Split (1)

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� {y0} ∪ {x0} ∗ Y ) and S(X ∧ Y )/S({x0} ∧ {y0}) are homeomorphic, and deduce that X ∗ Y ≃ S(X ∧ Y ). 25. homotopy equivalent to Y If X is a CW complex with components Xα, show that the suspension SX is In the case that X is a ﬁnite α SXα for some graph Y. graph, show that SX is homotopy equivalent to a wedge sum of ci...
rings, modules, and algebras also arise. The mechanisms that create these images — the ‘lanterns’ of algebraic topology, one might say — are known formally as functors and have the characteristic feature that they form images not only of spaces but also of maps. Thus, continuous maps between spaces are projected onto ...
linking A n times, where by ‘circle’ we mean a curve homeomorphic to a circle. To complete the scheme, we could let B0 be a circle not linked to A at all. Now, integers not only measure quantity, but they form a group under addition. Can the group operation be mimicked geometrically with some sort of addition op- erat...
two, A and B, and we ask whether C can be continuously deformed to unlink it completely from A and B, always staying in the complement of A and B during the deformation. We can redraw the picture by pulling A and B apart, dragging C along, and then we see C winding back and forth between A and B as shown in the second...
and B. But from the picture on the left it is apparent that C can actually be unlinked completely from A and B. So in this case the product aba−1b−1 should be trivial. The fundamental group of a space X will be deﬁned so that its elements are loops in X starting and ending at a ﬁxed basepoint x0 ∈ X, but two such loop...
consequence of the deﬁnition of the fundamental group that homeomorphic spaces have isomorphic fundamental groups. Basic Constructions Section 1.1 25 This ﬁrst section begins with the basic deﬁnitions and constructions, and then proceeds quickly to an important calculation, the fundamental group of the circle, using n...
linearly parametrized as f0(s) + t[f1(s) − f0(s)] = (1 − t)f0(s) + tf1(s), with the segment from f0(s) to f1(s) covered by t values in the interval from 0 to 1. If f1(s) happens to equal f0(s) then this segment degenerates to a point and ft(s) = f0(s) for all t. This occurs in particular for s = 0 and s = 1, so each f...
ﬁned on the union of two closed sets is continuous if it is continuous when restricted to each of the closed sets separately. In the case at hand we have H(s, t) = F (s, 2t) for 0 ≤ t ≤ 1/2 and H(s, t) = G(s, 2t − 1) for 1/2 ≤ t ≤ 1 where F and G are the maps I × I→X associated to the homotopies ft and gt. Since H is c...
in a sequence of groups πn(X, x0), called homotopy groups, which are deﬁned in an entirely analogous fashion using the n dimensional cube In in place of I. Basic Constructions Section 1.1 27 Proof: By restricting attention to loops with a ﬁxed basepoint x0 ∈ X we guarantee that the product f g of any two such loops is...
reparametrization of f via the function ϕ whose graph is shown in the ﬁrst ﬁgure at the right, so f c ≃ f. Similarly, c f ≃ f where c is now the constant path at f (0), using the reparametrization function in the second ﬁgure. Taking f to be a loop, we deduce that the homotopy class of the constant path at x0 is a two...
1 − t)f0(s) + tf1(s), as described in Example 1.1. 28 Chapter 1 The Fundamental Group It is not so easy to show that a space has a nontrivial fundamental group since one must somehow demonstrate the nonexistence of homotopies between certain loops. We will tackle the simplest example shortly, computing the fundamental ...
g h] = βh[f ]βh[g]. Further, βh is an isomorphism with inverse βh since βhβh[f ] = βh[h f h] = [h h f h h] = [f ], and similarly βhβh[f ] = [f ]. ⊔⊓ Thus if X is path-connected, the group π1(X, x0) is, up to isomorphism, independent of the choice of basepoint x0. In this case the notation π1(X, x0) is often abbreviate...
1, 0). Note that [ω]n = [ωn] where ωn(s) = (cos 2π ns, sin 2π ns) for n ∈ Z. The theorem is therefore equivalent to the statement that every loop in S 1 based at (1, 0) is homotopic to ωn for a unique n ∈ Z. To prove this the idea will be to compare paths in S 1 with paths in R via the map p : R→S 1 given by p(s) = (co...
p : (a) For each path f : I→X starting at a point x0 ∈ X and each X starting at (b) For each homotopy ft : I→X of paths starting at x0 and each x0. e Before proving these facts, let us see how they imply the theorem. e e X of paths starting at is a unique lifted homotopy ft : I→ f : I→ x0. e e e e x0 ∈ p−1(x0) there e...
F \|Y × {0}, then there is a unique map F : Y × I→ X lifting F and restricting to the given F on Y × {0}. e e e Statement (a) is the special case that Y is a point, and (b) is obtained by applying (c) with Y = I in the following way. The homotopy ft in (b) gives a map F : I × I→X by setting F (s, t) = ft(s) as usual. A...
N × {ti} by its intersection with F (N × {ti}) is contained in F on N × [ti, ti+1] to be the composition of F Ui. After a ﬁnite number of steps we eventually that e F \|\| N × {ti})−1( ( with the homeomorphism p−1 : Ui→ e get a lift Ui). Now we can deﬁne Ui ⊂ F : N × I→ X for some neighborhood N of y0. e Next we show th...
I. e Now we turn to some applications of the calculation of π1(S 1), beginning with a proof of the Fundamental Theorem of Algebra. Theorem 1.8. Every nonconstant polynomial with coeﬃcients in C has a root in C. Proof: We may assume the polynomial is of the form p(z) = zn + a1zn−1 + ··· + an. If p(z) has no roots in C,...
1.9. Every continuous map h : D2→D2 has a ﬁxed point, that is, a point x ∈ D2 with h(x) = x. Here we are using the standard notation Dn for the closed unit disk in Rn, all vectors x of length \|x\| ≤ 1. Thus the boundary of Dn is the unit sphere S n−1. 32 Chapter 1 The Fundamental Group Proof: Suppose on the contrary th...
which Brouwer deﬁned directly in more geometric terms. These proofs are all arguments by contradiction, and so they show just the exis- tence of ﬁxed points without giving any clue as to how to ﬁnd one in explicit cases. Our proof of the Fundamental Theorem of Algebra was similar in this regard. There exist other proo...
⊂ R3 by η(s) = (cos 2π s, sin 2π s, 0), and let h : I→S 1 be the composed loop gη. Since g(−x) = −g(x), we have the relation h(s + 1/2) = −h(s) for all s in the interval [0, 1/2]. As we showed in the calculation of π1(S 1), the loop h can be lifted to a path h : I→R. The equation h(s + 1/2) = −h(s) implies that h(s) +...
2(x) = d2(−x). If either of these two distances is zero, then x and −x both lie in the same set A1 or A2 since these are closed sets. On the other hand, if the distances from x and −x to A1 and A2 are both strictly positive, then ⊔⊓ x and −x lie in neither A1 nor A2 so they must lie in A3. d1(x), d2(x) \|x − y\|. This To...
t of the corresponding loops ≈ π1(X, x0)× π1(Y, y0), in X and Y. Thus we obtain a bijection π1 [f ] ֏ ([g], [h]). This is obviously a group homomorphism, and hence an isomor⊔⊓ phism. X × Y, (x0, y0) Example 1.13: The Torus. By the proposition we have an isomorphism π1(S 1 × S 1) ≈ Z× Z. Under this isomorphism a pair (p...
f ) (ϕg), both functions having the value ϕf (2s) for 0 ≤ s ≤ 1/2 and the value ϕg(2s − 1) for 1/2 ≤ s ≤ 1. Two basic properties of induced homomorphisms are: ψ-----→ (Y, y0) (ϕψ)∗ = ϕ∗ψ∗ for a composition (X, x0) 11∗ = 11, which is a concise way of saying that the identity map 11 : X→X induces the identity map 11 : π1...
of I such that each subinterval [si−1, si] is mapped by f to a single Aα. Namely, since f is continuous, each s ∈ I has an open neighborhood Vs in I mapped by f to some Aα. We may in fact take Vs to be an interval whose closure is mapped to a single Aα. Compactness of I implies that a ﬁnite number of these intervals c...
is a homeomorphism. The case n = 1 is easily disposed of since R2 − {0} is path-connected but the homeomorphic space Rn − {f (0)} is not path-connected when n = 1. When n > 2 we cannot distinguish R2 − {0} from Rn − {f (0)} by the number of path-components, but we can distinguish them by their fundamental groups. Name...
so i∗ is also surjective. This gives another way of seeing that S 1 is not a retract of D2, a fact we showed earlier in the proof of the Brouwer ﬁxed point theorem, since the inclusion-induced map π1(S 1)→π1(D2) is a homomorphism Z→0 that cannot be injective. The exact group-theoretic analog of a retraction is a homom...
= ϕ1∗[f ], the middle equality coming from the homotopy ϕtf. There is a notion of homotopy equivalence for spaces with basepoints. One says (X, x0) ≃ (Y, y0) if there are maps ϕ : (X, x0)→(Y, y0) and ψ : (Y, y0)→(X, x0) with homotopies ϕψ ≃ 11 and ψϕ ≃ 11 through maps ﬁxing the basepoints. In this case the induced map...
ht is still [0, 1]. Explicitly, we can take ht(s) = h(ts). Then if f is a loop in X at the basepoint x0, the product ht (ϕtf ) ht gives a homotopy of loops at ϕ0(x0). Restricting this homotopy to t = 0 and t = 1, we see that ϕ0∗([f ]) = ⊔⊓ βh. ϕ1∗([f ]) Proof of 1.18: Let ψ : Y →X be a homotopy-inverse for ϕ, so that ...
is said to be star-shaped if there is a point x0 ∈ X such that, for each x ∈ X, the line segment from x0 to x lies in X. Show that if a subspace X ⊂ Rn is locally star-shaped, in the sense that every point of X has a star-shaped neighborhood in X, then every path in X is homotopic in X to a piecewise linear path, that...
f restricts to the identity on the two boundary circles of S 1 × I. Show that f is homotopic to the identity by a homotopy ft that is stationary on one of the boundary circles, but not by any homotopy ft that is stationary on both boundary circles. [Consider what f does to the path s ֏ (θ0, s) for ﬁxed θ0 ∈ S 1.] 8. D...
π1(X)× π1(Y ) in Proposition 1.12 is given by [f ] ֏ (p1∗([f ]), p2∗([f ])) where p1 and p2 are the projections of X × Y onto its two factors. 15. Given a map f : X→Y and a path h : I→X from x0 to x1, show that f∗βh = βf hf∗ in the diagram at the right. 16. Show that there are no retractions r : X→A in the following c...
winding once 40 Chapter 1 The Fundamental Group around the circle. The more diﬃcult part of the calculation of π1(S 1) is therefore the fact that no iterate of this loop is nullhomotopic.] 20. Suppose ft : X→X is a homotopy such that f0 and f1 are each the identity map. Use Lemma 1.19 to show that for any x0 ∈ X, the ...
, for example (ab2a−3b−4)−1 = b4a3b−2a−1. It would be very nice if such words in a and b corresponded exactly to elements of π1(X), so that π1(X) was isomorphic to the group Z ∗ Z. The van Kampen theorem will imply that this is indeed the case. Similarly, if X is the union of three circles touching at a single point, t...
m ≥ 0, where each letter gi belongs to, and adjacent letters gi and gi+1 a group Gαi belong to diﬀerent groups Gα, that is, αi ≠ αi+1. Words satisfying these conditions are called reduced, the idea being that unreduced words can always be simpliﬁed to reduced words by writing adjacent letters that lie in the same Gαi ...
gLg′ implies that Lg is invertible with inverse Lg−1. Therefore the association g ֏ Lg deﬁnes a homomorphism from Gα to the group P (W ) of all permutations of W. More generally, we can deﬁne L : W→P (W ) by L(g1 ··· gm) = Lg1 for each reduced word g1 ··· gm. This function L is injective since the permutation L(g1 ··· ...
� is surjective, and its kernel consists of the words of even length. These form an inﬁnite cyclic subgroup generated by ab since ba = (ab)−1 in Z2 ∗ Z2. In fact, Z2 ∗ Z2 is the semi-direct product of the subgroups Z and Z2 generated by ab and a, with the conjugation relation a(ab)a−1 = (ab)−1. This group is sometimes ...
G× H. must be ϕα1 (g1) ··· ϕαn Van Kampen’s Theorem Section 1.2 43 The van Kampen Theorem Suppose a space X is decomposed as the union of a collection of path-connected open subsets Aα, each of which contains the basepoint x0 ∈ X. By the remarks in the preceding paragraph, the homomorphisms jα : π1(Aα)→π1(X) induced by...
s. α Xα of a collection of spaces Xα with basepoints xα ∈ Xα to be the quotient space of the α Xα in which all the basepoints xα are identiﬁed to a single point. disjoint union If each xα is a deformation retract of an open neighborhood Uα in Xα, then Xα is a deformation retract of its open neighborhood Aα = Xα β≠α Uβ....
≈ ∗α π1(Aα). Each Aα deformation retracts onto a circle, so π1(X) is free on ﬁve generators, as claimed. As explicit generators we can choose for each edge eα of X − T a loop fα that starts at a basepoint in T, travels in T to one end of eα, then across eα, then back to the basepoint along a path in T. Van Kampen’s th...
kernel of an element [f ] ∈ π1(X) we shall mean a formal product [f1] ··· [fk] where: is N, we ﬁrst introduce some terminology. By a factorization of Φ Φ Each fi is a loop in some Aα at the basepoint x0, and [fi] ∈ π1(Aα) is the homotopy class of fi. The loop f is homotopic to f1 ··· fk in X. A factorization of [f ] i...
of [f ]. The composed ℓ are then homotopic, so let F : I × I→X be a homopaths f1 ··· fk and f ′ topy from f1 ··· fk to f ′ ℓ. There exist partitions 0 = s0 < s1 < ··· < sm = 1 and 0 = t0 < t1 < ··· < tn = 1 such that each rectangle Rij = [si−1, si]× [tj−1, tj] is mapped by F into a single Aα, which we label Aij. These...
(v) that lies in the intersection of the two or three Aij ’s corresponding to the Rr ’s containing v, since we assume the intersection of any two or three Aij ’s is path-connected. Then we obtain a factorization of [F \|\| γr ] by inserting the appropriate paths gv gv into F \|\| γr at successive vertices, in Lemma 1.15. ...
′ Example 1.23: Linking of Circles. We can apply van Kampen’s theorem to calculate the fundamental groups of three spaces discussed in the introduction to this chapter, the complements in R3 of a single circle, two unlinked circles, and two linked circles. The complement R3 −A of a single circle A deformation retracts...
and B are linked, then R3 − (A ∪ B) deformation retracts onto the wedge sum of S 2 and a torus S 1 × S 1 separating A and B, ≈ as shown in the ﬁgure to the left, hence π1 π1(S 1 × S 1) ≈ Z× Z. R3 − (A ∪ B) Van Kampen’s Theorem Section 1.2 47 Example 1.24: Torus Knots. For relatively prime positive integers m and n, th...
− K) by showing that it deformation retracts onto a 2 dimensional complex X = Xm,n homeomorphic to the quotient space of a cylinder S 1 × I under the identiﬁcations (z, 0) ∼ (e2π i/mz, 0) and (z, 1) ∼ (e2π i/nz, 1). If we let Xm and Xn be the two halves of X formed by the quotients of S 1 × [0, 1/2] and S 1 × [1/2, 1]...
n in the second solid torus. 48 Chapter 1 The Fundamental Group The complement of K in the ﬁrst solid torus deformation retracts onto Xm by ﬂowing within each meridian disk as shown. In similar fashion the complement of K in the second solid torus deformation retracts onto Xn. These two deformation retractions do not a...
consisting of elements that commute with all elements of Gm,n. The element am = bn commutes with a and b, so the cyclic subgroup C generated by this element lies in the center. In particular, C is a normal subgroup, so we can pass to the quotient group Gm,n/C, which is the free product Zm ∗ Zn. According to Exercise 1...
for computing a presentation for π1(R3−K) for an arbitrary smooth or piecewise linear knot K is described in the exercises, but the problem of determin- ing when two of these fundamental groups are isomorphic is generally much more diﬃcult than in the special case of torus knots. Example 1.25: The Shrinking Wedge of C...
For a complete description of π1(X) see [Cannon & Conner 2000]. Q It is a theorem of [Shelah 1988] that for a path-connected, locally path-connected compact metric space X, π1(X) is either ﬁnitely generated or uncountable. Applications to Cell Complexes For the remainder of this section we shall be interested in cell ...
. (c) For a path-connected cell complex X the inclusion of the 2 skeleton X 2 ֓ X induces an isomorphism π1(X 2, x0) ≈ π1(X, x0). It follows from (a) that N is independent of the choice of the paths γα, but this can also be seen directly: If we replace γα by another path ηα having the same endpoints, then γαϕαγα change...
at z0 representing the elements of π1(A, z0) corresponding to [γαϕαγα] ∈ π1(A, x0) under the basepoint-change isomorphism βh for h the line segment connecting z0 to x0 in the intersection of the Sα ’s. To ﬁnish the proof of part (a) we just need to check that π1(A ∩ B, z0) is generated by the loops δα. This can be don...
), we conclude that f is nullhomotopic in X 2. ⊔⊓ As a ﬁrst application we compute the fundamental group of the orientable surface Mg of genus g. This has a cell structure with one 0 cell, 2g 1 cells, and one 2 cell, as we saw in Chapter 0. The 1 skeleton is a wedge sum of 2g circles, with fundamental group free on 2g ...
1(Ng) ≈. This abelianizes to the direct sum of Z2 with g − 1 copies of Z since in the abelian- ization we can rechoose the generators to be a1, ···, ag−1 and a1 + ··· + ag, with 2(a1 + ··· + ag) = 0. Hence Ng is not homotopy equivalent to Nh if g ≠ h, nor is Ng homotopy equivalent to any orientable surface Mh. a1, ···,...
larger n one would use an n pointed ‘asterisk’ and a 1/n twist. instead of Exercises 1. Show that the free product G ∗ H of nontrivial groups G and H has trivial center, and that the only elements of G ∗ H of ﬁnite order are the conjugates of ﬁnite-order elements of G and H. 2. Let X ⊂ Rm be the union of convex open s...
′ k obtained from the closed surfaces Mh and Mk by deleting an open disk from each. Show that M ′ not retract onto C. nonseparating circle C ′ in the ﬁgure. 10. Consider two arcs α and β embedded in D2 × I as shown in the ﬁgure. The loop γ is obviously nullhomotopic in D2 × I, but show that there is no nullhomotopy of...
Show that the other way yields a space Z with π1(Z) not isomorphic to π1(Y ). [Abelianize the fundamental groups to show they are not isomorphic.] 14. Consider the quotient space of a cube I3 obtained by identifying each square face with the opposite square face via the right-handed screw motion consisting of a transl...
a point, is suspension the shrinking wedge of circles in Example 1.25, with an uncountable fundamental group. Σ (b) Let C be the mapping cone of the quotient map SX→ countable by constructing a homomorphism from π1(C) onto X. Show that π1(C) is un∞ Z. Note ∞ Z/ Σ Q L Van Kampen’s Theorem Section 1.2 55 that C is the r...
ched so that the two long edges of Ri are identiﬁed with points of T, as in the second ﬁgure. Any arcs βℓ that cross over αi are positioned to lie in Ri. Finally, over each arc βℓ put a square Sℓ, bent downward along its four edges so that these edges are identiﬁed with points of three strips Ri, Rj, and Rk as in the t...
the Galois group. Let us recall the deﬁnition. A covering space of a space X is a space X together X→X satisfying the following condition: Each point x ∈ X has an X, with a map p : open neighborhood U in X such that p−1(U) is a union of disjoint open sets in each of which is mapped homeomorphically onto U by p. Such a...
embed S 1 in the boundary torus of a solid torus S 1 × D2 so that it winds n times Covering Spaces Section 1.3 57 monotonically around the S 1 factor without self-intersections, then restrict the projection S 1 × D2→S 1 × {0} to this embedded circle. The ﬁgure for Example 1.29 in the preceding section illustrates the ...
Conversely, every covering space of X is a graph that inherits a 2 orientation from X. As the reader will discover by experimentation, it seems that every graph having four edge ends at each vertex can be 2 oriented. This can be proved for ﬁnite graphs as follows. A very classical and easily shown fact is that every ﬁ...
is called the universal cover of X because, as our general theory will show, it is a covering space of every other connected covering space of X. The covering spaces (1)–(14) in the table are all nonsimply-connected. Their fun- e e words in a and b, starting at the basepoint damental groups are free with bases represe...
)→(X, x0) and actual subgroups of π1(X, x0), not just conjugacy classes. e Of course, for these statements to make sense one has to have a precise notion of If one keeps track of the basepoint p : ( X, e e e e when two covering spaces are the same, or ‘isomorphic’. In the case at hand, an iso- e e 60 Chapter 1 The Fund...
f : Y →X is a map properties of covering spaces and derive a few applications of these. f = f. We will describe three special lifting X such that p f : Y → e e e First we have the homotopy lifting property, also known as the covering homo- topy property: Proposition 1.30. Given a covering space p : map lifts ft. e f0 ...
ft : I→X of f0 = p proposition, there is a lifted homotopy of loops X with a f0 to the trivial loop f1. By the remarks preceding the ft starting with f0 and ending with x0) and p∗ is injective. e f0] = 0 in π1( a constant loop. Hence [ X, e e e e For the second statement of the proposition, loops at x0 lifting to loop...
lifts of general maps, not just lifts of homotopies. For the existence question an answer is provided by the following lifting criterion: Proposition 1.33. Suppose given a covering space p : ( x0)→(X, x0) and a map f : (Y, y0)→(X, x0) with Y path-connected and locally path-connected. Then a lift f : (Y, y0)→( π1(Y, y0...
(1). This shows that g well-deﬁned. g g f γ(1) = f is g e e e U ⊂ To see that f (y) such that p : X containing e f is continuous, let U ⊂ X be an open neighborhood of f (y) having U→U is a homeomorphism. Choose a a lift path-connected open neighborhood V of y with f (V ) ⊂ U. For paths from y0 to points y ′ ∈ V we can ...
2 be the sheets containing f2 there is a neighborhood N of y mapped into f2 and p is injective on e f1(y) ≠ e f2 on N since p f2(y) then e f2. If e f1(y) and U2, hence e f1 = f1 = p f1 and U1 by U2 by U1 ≠ Covering Spaces Section 1.3 63 The Classiﬁcation of Covering Spaces We consider next the problem of classifying al...
∈ X has a neighborhood U such that the inclusion-induced map π1(U, x)→π1(X, x) is trivial; one says X is semilocally simply-connected if X→X is a covering this holds. To see the necessity of this condition, suppose p : X simply-connected. Every point x ∈ X has a neighborhood U having a space with U ⊂ lift e e e e X pr...
semilocally simply-connected X purely in terms of X. space X with a basepoint x0 ∈ X, we are therefore led to deﬁne X = [γ] \|\|\|\| γ is a path in X starting at x0 that ﬁx the endpoints γ(0) and γ(1). The function p : where, as usual, [γ] denotes the homotopy class of γ with respect to homotopies X→X sending [γ] to γ(1) ...
�′ = γ η then elements of U[γ′] have the form [γ η µ] and hence lie in U[γ], while elements of U[γ] have the form [γ µ] = [γ η η µ] = [γ′ η µ] and hence lie in U[γ′]. X. For if This can be used to show that the sets U[γ] form a basis for a topology on we are given two such sets U[γ], V[γ′] and an element [γ′′] ∈ U[γ] ∩...
π1( X let γt be the path in X that equals γ on [0, t] and is stationary at γ(t) on [t, 1]. Then the function t ֏ [γt] is a path in X lifting γ that starts at [x0], the homotopy class of X, this the constant path at x0, and ends at [γ]. Since [γ] was an arbitrary point in X, [x0]) = 0 it suﬃces to show shows that that ...
⊂ X and B ⊂ X be the quotients of S 1 × [0, 1/2] and S 1 × [1/2, 1], so A and B are the mapping cylinders of z ֏ zm and z ֏ zn, with A ∩ B = S 1. The simplest case is m = n = 2, when A and B are M¨obius bands and X2,2 is the Klein bottle. We encountered the complexes Xm,n previously in analyzing torus knot complements...
Cm or Cn. Each of these ﬁnite subgraphs deformation retracts onto the just like the construction of e preceding one. The inﬁnite concatenation of these deformation retractions, with the k th graph deformation retracting to the previous one during the time interval [1/2k, 1/2k−1], gives a deformation retraction of Tm,n...
For points [γ], [γ′] in the simply-connected covering space X constructed above, deﬁne [γ] ∼ [γ′] to mean γ(1) = γ′(1) and [γ γ′] ∈ H. It is easy to see that this is an equivalence relation since H is a subgroup: it is reﬂexive since H contains e the identity element, symmetric since H is closed under inverses, and tr...
−1 structures, taking p−1 isomorphism, and the composition of two isomorphisms is an isomorphism, so we X2→X is a homeomorphism f : X1→X and p2 : X1→ e e e have an equivalence relation. e e e X2, = p2∗ e x2) 1 (x0) to a basepoint X1→ x1) X1, e e X1→X and p2 : x1 ∈ p−1 X2 taking a basepoint. π1( Proposition 1.37. If X i...
0)→(X, x0) and the x0) x0). If basepoints are ignored, this correspondence gives a e e X→X isomorphism classes of path-connected covering spaces p : ( set of subgroups of π1(X, x0), obtained by associating the subgroup p∗ to the covering space ( bijection between isomorphism classes of path-connected covering spaces p ...
�� e e e e e e e A consequence of the lifting criterion is that a simply-connected covering space of a path-connected, locally path-connected space X is a covering space of every other path-connected covering space of X. A simply-connected covering space of X is therefore called a universal cover. It is unique up to is...
and b lift to transpositions of diﬀerent pairs of the three vertices, while in (5) and (6) they lift to cyclic permutations of the vertices. In (11) the vertices can be labeled by Z, with a lifting to the identity permutation and b lifting to the shift n ֏ n + 1. Indeed, one can see from these Covering Spaces Section ...
F = p−1(x0), assuming that X is path-connected, e locally path-connected, and semilocally simply-connected, so it has a universal cover X0→X. We can take the points of X0 to be homotopy classes of paths in X starting X0× F→ at x0, as in the general construction of a universal cover. Deﬁne a map h : X e e x0. Then h se...
) to the permutation group of F speciﬁed by the action. e e e e Notice that the deﬁnition of Xρ makes sense whenever we are given an action ρ of π1(X, x0) on a set F regarded as a space with the discrete topology. There is a x0) to γ(1), and this is a covering space since natural projection X0 is a product U × π1(X, x0...
such that the compositions of these two maps, Xρ1→ Xρ2 e e in either order, are the identity. This shows that n sheeted covering spaces of X are classiﬁed by equivalence classes of homomorphisms π1(X, x0)→ n is the symmetric group on n symbols and the equivalence relation identiﬁes a homomorphism ρ with each of its Σ ...
Z2 × Z2. Sometimes normal covering spaces are called regular covering spaces. The term ‘normal’ is motivated by the following result. X, Proposition 1.39. Let p : ( x0)→(X, x0) be a path-connected covering space of the path-connected, locally path-connected space X, and let H be the subgroup p∗ π1( (a) This covering s...
deck transformation τ ′ taking γ (τ( e x0), so ττ ′ is the deck transformation corresponding a path from to [γ][γ′]. By the preceding paragraph ϕ is surjective. Its kernel consists of classes e = H. ⊔⊓ X. These are exactly the elements of p∗ [γ] lifting to loops in 1 then γ γ′ lifts to x′ 1) = ττ ′( x′ x0 to τ( x0 to ...
ﬁxes this point, so g−1 Then g−1 1 g2 = 11 and g1 = g2. Note that in (∗) it suﬃces to take g1 to be the identity since g1(U) ∩ g2(U) ≠ ∅ 1 g2(U) ≠ ∅. Thus we have the equivalent condition that e U in only one point, we must have e U ) ∩ g2( If g1( e x2 ∈ U. Since X satisﬁes (∗). To see this, U ) ≠ ∅ for some x2 must l...
from g(U) onto p(U) for each g ∈ G so we have a covering space. Each element of G acts as a deck transformation, and the covering space is normal since g2g−1 takes g1(U) to g2(U). The deck transformation group contains G as a subgroup, and equals this subgroup if Y is path-connected, since if f is any deck transformat...
only when g1 = g2, since g1(y) = g2(y) is equivalent to g−1 1 g2(y) = y. Though condition (∗) implies freeness, the converse is not always true. An example is the action of Z on S 1 in which a generator of Z acts by rotation through an angle α that is an irrational multiple of 2π. In this case each orbit Zy is dense i...
of a group G on a simply-connected locally path-connected space Y, the orbit space Y /G has fundamental group isomorphic to G. Under this isomorphism an element g ∈ G corresponds to a loop in Y /G that is the projection of 74 Chapter 1 The Fundamental Group a path in Y from a chosen basepoint y0 to g(y0). Any two such...
. Thus we see that the fundamental groups of the torus and the Klein bottle are the symme- try groups G in the two cases. In the second case the subgroup of G formed by the translations has index two, and the orbit space for this subgroup is a torus forming a two-sheeted covering space of the Klein bottle. Example 1.43...
of spaces called lens spaces deﬁned in Example 2.43. There are also noncyclic ﬁnite groups that act freely as rotations of S n for odd n > 1. These actions are classiﬁed quite explicitly in [Wolf 1984]. Examples in the simplest case n = 3 can be produced as follows. View R4 as the quaternion algebra H. Multiplication ...
symmetries of a regular tetrahedron, octahedron (or cube), and icosahedron (or dodecahedron). In fact, it is not hard to see that the homomorphism S 3→SO(3) sending u ∈ S 3 ⊂ H to the isometry v→u−1vu of R3, viewing R3 as the ‘pure imaginary’ quaternions v = ai + bj + ck, is surjective with kernel {±1}. Then 48, I∗ th...
,n). From van Kampen’s theorem applied to the decomposition of Xm,n into the two mapping cylinders we have the presentation for this group Gm,n = π1(Xm,n). It is interesting to look at the action of Gm,n on Xm,n more closely. Xm,n into rectangles, with Xm,n the quotient of We described a decomposition of e Xm,n lifting...
. There is an induced action of the quotient group Zm ∗ Zn on Tm,n, but this is not a free action since the elements a and b and all their conjugates ﬁx vertices of Tm,n. On the other hand, if we restrict the action of Gm,n on Tm,n to the kernel K of the map Gm,n→Z given by the action of Gm,n on the R factor of Xm,n, t...
gα. The action extends to 2 cells in the obvious way. This is clearly a covering space action, and the orbit space is just XG. In fact XG is the universal cover of XG since it is simply-connected. This can be seen by considering the homomorphism ϕ : π1(XG)→G deﬁned in the proof of Proposition 1.39. For an edge eα in XG...
of disks by sending Di to Di+1 via a 2π /n rotation, the subscript i being taken mod n. The common boundary circle of the disks is rotated by 2π /n., XG is RP2 and e e Example 1.48. If G = Z2 ∗ Z2 = an inﬁnite sequence of circles each tangent to its two neighbors. a, b \|\|\|\| a2, b2 then the Cayley graph is a union of e...
a diameter. Do the same when X is the union of a sphere and a circle e e e e intersecting it in two points. 5. Let X be the subspace of R2 consisting of the four sides of the square [0, 1]× [0, 1] together with the segments of the vertical lines x = 1/2, 1/3, 1/4, ··· inside the square. X→X there is some neighborhood ...
common ﬁnite-sheeted covering space X1 = X2, but such that there is no space having both X1 and X2 as covering spaces. 12. Let a and b be the generators of π1(S 1 ∨ S 1) corresponding to the two S 1 e summands. Draw a picture of the covering space of S 1 ∨ S 1 corresponding to the normal subgroup generated by a2, b2, ...
≈ G, π1( covering space X) ≈ G/N. G( e e e e 18. For a path-connected, locally path-connected, and semilocally simply-connected X→X abelian if it is normal and has abelian deck transformation group. Show that X has an abelian covering space that is space X, call a path-connected covering space a covering space of ever...
, and that (X1 × X2)/(G1 × G2) is homeomorphic to X1/G1 × X2/G2. 23. Show that if a group G acts freely and properly discontinuously on a Hausdorﬀ space X, then the action is a covering space action. (Here ‘properly discontinuously’ means that each x ∈ X has a neighborhood U such that { g ∈ G \| U ∩ g(U) ≠ ∅ } is ﬁnite....
Under the Galois correspondence between connected covering spaces of X and subgroups of π1(X, x0), the subgroup corresponding to the component of X e 82 Chapter 1 The Fundamental Group containing a given lift x0 of x0 is the stabilizer of x0, the subgroup consisting of elements whose action on the ﬁber leaves x0 ﬁxed....

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