covering mas to obtain unique lifting of aths and homotoies. The case for grouoids is more sartan: the analogue of a homotoy class of aths between two oints b and b is simly a morhism between them, and given a choice of source e F b,this morhism lifts uniquely by definition. We can hence immediately rove Theorem 8.7 (Lifting Criterion). Let : (E,e 0 ) (B,b 0 ) be a covering of grouoids, and f :(X,x 0 ) (B,b 0 ) any ma with X connected. Then a lift f :(X,x0 ) (E,e 0 ) exists if and only if f(π(x,x 0 )) (π(e,e 0 )). Furthermore, if such a lift exists, it is unique. Proof. If f exists, then by definition, we have f = f, so the image of f is contained in that of. To rove the converse, let x be an object of X, and let α: x 0 x be a morhism in X. Let α be the unique morhism with source e 0 rojecting down to f(α). We define f by setting f(x) to be the target of α, and check that it is a well-defined functor. Suose we have another morhism α : x 0 x. Then there is a unique lift of f(α 1 α ) to a morhism β in E with source e 0. But by assumtion, β is an element of π(e,e 0 ) and thus also has target e 0. We then have the following chain of equalities: ( αβ) =( α)(β) =f(α)f(α 1 α )=f(α ) which imlies that by definition, α = αβ. As such, α has the same target as α,so f is well-defined on objects and morhisms. Finally, note that f is functorial because every morhism in X can be written in terms of elements of St X (x 0 ), and on these, we can write f as a comosition of functors, namely f =( StB (b 0)) 1 f. As with covering saces, this theorem allows us to rove the analogues of our results from Sections 4 and 5: we can classify covers by associating them with subgrous of π(b,b 0 ). Since the roofs are similar to those for covering saces, we leave these to the reader. Definition 8.8. Let : E Band : E Bbe coverings of grouoids, and let g : E E be a functor such that the following diagram commutes: E g E B

16 16 YAN SHUO TAN We then say that g is a ma of coverings over B. If g is in addition a grouoid isomorhism, we say that it is an isomorhism of coverings. Theorem 8.9. Let : (E,e 0 ) (B,b 0 ) and : (E,e 0 ) (B,b 0 ) be covering mas. There is an isomorhism of coverings h :(E,e 0 ) (E,e 0 ) if and only if the grous H = (π(e,e 0 )) and H = (π(e,e 0)) of π(b,b 0 ) are equal. If h exists, it is unique. Lemma Let : E Bbe a covering of grouoids. Let e 0 and e 1 be oints of 1 (b 0 ), and let H i = (π(e,e i )) for i =0, 1. Then H 0 and H 1 are conjugate. Conversely, given e 0, and given a subgrou H of π(b,b 0 ) conjugate to H 0, there exists a oint e 1 of 1 (b 0 ) such that H 1 = H. Theorem Let : E Band : E Bbe covering mas, and let (e 0 )= (e 0 )=b 0. There is an isomorhism of coverings h : E E if and only if the subgrous H = (π(e,e 0 )) and H = (π(e,e 0)) of π(b,b 0 ) are conjugate. Proosition Let B be a small grouoid. Then for every subgrou H π(b,b 0 ), there is a covering : E H Bsuch that (π(e H,e 0 )) = H for a suitably chosen base object e 0 E H. Theorem 8.13 (Classification of connected coverings of grouoids). Let B be a grouoid. Fix a base object b 0. Then (1) There is a bijection between the set of isomorhism classes of based covering grouoids : (E,e 0 ) (B,b 0 ) and the set of subgrous of π(b,b 0 ) obtained by associating the covering sace : E Bwith the subgrou (π(e,e 0 )). (2) This association also induces a bijection between the set of isomorhism classes of covering saces : E Band conjugacy classes of subgrous of π(b,b 0 ). Remark It is ossible to first classify covering grouoids and then use some of the results to rovide immediate analogues for the theory of covering saces. Indeed, this is the aroach of [6]. We have chosen to resent the concets in the inverse order, however, because it is instructive to see the secific instances where considering grouoids rather than saces simlifies the theory. 9. Grou Actions and Grouoid Actions Having classified covering saces and covering grouoids u to isomorhism tye, we would next like to classify the mas between these objects. To do so, we need to make a brief detour by considering the theory of grou actions and how we can generalize this to obtain the notion of a grouoid action. Let us first recall some basic definitions. Definition 9.1. Let G be a grou, and S aset. (1) A (left) action of a grou G on a set S is a function : G S S such that e s = s and (g g) s = g (gs) for all s S. The axiom of associativity allows us to omit writing without causing ambiguity. (2) The stabilizer or isotroy grou G s of a oint s is the subgrou of G with elements {g gs = s}. (3) An action is free or semiregular if all stabilizers are trivial. (4) An action is transitive if for every air of set elements s and s S, there is a grou element g G such that gs = s. (5) An action is regular if it is both free and transitive.

17 A SKELETON IN THE CATEGORY: THE SECRET THEORY OF COVERING SPACES 17 Remark 9.2. We call a set S with a G-action a G-set, and we often refer to it by simly S. This is not entirely rigorous, for we can ut two different G-actions on the same set, so one should secify the associated ma G S S. Nonetheless, we seldom do this in ractice. Definition 9.3. Let G be a grou, and let S and S be G-sets. A G-ma φ: S S is a set function that resects the G-set structure, i.e. such that φ(gs) =gφ(s) for all s S and g G. Ifφ is also a bijection, we call it an isomorhism of G-sets. Examle 9.4. Let H be a subgrou of G. Left multilication by elements of G makes the set of cosets G/H into a transitive G-set. If we fix a coset gh, its stabilizer is simly ghg 1. It turns out that this examle, in a sense, accounts for all ossible transitive G-sets. More recisely, we have Proosition 9.5. Let G act transitively on a set S. Pick any element s S, and let G s denote its stabilizer. Then S is isomorhic to G/G s as a G-set. Proof. Define φ: S G/G s by sending gs to gg s for each g G. We then have φ(gs) =gg s = gφ(s). The ma is clearly surjective. It is injective as well because if gg s = hg s,thengh 1 G s,sogs = hs. A G-set can be interreted as a functor from G seen as a category with one object to the category of sets, so studying grou actions of G is equivalent to studying the functor category Set G. If we restrict our attention to the subcategory consisting of transitive actions, then Proosition 9.5 tells us recisely that the following full subcategory contains a skeleton: Definition 9.6. Let G be a grou. The orbit category of G, denoted O(G) is defined to be the category with orbits the G-sets G/H and morhisms the G-mas between them. These G-sets are called canonical orbits. Having dealt with the objects, we now study the morhisms in more detail. It is easy to see that a G-ma of transitive G-sets is fully determined by where it sends a single element. We then have Proosition 9.7. Let φ: G/H G/K be a G-ma, and let γ G be such that φ(h) =γ 1 K. Then γhγ 1 is a subgrou of K. Conversely, for every γ G such that γhγ 1 is a subgrou of K, themaφ γ : gh gγ 1 K defines a G-ma. Finally φ γ = φ γ if and only if both γ and γ belong to the same coset of K. Proof. For all h H, wehaveγ 1 K = φ(h) =φ(hh) =hφ(h) =hγ 1 K,which imlies that γhγ 1 K, so the first statement follows. The second statement and third statements are obvious. As is the case with any other mathematical object, the automorhisms of a G-set form a grou. The following roosition characterizes its structure. Proosition 9.8. The automorhism grou of the G-set G/H is isomorhic to N G (H)/H, the quotient of the normalizer of H in G by H. Proof. We define a ma Φ: N G (H)/H Aut G (G/H) byγh φ γ,whereφ γ is the G-ma defined in Proosition 9.7. This is indeendent of our choice of coset

18 18 YAN SHUO TAN reresentative. Indeed, if we ick another reresentative γ,thenγ γ 1 H, and we have φ γ (H) =γ 1 H = γ 1 (γ γ 1 )H = γ 1 H = φ γ (H). Proosition 9.7 tells us that Φ is a well-defined bijection. Furthermore, given two elements αh and βh of N G (H)/H, wehave φ βα (H) =(βα) 1 H = α 1 β 1 H = α 1 φ β (H) =φ β (α 1 H)=φ β (φ α (H)) which roves that Φ is a homomorhism. Remark 9.9. Such grous N G (H)/H are sometimes denoted W (H) and called Weyl grous. It is ossible to generalize the notion of grou actions to grouoids. Since a grou action is a functor from a grou G to the category Set, it is natural to define a grouoid action as a functor Ψ from a grouoid G to Set. Studying grouoid actions is then tantamount to studying the functor category Set G. For each object x G,restrictingΨtoπ(G,x) gives a grou action on Ψ(x). Recall that if objects x and x are in the same connected comonent of G, then π(g,x) and π(g,x ) are isomorhic. We can then ask whether we can make the resective actions of both grous comarable by generalizing the notion of G-ma. Let us formalize this notion as follows: Definition Form the category A with objects all triles (G, S, ), where G is a grou, S a set, and : G S S is a grou action. Given two objects (G, S, ) and (G,S, ), we define a morhism (α, φ): (G, S, ) (G,S, ) wheneverα: G G is a homomorhism and φ: S S is a G-ma satisfying φ(g s) =α(g) φ(s) for all g G and s S. We call a morhism in this category a ma of grou actions, and an isomorhism, an isomorhism of grou actions. One can check that this is a well-defined category. Furthermore, a G-ma is then the secial case of a ma of grou actions in which the source and target objects have the same first coordinate G. Proosition Let Ψ: G Set be a grouoid action, and let x and x be objects in G. Then the induced actions of π(g,x) on Ψ(x) and π(g,x ) on Ψ(x ) are isomorhic grou actions. Proof. First, write G = π(g,x) and G = π(g,x ). Let γ : x x be a morhism in G, and let c γ : G G denote the conjugation ma defined by g γgγ 1 for all g G. We claim that (c γ, Ψ(γ)): (G, Ψ(x), ) (G, Ψ(x ), ) is an isomorhism of grou actions. But given g G and s Ψ(x), we have Ψ(γ)(g s) =Ψ(γg)(s) =Ψ(γgγ 1 )Ψ(γ)(s) =γgγ 1 Ψ(γ)(s) =c γ (g) Ψ(γ)(s) which roves the claim. Examle Let : E Bbe a covering of grouoids. We can view E as the collection of fibers F b as b runs over objects of B. ThenB acts on E via the functor T defined on objects by T : b F b and on morhisms as follows: Given γ : b b, consider any e F b. Let γ be the unique lift of γ with source e, and let e be the target of γ. Wethensetγ e = e. One easily checks that this gives a well-defined functor. We call T a fiber translation functor. Since B is connected, Proosition 9.11 tells us that every action of π(b,b) on its corresonding fiber F b is isomorhic. In articular, there is a bijection between