32 II SurfacesII.2 Fundamental GroupIn this section we talk about paths and loops, trying to get around on surfacesand elsewhere without getting confused. We use this topic as an excuse tointroduce the first substantially algebraic tool used in topology.Paths and products. Recall that a path in a topological space is a con-tinuous map from the unit interval. Call two paths p, q : [0, 1] → X fromx = p(0) = q(0) to y = p(1) = q(1) equivalent if there is a continuous mapH : [0, 1]×I → X, where I = [0, 1], such that H(s, 0) = p(s) and H(s, 1) = q(s)for all s ∈ [0, 1] as well as H(0, t) = x and H(1, t) = y for all t ∈ I. We writep ∼ q if p and q are equivalent and think of H as a continuous deformationof p to q that keeps the endpoints fixed. The deformation happens within Xbut is not hindered by self-intersections. Recall that an equivalence relationpartitions. In this case, it partitions the set of paths from x to y. Given sucha path p, we denote the class of paths equivalent to p by [p]. The product oftwo paths p and q is defined if p(1) = q(0), namelypq(s) =p(2s) for 0 ≤ s ≤12,q(2s − 1) for12≤ s ≤ 1.We are interested in equivalence classes more than in individual paths. For-tunately, the product operation is compatible with the concept of equivalence,p0∼ q0and p1∼ q1implies p0p1∼ q0q1. In other words, [p][q] = [pq] is welldefined because it does not depend on the representatives p and q of the twoclasses. Defining the product for classes of paths, we check the group axiomsfor this operation.Associativity. Note that p(qr) 6= (pq)r, even if all products are defined, butwe can reparametrize to make them equal so they are certainly equivalent.Hence α(βγ) = (αβ)γ, where α = [p], β = [q], and γ = [r].Neutral element. Write 1x(s) = x for the constant path at x ∈ X and letεx= [1x]. Clearly εxα = α and αεy= α assuming α = [p] with x = p(0)and y = p(1).Inverse element. Let p−1(s) = p(1−s) be the same path but going backwardsand define α−1= [p−1]. Then every path in αα−1is equivalent to 1xandevery path in α−1α is equivalent to 1y.In summary, we almost have a group but the product operation is not alwaysdefined.II.2 Fundamental Group 33Loops. As a remedy we consider paths that start and end at the same point.A loop is a path p : [0, 1] → X with x = p(0) = p(1); it is said to be based at x.Now the product operation is always defined and αα−1= α−1α = εx.Definition. The fundamental group, denoted as π(X, x), consists of allequivalence classes of loops based at x and the product operation betweenthem.In most cases, the point x is not important. Indeed suppose points x, y ∈ X areconnected by a path m. Then we can map each loop p based at x to the loopm−1pm based at y. In the other direction we map the path q based at y tomqm−1based at x; see Figure II.6. Extending the construction to equivalencexypqmFigure II.6: The loops p and mqm−1are based at x and the loops q and m−1pm arebased at y. Since q goes around the hole and p does not, the two are not equivalent.classes we get u : π(X, x) → π(X, y) defined by u(α) = µ−1αµ, where µ = [m]and α = [p]. It is easy to see that u commutes with the product operation,that is, u(α)u(β) = u(αβ). In words, u is a homomorphism. Symmetrically,we get a homomorphism v : π(X, y) → π(X, x) defined by v(γ) = µγµ−1. Thecomposition v ◦ u maps α to µ(µ−1αµ)µ−1= α and is therefore the identityon π(X, x). Symmetrically, u ◦ v is the identity on π(X, y). Hence u and v areisomorphisms. To summarize, if X is path-connected then π(X, x) and π(X, y)are isomorphic for any two points x, y ∈ X. It thus makes sense to leave outthe base point and write π(X), but we need to keep in mind that there is nocanonical isomorphism between the fundamental groups since it depends onthe connecting path.Induced homomorphisms. An important idea in algebraic topology is thatcontinuous maps between spaces induce homomorphisms between groups. Letf : X → Y be such a map. As illustrated in Figure II.7, any two equiva-lent paths p, q : [0, 1] → X map to equivalent paths f(p), f (q) : [0, 1] → Y.34 II Surfacesfxpf (p)f (x )Figure II.7: The annulus on the left is injectively mapped into the annulus on theright. Since this operation fills the hole, all non-trivial loops on the left map to εf (x)on the right.Hence f induces the homomorphism f∗: π(X, x) → π(X, f (x)) which maps theclass α = [p] to the class f∗(α) = [f(p)], which contains all paths f(p) withp ∈ α but possibly more. In Figure II.7, each class in π(X, x) is mapped tothe neutral element in π(X, f (x)). Even though both groups are isomorphic,this particular homomorphism is not an isomorphism. It is clear that if f is ahomeomorphism then f∗is an isomorphism. On the other hand, the exampleshows that f being injective does not imply that f∗is a monomorphism (in-jective homomorphism). Similarly, f being surjective does not imply that f∗isan epimorphism (surjective homomorphism).Retracts. The fundamental group of the annulus is infinitely cyclic, consist-ing of all classes αk, k and integer, where α goes around the annulus once. Incontrast, the fundamental group of the disk is trivial, consisting only of oneelement. We can use this difference to prove Brouwer’s Theorem in the plane.Call a subset A ⊆ X a retract of X if there is a continuous map r : X → Awith r(a) = a for all a ∈ A. The map r is called a retraction. A simple exampleis the torus and a closed curve in the torus. Letting i : A → X be the inclusionmap, i(a) = a for all a ∈ A, we have two induced homomorphisms for eachpoint a ∈ A,i∗: π(A, a) → π(X, a),r∗: π(X, a) → π(A, a).Since r ◦ i : A → A is the identity, r∗◦ i∗: π(A, a) → π(A, a) is the iden-tity homomorphism. This implies that i∗is a monomorphism and r∗is anepimorphism.II.2 Fundamental Group 35Brouwer’s Theorem. A continuous map f : B2→ B2has at least onefixed point x = f(x).Proof. To get a contradiction assume x 6= f(x) for all x ∈ B2. Consider thehalf-line of points x + λ(x − f(x)), for all λ ≥ 0, and let r(x) be the point atwhich this half-line intersects the boundary circle; see Figure II.8. The map r isxf (x )(r x )Figure II.8: Mapping a point x of the disk to a point r(x) on the circle bounding thedisk. Note that r(x) = x if x belongs to the circle.continuous because f is continuous and without fixed point.
View Full Document