MATH8808: ALGEBRAIC TOPOLOGY

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1 MATH8808: ALGEBRAIC TOPOLOGY DAWEI CHEN Contents 1. Underlying Geometric Notions Homotopy Cell Complexes Operations on Cell Complexes Criteria for Homotopy Equivalence The Homotopy Extension Property 6 2. The Fundamental Group Basic Constructions The Fundamental Group of S Induced Homomorphisms Van Kampen Theorem Free Products of Groups The Van Kampen Theorem Applications to cell complexes Covering Spaces Lifting Properties Classification of Covering Spaces Deck Transformations and Group Actions Simplicial and Singular Homology complexes Simplicial Homology Singular Homology Homotopy Invariance Exact Sequences and Excision Exact Sequences Relative Homology Groups Excision Equivalence of Simplicial and Singular Homology Computations and Applications Degree Cellular Homology Euler Characteristic Mayer-Vietoris Sequences Homology with Coefficients Cohomology Groups 36 Date: Fall Most of the material is from [H]. 1

2 The Universal Coefficients Theorem Cohomology of Spaces Cup Product Cohomology Ring Künneth Formula Poincaré Duality Orientations The Duality Theorem Cohomology with Compact Supports Duality for Noncompact Manifolds Connection with Cup Product 52 References Homotopy. 1. Underlying Geometric Notions Definition 1.1. Let X be a (topological) space and A a subspace of X. A deformation retraction of X onto A is a (continuous) family of maps f t : X X for t I = [0, 1], such that f 0 = 1, f 1 (X) = A, and f t A = 1 for all t. The family f t is continues in the sense that the map X I X by (x, t) f t (x) is continuous. Example 1.2. Thick letters, Möbius strap, punctured disks. A deformation retraction is a special case of the notion of a homotopy. Definition 1.3. A homotopy is a family of maps f t : X Y for t I such that F : X I Y by F (x, t) = f t (x) is continuous. Two maps f 0, f 1 : X Y are homotopic if there exists a homotopy f t connecting them, and in this case we write f 0 f 1. Suppose X deformation retracts onto A. Denote by r = f 1 : X A the retraction and i : A X the inclusion. Then r i = 1 and i r 1, where the latter is given by f t. Hence one can generalize the situation in a more symmetric way as follows. Definition 1.4. Two spaces X and Y are homotopy equivalent (or simply homotopic) if there exist maps f : X Y and g : Y X such that f g 1 and g f 1. In this case we write X Y. It is easy to check that homotopy equivalence is an equivalence relation. A space A homotopic to a point a is called contractible. Namely, the map p : A a A is homotopic to the identity map 1 on A. In other words, there exists a map f : A I A such that f 0 = 1 and f 1 = p. In particular, one can visualize the contraction from A to a A by changing A to f t (A) A at time t. Example 1.5. The graphs given by a pair of glasses, figure 8, and a capsule are all homotopic, since they are deformation retracts of a 2-punctured disk. Example 1.6. An n-disk is contractible, since it deformation retracts to its center.

3 Cell Complexes. Denote an n-disk by D n. The boundary of D n is an (n 1)- dimensional sphere S n 1. Example 1.7. Construct a torus by gluing one 0-cell, two 1-cells, and one 2-cell. In general, one can construct a space by the following inductive procedure: (1) Start with a discrete set X 0, whose points are 0-cells. (2) Suppose we have constructed X 0,..., X n 1. Let {Dα} n be a collection of n-disks and {ϕ α } a collection of maps ϕ α : S n 1 X n 1. Define X n to be the quotient space of X n 1 α Dα n under the identifications x ϕ α (x) for all x Dα. n Hence as a set, X n = X n 1 α e n α where e n α is an open n-disk. A space X constructed in the above way is called a cell complex or CW complex. 1 Each X n is called the n-skeleton of X and each e n α is called an n-cell of X. If X = X n for some n Z 0, X is called finite-dimensional and dim X = n. Example 1.8. A 1-dimensional cell complex corresponds to a graph, where vertices are 0-cells and edges are 1-cells. Example 1.9. The sphere S n is a cell complex. One can glue two cells e 0 and e n by identifying the boundary of e n as a single point. Example We describe a cell complex structure for the real projective space RP n = R n+1 {0}/, where v λv for scalars λ 0. Note that RP n = S n /v ( v), which is equivalent to saying that RP n is the quotient space of a hemisphere D n with opposite points of D n identified. Since D n = S n 1, we see that RP n is obtained from RP n 1 by attaching an n-cell, where the attaching map is given by S n 1 RP n 1. By induction, we thus conclude that RP n has a cell complex structure with one call e i in each dimension 0 i n. Definition If a closed space A X is a union of cells of X, we say that A is a subcomplex of X and the pair (X, A) is called a CW pair. Example Each skeleton X n is a subcomplex of X. Example In the above construction of RP n, each RP i is a subcomplex of RP n for i n. Remark In general, the closure of a cell may not be a subcomplex. For example, start with S 1 with its minimal cell structure. Attach to it a 2-cell by a map S 1 S 1 whose image is a nontrivial arc of S 1 (say, a semicircle by folding). Then the closure of the 2-cell is not a subcomplex since it contains only part of the 1-cell Operations on Cell Complexes. We describe several standard operations on cell complexes. Products. Suppose X and Y are cell complexes with cells e m α and e n β respectively. Then the products e m α e n β give a cell complex structure for X Y. For example, the cell complex structure of a torus described before is the product S 1 S 1 where S 1 is with its standard cell structure. Quotients. Suppose (X, A) is a CW pair. Then the quotient space X/A has a cell structure whose cells are those of X A plus one new 0-cell as the image of A in 1 The C stands for closure-finite, and the W for weak topology.

4 4 X/A. For a cell e n α of X A with attaching map S n 1 X n 1, the corresponding cell in X/A is attached by the composition S n 1 X n 1 X n 1 /A n 1. For example, given S n 1 with any cell structure, the quotient D n /S n 1 is S n with its standard cell structure. For another example, let X = S 1 S 1 be the torus with the product structure, and A X be the complement of the unique 2-cell. Then X/A consists of a 0-cell and a 2-cell, and hence X/A is S 2. Cone and Suspension. For a cell complex X, define the cone CX = (X I)/(X {0}) with the standard product and quotient structure. One can visualize CX as a cone, say, if X = S 1. Define the suspension SX to be the quotient of X I by collapsing X {0} to a point and X {1} to another point. In other words, SX is the union of two copies of the cone CX. One can visualize SX as a suspension, say, if X = S 1. Join. Note that a cone CX is the union of all line segments joining points of X to a vertex, and a suspension SX is the union of all line segments joining points of X to two vertices. In general, given two spaces X and Y, define the join X Y to be the space of line segments joining points in X to points in Y. More precisely, X Y is defined as the quotient space X Y I under the identifications (x, y 1, 0) (x, y 2, 0), (x 1, y, 1) (x 2, y, 1). In other words, we collapse X Y {0} to X and X Y {1} to Y. For example, if X and Y are both closed intervals, then X Y becomes a tetrahedron. A nice way to write X Y is as formal linear combinations t 1 x + t 2 y with 0 t i 1, t 1 + t 2 = 1, and the identifications 0x + 1y = y and 1x + 0y = x. Similarly, an iterated join X 1 X n can be viewed as the space of formal linear combinations t 1 x t n x n with 0 t i 1, t t n = 1, and 0x i being omitted. An important case is when each X i is a point. The join of n points can be viewed as a convex polyhedron of dimension n 1, called a simplex: n 1 = {(t 1,..., t n ) R n t t n = 1 and t i 0}. For example, the join of two points is a line segment, the join of three points is a triangle, and the join of four points is a tetrahedron. Wedge Sum. Given x X and y Y, define the wedge sum X Y to be the quotient X Y/x y. For example, S 1 S 1 is the figure 8. Smash Product. Given x X and y Y, consider the subspaces X {y} and {x} Y in the product X Y. They only intersect at one point {x} {y}, hence their union can be identified with the wedge sum X Y. Then we define Smash product X Y to be the quotient X Y/X Y. If x and y are 0-cells in X and Y respectively, then X Y is also a cell complex. For example, S m S n has a standard cell structure with only two cells, of dimension 0 and m+n, hence S m S n = S m+n. One can clearly visualize this when m = n = 1 by collapsing the boundary of a square to a point to produce a 2-sphere Criteria for Homotopy Equivalence. One way to produce homotopy equivalent spaces is via collapsing subspaces. Proposition If (X, A) is a CW pair with A contractible, then X X/A. Before we prove it, let us look at some examples first.

5 5 Example A pair of glasses and a capsule are homotopic to the figure 8, since their middle bars are contractible subcomplexes. More generally, suppose X is a graph. If the two endpoints of an edge of X are distinct, then it is contractible, hence we can collapse it to a point and produce a homotopy equivalent graph with fewer edges. Consequently any graph is homotopy equivalent to a graph whose edges are all loops. In particular, any connected component of X is either an isolated vertex or a wedge sum of circles. Example Let X be S 2 attached with an external arc A connecting its north and south poles. Let B S 2 be an internal arc connecting the two poles. Since A and B are both contractible, we conclude that X/A X X/B. Note that X/A is a nodal sphere and X/B S 2 S 1, and it may not be directly obvious that X/A X/B. Another way to produce homotopy equivalent spaces is via attaching one space to another and continuously varying how the parts are attached together. Let A X 1 be a subspace, and f : A X 0 a map to another space. Define X 0 f X 1 as the quotient space X 0 X 1 by identifying each point a A with f(a) X 0. We say that X 0 f X 1 is the space X 0 with X 1 attached along A via f. Example If (X 1, A) = (D n, S n 1 ), this is the case of attaching an n-cell to X 0 via f : S n 1 X 0. Example Let f : X Y be a map, and A = X {1} X I is a copy of X on the bottom of the cylinder X I. Still denote by f the map A Y by (x, 1) f(x). Then M f = (X I) f Y is called the mapping cylinder of f. In particular, M f deformation retracts to Y by sliding each point (x, t) along the segment {x} I to the endpoint f(x) Y (draw a picture). The quotient space C f = M f /X {0} is called the mapping cone of f. Equivalently, C f = CX f Y where CX is the cone (X I)/(X {0}) (draw a picture). Proposition If (X 1, A) is a CW pair and the two attaching maps f, g : A X 0 are homotopic, then X 0 f X 1 X 0 g X 1. Before we prove it, let us again look at some examples first. Example Recall that a nodal sphere is homopotic to S 1 S 2. Let f : A S 1 be the map that wraps an arc A S 2 around S 1. Since A is contractible, f is homotopic to a constant map g : A { }. Hence S 1 f S 2 S 1 g S 2, where the former is a nodal sphere and the latter is S 1 S 2. Example Let (X, A) be a CW pair. Denote by X CA the mapping cone of the inclusion A X. Then we have X/A = (X CA)/CA X CA, where the latter homotopy equivalence is by Proposition 1.15 since CA is contractible. Suppose A is contractible in X, i.e., A X is homotopic to a constant map A x X. Then by Proposition 1.20, the mapping cone X CA of the inclusion A X is homotopic to the mapping cone of the constant map, which is X SA where SA is the suspension of A (draw a picture). Hence by the previous paragraph we have X/A X CA X SA in this case. For example, S 0 (as a discrete set of two points) is contractible in S 2, hence we see again that S 2 /S 0 S 2 S 1, where S 2 /S 0 is the nodal sphere and S 1 is the suspension of S 0.

6 The Homotopy Extension Property. Suppose f 0 : X Y is a map, and A X is a (closed) subspace. Given a homotopy f t : A Y of f 0 A, we would like to extend it to a homotopy f t : X Y of the given f 0. If this extension problem can always be solved, we say that (X, A) has the homotopy extension property. In other words, (X, A) has the homotopy extension property if every pair of maps X {0} Y and A I Y that agree on A {0} can be extended to a map X I Y, i.e., if any map X {0} A I Y can be extended to a map X I Y. A map r : X X is called a retraction of X onto A if r(x) = A and r A = 1. Then a deformation retraction of X onto A is thus a homotopy from the identity map of X to a retraction of X onto A. Proposition A pair (X, A) has the homotopy extension property if and only if X {0} A I is a retraction of X I. Proof. If (X, A) has the homotopy extension property, then by definition the identity map X {0} A I X {0} A I can be extended to a map X I X {0} A I, hence X I retracts onto X {0} A I. Conversely, suppose X {0} A I is a retraction of X I. Given a map X {0} A I Y, composing it with the retraction map X I X {0} A I, we thus obtain the desired extension X I Y. Remark Even if A is not closed in X, Proposition 1.23 still holds. Proposition If (X, A) is a CW pair, then X {0} A I is a (deformation) retraction of X I, hence (X, A) has the homotopy extension property. Proof. See the proof of [H, Proposition 0.16]. Recall that Proposition 1.15 says that collapsing a contractible subcomplex is a homotopy equivalence. Now we prove a generalization of this assertion. Proposition If the pair (X, A) satisfies the homotopy extension property and A is contractible, then the quotient map q : X X/A is a homotopy equivalence. Proof. We need to find an inverse homotopy equivalent map for q. For t I, suppose f t : X X is a homotopy extending a contraction of A, where f 0 = 1. Since f t (A) A for all t, let f t : X/A X/A be the map induced by f t. Then we have qf t = f t q (draw a commutative diagram). Note that f 1 (A) is a point, so f 1 induces a map g : X/A X such that f 1 = gq (draw a commutative diagram). It follows that qg = f 1. Therefore, g and q are inverse homotopy equivalences, since gq = f 1 f 0 = 1 via f t and qg = f 1 f 0 = 1 via f t. Next we generalize Proposition First, for two pairs (W, Y ) and (Z, Y ), if there exist maps ϕ : W Z and ψ : Z W, restricting to the identity on Y, such that ϕψ 1 and ψϕ 1 via homotopies that restrict to the identity on Y at all times, we say that W and Z are homotopic relative to Y and denote it by W Z rel Y. Proposition If (X 1, A) is a CW pair and we have attaching maps f, g : A X 0 that are homotopic, then X 0 f X 1 X 0 g X 1 rel X 0. Proof. Let F : A I X 0 be a homotopy from f to g. Consider the space X 0 F (X 1 I) that glues X 0 and X 1 I with the attaching map F. It contains

7 7 both X 0 f X 1 and X 0 g X 1 as subspaces. Applying Proposition 1.25 to the CW pair (X 1, A), there is a deformation retraction of X 1 I onto X 1 {0} A I, which induces a deformation retraction of X 0 F (X 1 I) onto X 0 F 0 X 1 = X 0 f X 1 (why?). Similarly, X 0 F (X 1 I) deformation retracts onto X 0 g X 1. Note that both deformation retractions restrict to the identity on X 0, hence together they give a homotopy equivalence X 0 f X 1 X 0 g X 1 rel X The Fundamental Group 2.1. Basic Constructions. A path in a space X is a continuous map f : I X, where I = [0, 1]. A homotopy of paths is a family f t : I X, 0 t 1, such that (1) The endpoints f t (0) = x 0 and f t (1) = x 1 are independent of t. (2) The map F : I I X given by F (s, t) = f t (s) is continuous. When two paths f 0 and f 1 are connected by such a homotopy f t, we say that they are homotopic (rel the endpoints) and denote it by f 0 f 1. Geometrically speaking, it means that f 0 can continuously deform to f 1. Example 2.1. Any two paths f 0 and f 1 in R n (with the same endpoints) are homotopic via the linear homotopy f t (s) = (1 t)f 0 (s) + tf 1 (s). Proposition 2.2. The relation of homotopy on paths with fixed endpoints is an equivalence relation. Consequently for a path f, we denote by [f] the homotopy class of f. Proof. A path f f via the constant homotopy f t = f for all 0 t 1, which verifies reflexivity. If f 0 f 1 via the homotopy f t (s), then f 1 f 0 via f 1 t (s), which verifies symmetry. Finally suppose f 0 f 1 via f t and g 0 g 1 via g t, where f 1 = g 0. Define h t = f 2t for 0 t 1/2 and h t = g 2t 1 for 1/2 t 1 (draw a picture). Then f 0 g 1 via h t, which verifies transitivity. Given two paths f, g : I X such that f(1) = g(0), define the composition f g (from left to right) by f g(s) = f(2s) for 0 s 1/2 and f g(s) = g(2s 1) for 1/2 s 1. If f 0 f 1 via f t and g 0 g 1 via g t such that f 0 (1) = g 0 (0) and f 1 (1) = g 1 (0), then f 0 g 0 f 1 g 1 via f t g t (draw a picture). Therefore, this operation preserves homotopy classes. For a path f : I X with f(0) = f(1) = x 0 X, we say that f is a loop (or a closed path) and call x 0 the base point. The set of all homotopy classes [f] of loops at the base point x 0 is denoted by π 1 (X, x 0 ). Proposition 2.3. π 1 (X, x 0 ) is a group with respect to the product [f][g] = [f g]. This group π 1 (X, x 0 ) is called the fundamental group of X at the base point x 0, which is not necessarily commutative. Proof. For two loops f, g with the same base point x 0, f g is obviously a loop with base point x 0. We have also verified that the product [f][g] = [f g] is well-defined. It remains to verify the group axioms for π 1 (X, x 0 ). Some preparation first. Suppose ϕ : I I is any continuous map such that ϕ(0) = 0 and ϕ(1) = 1. For a path f, we call the new path fϕ a reparameterization

8 8 of f. Note that fϕ f via the homotopy fϕ t where ϕ t : I I is defined by ϕ t (s) = (1 t)ϕ(s)+ts, so that ϕ 0 = ϕ and ϕ 1 = 1. In other words, we have ϕ 1 on I. Given paths f, g, h with f(1) = g(0) and g(1) = h(0), we want to verify that (f g) h f (g h). Note that and (f g) h = f (g h) = { (f g)(2s), 0 s 1/2 h(2s 1), 1/2 s 1 { f(2s), 0 s 1/2 (g h)(2s 1), 1/2 s 1 f(4s), 0 s 1/4 = g(4s 1), 1/4 s 1/2 h(2s 1), 1/2 s 1 f(2s), 0 s 1/2 = g(4s 2), 1/2 s 3/4 h(4s 3), 3/4 s 1. Take a piecewise linear function ϕ that maps the three subintervals in the latter to the three subintervals in the former, respectively. Composing with ϕ reparameterizes (f g) h as f (g h). This verifies that the product operation in π 1 (X, x 0 ) is associative. Given a path f : I X, define a constant path c at f(1) by c(s) = f(1) for all s I. Then one can similarly check as above that f c f c f. In particular, the homotopy class of the constant path at x 0 is a two-sided identity in π 1 (X, x 0 ). For a path f starting at x 0, define the inverse path f by f(s) = f(1 s). Then one checks that f f is homotopic to the constant path at x 0. In particular if f is a loop with base point x 0, then [f] is a two-sided inverse for [f] in π 1 (X, x 0 ). Example 2.4. Suppose X is a convex set in R n with a base point x 0 X. Then any two loops f 0 and f 1 based at x 0 are homotopic via the linear homotopy We thus conclude that π 1 (X, x 0 ) = 0. f t (s) = (1 t)f 0 (s) + tf 1 (s). For any two points x 0, x 1 X, if there exists a path h such that h(0) = x 0 and h(1) = x 1, we say that X is path connected. For any loop f based at x 1, we associate the loop h f h based at x 0 (draw a picture). It induces a change of base point map β h : π 1 (X, x 1 ) π 1 (X, x 0 ). Proposition 2.5. The map β h : π 1 (X, x 1 ) π 1 (X, x 0 ) is an isomorphism. Proof. Since [h][h] is the identity, we have β h [f g] = [h f g h] = [h f h h g h] = β h [f]β h [g], so β h is a homomorphism. Moreover, β h is the inverse of β h, since and similarly β h β h [f] = [f]. β h β h [f] = β h [h f h] = [h h f h h] = [f] Therefore, if X is path connected, then the group π 1 (X, x 0 ) is independent of the base point, up to isomorphism. In this case we simply write π 1 (X) instead. If X is path connected and if π 1 (X) = 0, we say that X is simply connected. Proposition 2.6. A space X is simply connected iff there is a unique homotopy class of paths connecting any two points in X.

9 9 Proof. If X is simply connected, then any two points x 0 and x 1 in X can be connected by a path. If f and g are two paths joining x 0 and x 1, then f f (g g) (f g) g g, since f g is homotopic to a constant loop by the assumption on π 1 (X). Conversely, suppose for any two points in X there is a unique homotopy class of paths connecting them. Then X is path connected. Moreover, any two loops connecting x 0 to itself are homotopic, hence π 1 (X, x 0 ) = The Fundamental Group of S 1. In this section we will prove the following result. Theorem 2.7. π 1 (S) = Z generated by the homotopy class of the loop ω(s) = (cos 2πs, sin 2πs) based at (1, 0). For n Z, set ω n (s) = (cos 2πns, sin 2πns). Note that [ω] n = [ω n ]. Then the theorem says that every loop in S 1 based at (1, 0) is homotopic to ω n for a unique n. To prove the theorem, we will use the map p : R S 1 given by p(s) = (cos 2πs, sin 2πs) as a cover of S 1. So we first introduce the notion of covering space in more generality. Give a space X, a covering space of X is a space X with a map p : X X satisfying the following condition: For each point x X, there is an open neighborhood U of x in X such that p 1 (U) is a union of disjoint open sets each of which maps homeomorphically onto U via p. Such U in the above is called evenly covered. For example, any open arc in S 1 is evenly covered by p. Given a covering p : X X, we have the following facts: (a) For each path f : I X starting at x 0 X and each x 0 p 1 (x 0 ), there is a unique lift f : I X starting at x 0. (b) For each homotopy f t : I X of paths starting at x 0 X and each x 0 p 1 (x 0 ), there is a unique lifted homotopy f t : I X of paths starting at x 0. These facts can be deduced from a more general fact (c), see [H, p. 30]. Proof of Theorem 2.7. Suppose f : I S 1 is a loop at the base point x 0 = (1, 0), representing an element of π 1 (S 1, x 0 ). By (a) there is a lift f to R starting at 0 p 1 (1, 0) = Z. Suppose f ends at some integer n. Note that ω n : I R given by ω n (s) = ns is another path joining 0 to n, and f ω n via the linear homotopy. Composing it with p gives a homotopy f ω n, so [f] = [ω n ]. Now suppose ω n ω m. Let f t be a homotopy from ω m = f 0 to ω n = f 1. By (b) this homotopy lifts to a homotopy f t of paths starting at 0. The uniqueness part of (a) implies that ω m = f 0 and ω n = f 1. Since f t is a homotopy of paths, the endpoint of f t (1) is independent of t. For t = 0 the endpoint is m and for t = 1 the endpoint is n, so m = n. We discuss an application. Theorem 2.8 (Brouwer Fixed Point Theorem). Every continuous map h : D 2 D 2 has a fixed point, where D 2 is the closed unit disk.

10 10 The upshot is that there is no retraction of D 2 onto S 1. Proof. Suppose on the contrary that h(x) x for all x D 2. Define a map r : D 2 S 1 by sending x to the intersection of the ray h(x)x with D = S 1 (draw a picture). Clearly r is continuous, and r(x) = x for all x S 1, so r is a retraction of D 2 onto S 1. Suppose f 0 is any loop in S 1 based at x 0 S 1. Since D 2 is convex, the linear homotopy f t (s) = (1 t)f 0 (s)+tx 0 gives a homotopy from f 0 to the constant loop at x 0. Since r is the identity on S 1, the composition rf t is a homotopy from rf 0 = f 0 to rf 1 which is the constant loop at x 0. This contradicts that π 1 (S 1 ) 0. Now we study the fundamental group of a product space. Proposition 2.9. Suppose X and Y are path connected. isomorphic to π 1 (X) π 1 (Y ). Then π 1 (X Y ) is Proof. A path f : I X Y is of the form f(s) = (g(s), h(s)), hence it is equivalent to a pair of paths g in X and h in Y. Similarly a homotopy f t in X Y is equivalent to a pair of homotopies g t in X and h t in Y. Thus we obtain an isomorphism π 1 (X Y ) = π 1 (X) π 1 (Y ) given by [f] ([g], [h]). Example The torus has fundamental group π 1 (S 1 S 1 ) = Z Z. We mention a useful result for future use. Theorem 2.11 (Borsuk-Ulam Theorem). For every continuous map f : S n R n, there exists a pair of antipodal points x and x in S n such that f(x) = f( x). Remark For n = 1, the proof is simple since f(x) f( x) changes sign as x goes half-way around the circle, hence the difference must be zero for some x. In general, the theorem implies that there is no continuous one-to-one map from S n to R n, so S n is not homeomorphic to any subspace of R n. Proof. See the proof of [H, Theorem 1.10]. Corollary If S 2 R 3 is expressed as the union of three closed sets A 1, A 2, and A 3, then at least one of these sets contain a pair of antipodal points {x, x}. Proof. Let d i : S 2 R be the distance function to A i, i.e., d i (x) = inf y A i x y. Clearly d i is a continuous function, so the map S 2 R 2 by x (d 1 (x), d 2 (x)) is continous. By the Borsuk-Ulam theorem, there exist x S 2 such that d 1 (x) = d 1 ( x) and d 2 (x) = d 2 ( x). If either of these two distances is zero, then x and x both lie in the same set A 1 or A 2 since they are closed sets. If both distances are nonzero, then x and x are not in A 1 A 2, so they both lie in A Induced Homomorphisms. Suppose ϕ : (X, x 0 ) (Y, y 0 ) is a map sending x 0 to y 0. For any loop f : I X with base point x 0, define ϕ [f] = [ϕf], which induces a homomorphism ϕ : π 1 (X, x 0 ) π 1 (Y, y 0 ). One checks that ϕ is well-defined and satisfies the properties of being a homomorphism. Moreover for ψ : (Y, y 0 ) (Z, z 0 ), we have (ψϕ) = ψ ϕ. Finally if 1 : X X is the identity map, then 1 is the identity on π 1 (X, x 0 ).

11 11 As an application, if ϕ is a homeomorphism with inverse ψ, then ϕ ψ = (ϕψ) = 1 = 1 and similarly ψ ϕ = 1, so ϕ and ψ are isomorphisms. In other words, two (path connected) homeomorphic spaces have the same fundamental group. We now study the fundamental groups of higher dimensional spheres. Theorem For n 2, π 1 (S n ) = 0. We introduce the following lemma first, which will also be used later. Lemma Suppose {A α } is an open cover of a space X such that each A α contains the base point x 0, each A α is path connected, and each intersection A α A β is path connected. Then every loop in X at x 0 is homotopic to a product of loops each of which is contained in a single A α. Proof. Given a loop f : I X with base point x 0, we want to find a partition 0 = s 0 < s 1 < < s m = 1 of I such that each subinterval [s i 1, s i ] is mapped by f to a single A α. Since f is continuous, for each s I, there is a (closed) subinterval V s containing s mapped by f to some A α. Since I is compact (any open cover has a finite subcover), there are a finite number of these subintervals that cover I. Then the endpoints of these intervals define the desired partition of I. Denote by A i the subinterval containing f([s i 1, s i ]) in the above, and let f i be the restriction of f to [s i 1, s i ]. Then f = f 1 f m with f i a path in A i. Since A i A i+1 is path connected, we may choose a path g i in A i A i+1 from x 0 to f(s i ) A i A i+1. Consider the loop (f 1 g 1 ) (g 1 f 2 g 2 ) (g 2 f 3 g 3 ) (g m 1 f m ), which is homotopic to f (draw a picture). Each loop in this composition lies in a single A i. Proof of Theorem Take A 1 and A 2 to be the complements of two antipodal points in S n, so S n is a union of the two open sets A 1 and A 2. The intersection A 1 A 2 is homeomorphic to S n 1 R, and choose a base point x 0 in A 1 A 2. If n 2, then A 1 A 2 is path connected. By the above lemma, every loop in S n based at x 0 is homotopic to a product of loops in A 1 or A 2. Since A i is homeomorphic to R n for i = 1, 2, we have π 1 (A i ) = 0, and the claim thus follows. Corollary R 2 is not homeomorphic to R n for n 2. Proof. Suppose on the contrary f : R 2 R n is a homeomorphism. If n = 1, since R 2 {0} is path connected but R 1 {f(0)} is not, the claim follows. When n > 2, for a point x = f(0) R n, the complement R n {x} is homeomorphic to S n 1 R, so Proposition 2.9 implies that π 1 (R n {x}) = π 1 (S n 1 ) π 1 (R), which is trivial by using Theorem contradiction. But π 1 (R 2 {0}) = Z, leading to a Remark More generally for m n, R m is not homeomorphic to R n. This can be proved similarly by using either higher homotopy groups or homology groups. Now we study the behavior of fundamental groups under retractions. Proposition If a space X retracts onto a subspace A, then i : π 1 (A, x 0 ) π 1 (X, x 0 ) induced by the inclusion is injective. Moreover if A is a deformation retract of X, then i is an isomorphism.

12 12 Proof. Let r : X A be a retraction. Since ri = 1, we have r i = 1 on π 1 (A, x 0 ), hence i is injective. If r t : X X is a deformation retraction of X onto A, where t I, then r 0 = 1 on X, r t A = 1, and r 1 (X) = A. For any loop f : I X based at x 0 A, the composition r t f gives a homotopy of f to a loop in A, so in this case i is also surjective. Next we consider more generally the fundamental groups of two homotopic spaces. Proposition If ϕ : X Y is a homotopy equivalence, then ϕ : π 1 (X, x 0 ) π 1 (Y, ϕ(x 0 )) is an isomorphism for all x 0 X. Let us prove the following lemma first. Lemma If ϕ t : X Y is a homotopy and h is the path ϕ t (x 0 ) formed by the images of a base point x 0 X, then the following diagram commutes: ϕ 0 = β h ϕ 1 where β h = [h () h] is the change of base point map (draw a diagram). Proof. Let h t be the restriction of h to [0, t], reparameterized such that the domain of h t is still [0, 1]. More explicitly we can take h t (s) = h(ts). Note that h(s) = ϕ s (x 0 ), so h t (s) = h(ts) = ϕ ts (x 0 ). If f is a loop in X based at x 0, then h t (ϕ t f) h t gives a homotopy of loops based at ϕ 0 (x 0 ) (draw a picture). Its restrictions to t = 0 and t = 1 are ϕ 0 f and β h (ϕ 1 f), so we conclude that ϕ 0 ([f]) = β h (ϕ 1 ([f])). Proof of Proposition Let ψ : Y X be a homotopy inverse of ϕ, so that ϕψ 1 and ψϕ 1. Consider the maps π 1 (X, x 0 ) ϕ π 1 (Y, ϕ(x 0 )) ψ π 1 (X, ψϕ(x 0 )) ϕ π 1 (Y, ϕψϕ(x 0 )). Applying the above lemma to the homotopy ψϕ 1, we see that ψ ϕ = β h 1 = β h for some h, which is an isomorphism. It follows that ϕ is injective. Similarly ψ is injective. But their composition is an isomorphism, so ϕ (and ψ ) must be surjective as well. 3. Van Kampen Theorem The Van Kampen Theorem gives a method to compute the fundamental group of a space that can be decomposed into simpler spaces with known fundamental groups Free Products of Groups. Before we state the theorem, we first recall the definition of taking free products of groups. Given a collection of groups {G α }, we define the free product group α G α as follows: The elements are words g 1 g 2 g m of arbitrary finite length m 0 such that g i G αi and is not the identity element and that any two adjacent α i α i+1. Such words are called reduced. The group operation is (g 1 g m )(h 1 h n ) = g 1 g m h 1 h n if g m and h 1 are not in the same G α. Otherwise we combine (g m h 1 ) as a single letter. (g 1 g m ) 1 = gm 1 g1 1. The empty word is the identity element.

13 The Van Kampen Theorem. Suppose X is the union of path-connected open sets A α, such that all A α contain the base point x 0 X (omitted in the notations below). The inclusion A α X induces a homomorphism j α : π 1 (A α ) π 1 (X). They extend to a homomorphism Φ : α π 1 (A α ) π 1 (X). Denote by i αβ : π 1 (A α A β ) π 1 (A α ) induced by the inclusion A α A β A α. Note that j α i αβ = j β i βα, both induced by the inclusion A α A β X. It follows that ker Φ contains all the elements of the form i αβ (ω)i βα (ω) 1, because Φ(i αβ (ω)i βα (ω) 1 ) = j α (i αβ (ω)) j β (i βα (ω) 1 )) = (j α i αβ )(ω) ((j β i βα )(ω)) 1. Theorem 3.1. If X is the union of path-connected open sets A α each containing the base point x 0 X and if each intersection A α A β is path-connected, then the homomorphism Φ : α π 1 (A α ) π 1 (X) is surjective. If in addition each intersection A α A β A γ is path-connected, then the kernel of Φ is the normal subgroup N generated by all elements of the form i αβ (ω)i βα (ω) 1 for ω π 1 (A α A β ), and hence Φ induces an isomorphism π 1 (X) = α π 1 (A α )/N. Note that if N = ker Φ, then N is automatically a normal subgroup. The surjectivity part of the theorem has been proved already in Lemma The other part is more technical, see [H] for a detailed proof. Let us look at some examples. Example 3.2. In the setting of the Van Kampen Theorem, if there are only two open sets A α and A β in the cover of X, then the condition on triple intersections is redundant, and one obtains an isomorphism π 1 (X) = (π 1 (A α ) π 1 (A β ))/N, under the assumption that A α A β is path connected. Example 3.3. Consider the wedge sum α X α of a collection of path connected spaces X α with base points x α to be the quotient of the disjoint union α X α in which all the base points x α are identified to a single point. If each x α is a deformation retract of an open neighborhood U α in X α, then X α is a deformation retract of its open neighborhood A α = X α β α U β, which is path connected, since X α is. The intersection of two or more distinct A α s is α U α, which deformation retracts to a point, hence it (is path connected and) has trivial fundamental group. The Van Kampen Theorem thus implies that Φ : α π 1 (X α ) π 1 ( α X α ) is an isomorphism. For instance, a wedge sum α Sα 1 of circles has fundamental group isomorphic to the free product of copies of Z, one for each Sα. 1 Example 3.4. Let A be a circle in R 3. Consider the complement R 3 A. It can deformation retract to S 2 union a diameter, by pushing points outside S 2 to S 2 and points inside S 2 to S 2 or the diameter away from A (draw a picture). Since S 2 union a diameter is homotopic to S 2 S 1, by the Van Kampen Theorem, π 1 (R 3 A) = π 1 (S 2 ) π 1 (S 1 ) = Z, since π 1 (S 2 ) is trivial. Indeed when one bends the diameter, one can see directly that R 3 A deformation retracts to S 2 S 1 (draw a picture). Similarly if A and B are two unlinked circles in R 3, then R 3 A B deformation retracts to S 1 S 1 S 2 S 2, hence π 1 (R 3 A B) = Z Z. If A and B are two linked circles in R 3, then R 3 A B deformation retracts to the wedge sum of S 2 and a torus S 1 S 1 separating A and B (draw a picture), hence π 1 (R 3 A B) = π 1 (S 1 S 1 ) = Z Z.

14 Applications to cell complexes. Let us study how the fundamental group is affected by attaching 2-cells. Suppose we attach a collection of 2-cells e 2 α to a path connected space X via maps ϕ α : S 1 X, producing a space Y. If s 0 is a base point of S 1, then ϕ α determines a loop based at ϕ α (s 0 ) in X, which we still denote by ϕ α. Since ϕ α (s 0 ) may not coincide for all α, choose a base point x 0 X and a path γ α from x 0 to ϕ α (s 0 ). Then γ α ϕ α γ α is a loop at x 0. Let N π 1 (X, x 0 ) be the subgroup generated by γ α ϕ α γ α for varying α. Proposition 3.5. The inclusion X Y induces a surjection π 1 (X, x 0 ) π 1 (Y, x 0 ) whose kernel is N, i.e., π 1 (Y ) = π 1 (X)/N. It follows that N is independent of the choice of the paths γ α. One can also see this directly. If we replace γ α by another path η α, then γ α ϕ α γ α changes to η α ϕ α η α = (η α γ α )γ α ϕ α γ α (γ α η α ), so γ α ϕ α γ α and η α ϕ α η α define conjugate elements of π 1 (X, x 0 ). Proof. We first expand Y to a larger space Z that deformation retracts to Y. For each α, take a rectangular strip S α = I I and attach it to Y by gluing the lower edge I {0} along γ α, gluing the right edge {1} I along an arc in e 2 α and identifying the left edge {0} I to a common arc above x 0 for all α (draw a picture). We do not attach anything to the top edges. Clearly the resulting space Z deformation retracts to Y by shrinking the height of S α. In each e 2 α choose a point y α not in the arc attached to S α. Let A = Z α {y α } and B = Z X. Then A deformation retracts to X and B is contractible, hence π 1 (B) = 0. Applying Van Kampen s Theorem to the cover {A, B} of Z, we see that π 1 (Y ) = π 1 (Z) = π 1 (X)/N, where N is the (normal) subgroup generated by the image of the map π 1 (A B) π 1 (A) = π 1 (X). It remains to show that π 1 (A B) is generated by loops in A B that are homotopic to the loops γ α ϕ α γ α. We apply Van Kampen s Theorem one more time to the cover of A B = Z X α {y α } (the strips with bottom removed union the open punctured 2-cells) by the open sets A α = A B β α e 2 β. Since A α deformation retracts to a circle in e 2 α {y α } (a punctured open disk), we conclude that π 1 (A α ) = Z is generated by a loop homotopic to γ α ϕ α γ α. Example 3.6. Let M g be the orientable surface of genus g, constructed by using one 0-cell, 2g 1-cells, and one 2-cell (draw a picture). The 1-skeleton is a wedge sum of 2g circles, hence its fundamental group is the free product Z Z with 2g copies of Z. The 2-cell is attached along the loop given by the product of the commutators [a 1, b 1 ] [a g, b g ]. Hence by the preceding proposition, we have π 1 (M g ) = a 1, b 1,, a g, b g [a 1, b 1 ] [a g, b g ]. Note that the abelianization of π 1 (M g ) is the direct sum of 2g copies of Z. In particular for g h, it implies that M g is not homeomorphic, or even homotopic to M h. Corollary 3.7. For every group G there is a 2-dimensional cell complex X G with π 1 (X G ) = G.

15 15 Proof. Choose a presentation G = g α r β, which means G is generated by g α s with relation r β s. It exists because every group is a quotient of a free group. Now construct X G from α Sα 1 by attaching 2-cells e 2 β by the loops specified by the words r β s. 4. Covering Spaces Recall that a covering space of a space X is a space X together with a map p : X X such that For each point x X, there is an open neighborhood U in X such that p 1 (U) is a union of disjoint open sets in X, each of which is mapped homeomorphically onto U by p. Such a U is called evenly covered and the disjoint open sets in X that map homeomorphically onto U are called sheets of X over U Lifting Properties. A lift of a map f : Y X is a map f : Y X such that p f = f. Recall the following homotopy lifting property we discussed before. Proposition 4.1. Given a covering space p : X X, a homotopy ft : Y X and a map f 0 : Y X lifting f 0, then there exists a unique homotopy f t : Y X of f 0 that lifts f t. The special cases when Y is a point and Y = I give the path lifting property and path homotopy lifting property. Below is a simple application. Proposition 4.2. The map p : π 1 ( X, x 0 ) π 1 (X, x 0 ) induced by a covering space p : ( X, x 0 ) (X, x 0 ) is injective. The image subgroup p (π 1 ( X, x 0 )) in π 1 (X, x 0 ) consists of the homotopy classes of loops in X based at x 0 whose lifts to X starting at x 0 are loops. Proof. Suppose f 0 : I X is a loop at x 0 in ker p. Let f t : I X be a homotopy connecting f 0 = p f 0 to the trivial loop f 1 at x 0. By the homotopy lifting property, there is a lifted homotopy f t starting with f 0 and ending with a constant loop. Hence [ f 0 ] = 0 in π 1 ( X, x 0 ) and p is injective. For the other statement, loops at x 0 lifting to loops at x 0 certainly represent elements of the image of p. Conversely, if a loop represents an element of the image of p, then it is homotopic to a loop having such a lift, hence by the homotopy lifting property, the loop itself must have such a lift. Proposition 4.3. Suppose X and X are path connected. Then the number of sheets of the covering space p : ( X, x 0 ) (X, x 0 ) equals the index of p (π 1 ( X, x 0 )) in π 1 (X, x 0 ). Proof. For a loop g in X based at x 0, denote by g its lift to X starting at x 0. Let H = p (π 1 ( X, x 0 )). A product h g with [h] H has the lift h g ending at the same point as g since h is a loop by the preceding proposition. Thus we define a function Φ from all cosets H[g] to π 1 (x 0 ) by H[g] g(1). It suffices to show that Φ is an isomorphism.

16 16 Since X is path connected, x 0 can be joined to any point in p 1 (x 0 ) by a path g projecting to a loop g at x 0, hence Φ is surjective. If Φ(H[g 1 ]) = Φ(H[g 2 ]), then g 1 and g 2 end at the same point, hence g 1 g 2 lifts to a loop in X based at x 0, so [g 1 ][g 2 ] 1 H by the preceding proposition, and hence H[g 1 ] = H[g 2 ], which implies that Φ is injective. If for each point x X and each neighborhood U of x there exists an open neighborhood V U of x such that V is path connected, we say that X is locally path connected. Example 4.4. Consider the graph of y = sin(π/x) for 0 < x < 1, together with a closed arc joining (1, 0) to (0, 0). Then this space is path connected, but it is not locally path connected at (0, 0). To conclude this section, we mention a more general lifting criterion. Proposition 4.5. Given a covering space p : ( X, x 0 ) (X, x 0 ) and a map f : (Y, y 0 ) (X, x 0 ) with Y path connected and locally path connected, a lift f : (Y, y 0 ) ( X, x 0 ) of f exists if and only if f (π 1 (Y, y 0 )) p (π 1 ( X, x 0 )). Proposition 4.6. Given a covering space p : X X and a map f : Y X, if two lifts f 1, f 2 : Y X of f agree at one point of Y and if Y is connected, then f 1 and f 2 agree on all of Y Classification of Covering Spaces. Since we talk about paths all the time, here we focus on path connected and locally path connected spaces X as well as path connected covering spaces. The motivating question is whether every subgroup of π 1 (X, x 0 ) can be realized as p (π 1 ( X, x 0 )) for some covering space p : ( X, x 0 ) (X, x 0 ). In particular to realize the trivial subgroup, since p is injective, it amounts to asking whether X has a simply connected covering space. Let us first develop a necessary condition. Suppose p : X X is a covering space with X simply connected. Every point x X has a neighborhood U with a lift Ũ X projecting homeomorphically to U by p. Each loop in U lifts to a loop in Ũ, and the lifted loop has trivial homotopy class since π 1( X) = 0. Therefore, composing it with p implies that the original loop in U is trivial in π 1 (X). To summarize, if each x X has a neighborhood U such that the inclusion-induced map π 1 (U, x) π 1 (X, x) is trivial, we say that X is semilocally simply connected. Remark 4.7. A locally simply connected space is certainly semilocally simply connected. Moreover, CW complexes satisfy the stronger property of being locally contractible. Conversely, given a path connected, locally path connected and semilocally simply connected space X, we shall construct a simply connected covering space X of X. To motivate the construction, suppose p : ( X, x 0 ) (X, x 0 ) is a simply connected covering space. Then each x X can be joined to x 0 by a unique homotopy class of paths. Consequently by the homotopy lifting property, homotopy classes of paths in X starting at x 0 are the same as homotopy classes of paths in X starting at x 0, since such a lift has unique homotopy class. This gives a way of describing X in terms of X. More precisely, define X = {[γ] γ is a path in X starting at x 0 }.

17 17 The function p : X X given by [γ] γ(1) is well-defined, since homotopic paths have the same endpoints. Since X is path connected, γ(1) can be any point of X, so p is surjective. Next we want to define a topology on X. Some observations first. Let U be the collection of path connected open sets U X such that π 1 (U) π 1 (X) is trivial. Since U is path connected, if the map π 1 (U) π 1 (X) is trivial for one choice of base point in U, then it is trivial for all choices of base point. If V U is a path connected open subset, since the composition π 1 (V ) π 1 (U) π 1 (X) is trivial, it follows that V is also in U. Therefore, U is a basis for the topology on X. Given a set U in U and a path γ in X joining x 0 to a point in U, let U [γ] = {[γ η] η is a path in U with η(0) = γ(1)}. Clearly U [γ] depends only on the homotopy class [γ]. Since U is path connected, p : U γ U is surjective. Since π 1 (U) π 1 (X) is trivial, different paths η in U joining γ 1 (1) to a fixed x U are all homotopic in X, and hence p : U γ U is also injective. Moreover, we have U [γ] = U [γ ] if [γ ] U [γ]. This is because if [γ ] = [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 ]. The above property can be used to show that the sets U [γ] form a basis for a topology on X. Given two such sets U [γ] and V [γ ] and an element [γ ] U [γ] V [γ ], by the above we have U [γ] = U [γ ] and V [γ ] = V [γ ]. So if W U is contained in U V and contains γ (1), then W [γ ] U [γ ] V [γ ] = U [γ] V [γ ] and [γ ] W [γ ]. Now we show that the bijection p : U γ U is a homeomorphism, by showing that it gives a bijection between the subsets V [γ ] U [γ] and the sets V U contained in U. In one direction we have seen that p(v [γ ]) = V. In the other direction, for any [γ ] U [γ] with endpoint in V, since V [γ ] U [γ ] = U [γ] and V [γ ] maps onto V by the bijection p, we have p 1 (V ) U [γ] = V [γ ]. Next we show that p : X X is a covering space. For fixed U U, the sets U [γ] for varying [γ] partition p 1 (U), because if [γ ] U [γ] U [γ ], then U [γ] = U [γ ] = U [γ ] by the above property. We further show that X is path connected. For any point [γ] X, let γ t be the path in X such that it equals γ on [0, t] and is stationary at γ(t) on [t, 1]. Then the function t [γ t ] is a path in X lifting γ that starts at [x 0 ], the homotopy class of the constant path at x 0, and ends at [γ]. This shows that X is path connected. To show that X is simply connected, it suffices to show that p (π 1 ( X, [x 0 ])) = 0 since p is injective. Recall that elements in the image of p are represented by loops γ at x 0 that lift to loops in X at [x 0 ]. In the preceding paragraph, we have observed that the path t [γ t ] lifts γ and starts at [x 0 ]. For this lifted path to be a loop, it means that [γ 1 ] = [x 0 ]. Since γ 1 = γ, it implies that [γ] = [x 0 ] is trivial in π 1 (X, x 0 ). The hypotheses for constructing a simply connected covering space of X in fact suffice for constructing covering spaces realizing arbitrary subgroups of π 1 (X). Proposition 4.8. Suppose X is path connected, locally path connected and semilocally simply connected. Then for every subgroup H π 1 (X, x 0 ), there is a covering space p : X H X such that p (π 1 (X H, x 0 )) = H for some base point x 0 X H.

18 18 Proof. Recall the construction of a simply connected covering space X above. For points [γ], [γ ] X, define [γ] [γ ] if γ(1) = γ (1) and [γ γ ] H. Since H is a subgroup, it is easy to see that this is an equivalence relation. Let X H be the quotient space of X obtained by identifying [γ] with [γ ] if [γ] [γ ]. Note that if γ(1) = γ (1), then [γ] [γ ] if and only if [γ η] [γ η]. It means that if any two points in basic neighborhoods U [γ] and U [γ ] are identified in X H, then the whole neighborhoods are identified. Hence the projection X H X induced by [γ] γ(1) is a covering space. Choose x 0 X H to be the equivalence class of the constant path c at x 0. For a loop γ in X based at x 0, its lift to X starting at [c] ends at [γ], so the image of this lifted path in X H is a loop if and only if [γ] [c], i.e., [γ] H. It implies that the image of p : π 1 (X H, x 0 ) π 1 (X, x 0 ) is exactly H. We are interested in classifying covering spaces up to isomorphism. For two covering spaces p 1 : X 1 X and p 2 : X 2 X, if f : X 1 X 2 is a homeomorphism such that p 1 = p 2 f, then we say that f is an isomorphism between the two covering spaces. In that case f 1 is also an isomorphism, and the composition of two isomorphisms is an isomorphism, so we have an equivalence relation. Proposition 4.9. Suppose X is path connected and locally path connected. Then two path connected covering spaces p 1 : X1 X and p 2 : X2 X are isomorphic via an isomorphism f : X1 X 2 taking a base point x 1 p 1 1 (x 0) to a base point x 2 p 1 2 (x 0) if and only if p 1 (π 1 ( X 1, x 1 )) = p 2 (π 1 ( X 2, x 2 )). Proof. If there is an isomorphism f : ( X 1, x 1 ) ( X 2, x 2 ), then from the relation p 1 = p 2 f it follows that p 1 (π 1 ( X 1, x 1 )) = p 2 (f (π 1 ( X 1, x 1 ))) = p 2 (π 1 ( X 2, x 2 )). Conversely, suppose that p 1 (π 1 ( X 1, x 1 )) = p 2 (π 1 ( X 2, x 2 )). By the lifting criterion for maps, we may lift p 1 to a map p 1 : ( X 1, x 1 ) ( X 2, x 2 ) with p 2 p 1 = p 1. Similarly we obtain p 2 : ( X 2, x 2 ) ( X 1, x 1 ) with p 1 p 2 = p 2. Then p 1 ( p 2 p 1 ) = (p 1 p 2 ) p 1 = p 2 p 1 = p 1, hence both p 2 p 1 and 1 are lifts of p 1 fixing the base point x 1. By the unique lifting property, we have p 2 p 1 = 1, and similarly p 1 p 2 = 1, so p 1 and p 2 are inverse isomorphisms. We summarize the discussion by the following result. Theorem Let X be path connected, locally path connected and semilocally simply connected. Then there is a bijection between the set of base point preserving isomorphism classes of path connected covering spaces p : ( X, x 0 ) (X, x 0 ) and the set of subgroups of π 1 (X, x 0 ), obtained by associating the subgroup p (π 1 ( X, x 0 )) to the covering space ( X, x 0 ). If the base points are ignored, then this correspondence gives a bijection between isomorphism classes of path connected covering spaces p : X X and conjugacy classes of subgroups of π 1 (X, x 0 ). Proof. We have proved the first half of the result. For the other half, we will show that for a covering space p : ( X, x 0 ) (X, x 0 ), changing the base point x 0 within p 1 (x 0 ) corresponds exactly to changing p (π 1 ( X, x 0 )) to a conjugate subgroup of π 1 (X, x 0 ). Suppose that x 1 is another base point in p 1 (x 0 ). Let γ be a path from x 0 to x 1. Then γ projects to a loop γ in X representing an element g π 1 (X, x 0 ). Set